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(defined as above) is principal. See Exercise 6. Thus principal rings may be obtained in a natural way from rings which are not principal. As Dedekind found out, some form of unique factorization can be re covered in some cases, replacing unique factorization into prime elements by unique factorization of (nonzero) ideals into prime ideals. Example. There are cases when the nonzero ideals give rise to a group.
Let o be a subring of a field K such that every element of K is a quotient of elements of o; that is, of the form ajb with a, b E o and b =F 0. By a fractional ideal a we mean a nonzero additive subgroup of K such that o a c a (and therefore o a = a since o contains the unit element); and such th(!t there exists an element c E o, c =F 0, such that ca c o. We might say that a fractional ideal has bounded denominator. A Dedekind ring is a ring o as above such that the fractional ideals form a group under multiplication. As proved in books on algebraic number theory, the ring of algebraic integers in a number field is a Dedekind ring. Do Exercise 14 to get the property of unique factorization into prime ideals. See Exercise 7 of Chapter VII for a sketch of this proof. If a E K, a =F 0, then oa is a fractional ideal, and such ideals are called principal. The principal fractional ideals form a subgroup. The factor group is called the ideal class group , or Picard group of o, and is denoted by Pic( o). See Exercises 1 3  1 9 for some elementary facts about Dedekind rings. It is a basic problem to determine Pic(o) for various Dedekind rings arising in algebraic number theory and function theory. See my book Algebraic Num ber Theory for the beginnings of the theory in number fields. In the case of function theory, one is led to questions in algebraic geometry, notably the study of groups of divisor classes on algebraic varieties and all that this entails. The property that the fractional ideals form a group is essentially associated with the ring having " dimension 1 " (which we do not define here). In general one is led into the study of modules under various equiva lence relations; see for instance the comments at the end of Chapter III, §4. We return to the general theory of rings. By a ringhomomorphism one means a mapping f: A + B where A, B are rings, and such that f is a m onoidhomomorphism for the multiplicative structures on A and B, and also a monoidhomomorphism for the additive structure. In other words, f must satisfy :
f(a + a ' ) = f(a) + f(a ' ), f( 1) = 1,
f(aa') = f(a)f(a ' ), f(O) = 0,
for all a, a' E A. Its kernel is defined to be the kernel of f viewed as additive homomorphism.
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The kernel of a ringhomomorphism f: A __.. B is an ideal of A, as one verifies at once. Conversely, let a be an ideal of the ring A. We can construct the factor ring Aja as follows. Viewing A and a as additive groups, let A/a be the factor group. We define a multiplicative law of composition on A/a : If x + a and y + a are two cosets of a, we define (x + a) (y + a) to be the coset (x y + a). This coset is well defined, for if x 1 , y 1 are in the same coset as x, y respectively, then one verifies at once that x 1 y 1 is in the same coset as xy. Our multiplicative law of composition is then obviously associative, has a unit element, namely the coset 1 + a, and the distributive law is satisfied since it is satisfied for coset representatives. We have therefore defined a ring structure on Aja, and the canonical map f:
A __.. A/a
is then clearly a ringhomomorphism.
�\ I·
If g: A __.. A ' is a ringhomomorphism whose kernel contains a, then there exists a unique ringhomomorphism g * : Aja __.. A' making the following dia gram commutative : A
u
A'
Aja
Indeed, viewing f, g as grouphomomorphisms (for the additive struc tures), there is a unique grouphomomorphism g * making our diagram commutative. We contend that g * is in fact a ringhomomorphism. We could leave the trivial proof to the reader, but we carry it out in full. If x E A, then g(x) = g.f(x). Hence for x, y E A,
g.(f(x)f(y)) = g.(f(xy)) = g(xy) = g(x)g(y) = g.f(x)g.f(y). Given e, '1 E Aja, there exist x, y E A such that e = f(x) and '1 = f( y). Since f( 1) = 1, we get g.f( 1 ) = g(1) = 1, and hence the two conditions that g * be a multiplicative monoidhomomorphism are satisfied, as was to be shown. The statement we have j ust proved is equivalent to saying that the canonical map f: A __.. A/a is universal in the category of homomorphisms whose kernel contains a. Let A be a ring, and denote its unit element by e for the moment. The map A.: Z __.. A such that A.(n) = ne is a ringhomomorphism (obvious), and its kernel is an ideal (n), generated by an integer n > 0. We have a canonical injective homo morphism ZjnZ __.. A, which is a (ring) isomorphism between Z/nZ and a
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subring of A . If nZ is a prime ideal , then n = 0 or n = p for some prime number p . In the first case , A contains as a subring a ring which is isomorphic to Z , and which is often identified with Z. In that case , we say that A has characteristic 0 . If on the other hand n = p , then we say that A has characteristic p , and A contains (an isomorphic image of) Z/pZ as a subring . We abbreviate Z/pZ by FP . /
If K is a field, then K has characteristic 0 or p > 0. In the first case, K contains as a subfield an isomorphic image of the rational numbers, and in the second case, it contains an isomorphic image of FP . In either case, this subfield will be called the prime field (contained in K). Since this prime field is the smallest subfield of K containing 1 and has no automorphism except the identity, it is customary to identify it with Q or FP as the case may be. By the prime ring (in K) we shall mean either the integers Z if K has characteristic 0, or FP if K has characteristic p. Let A be a subring of a ring B. Let S be a subset of B commuting with A ; in other words we have as = sa for all a E A and s E S. We denote by A [S] the set of all elements i• . . . i a s sn" ' � i i · · · t n 1 � the sum ranging over a finite number of ntuples (i 1 , in ) of integers > 0, and a i . · · · i E A, s 1 , , sn E S. If B = A [S], we say that S is a set of generators " (or more precisely, ring generators) for B over A, or that B is generated by S over A. If S is finite, we say that B is finitely generated as a ring over A. One might say that A [S] consists of all notnecessarily commutative polynomials in elements of S with coefficients in A. Note that elements of S may not commute with each other. •
•
•
•
•
,
•
Example. The ring of matrices over a field is finitely generated over that
field, but matrices don't necessarily commute. As with groups, we observe that a homomorphism is uniquely determined by its effect on generators. In other words, let f : A __.. A' be a ring homomorphism, and let B = A [S] as above. Then there exists at most one extension of f to a ringhomomorphism of B having prescribed values on S. Let A be a ring, a an ideal, and S a subset of A. We write S if S c a. If x, y E A, we write
=
x =
0
(mod a)
y (mod a)
if x  y E a. If a is principal, equal to (a), then we also write
y (mod a). If f: A __.. Aja is the canonical homomorphism, then x y (mod a) means that f(x) = f(y). The congruence notation is sometimes convenient when we want to avoid writing explicitly the canonical map f x =
_
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The factor ring A/ a is also called a residue class ring. Cosets of a in A are called residue classes modulo a, and if x E A, then the coset x + a is called the residue class of x modulo a. We have defined the notion of an isomorphism in any category, and so a ringisomorphism is a ringhomomorphism which has a twosided inverse. As usual we have the criterion :
A ringhomomorphism f: A __.. B which is bijective is an isomorphism. Indeed, there exists a settheoretic inverse g: B __.. A, and it is trivial to verify that g is a ringhomomorphism. Instead of saying " ringhomomorphism " we sometimes say simply " homomorphism " if the reference to rings is clear. We note that rings form a category (the morphisms being the homomorphisms). Let f: A __.. B be a ringhomomorphism. Then the image f(A) of f is a subring of B. Proof obvious. It is clear that an injective ringhomomorphism f : A __.. B establishes a ringisomorphism between A and its image. Such a homomorphism will be called an embedding (of rings). Let f: A __.. A' be a ringhomomorphism, and let a' be an ideal of A'. Then f  1 (a ' ) is an ideal a in A, and we have an induced injective homo morphism A/a __.. A'/a ' . The trivial proof is left to the reader. Proposition 1.1.
Products exist in the category of rings. In fact, let {A i } i e i be a family of rings, and let A = 0 A i be their product as additive abelian groups. We define a multiplication in A in the obvious way : If (x i )i e J and (y i ) i e i are two elements of A, we define their product to be (x i y i ) i e i.e. we define multiplication componentwise, just as we did for addition. The multiplicative unit is simply the element of the product whose ith component is the unit element of A i . It is then clear that we obtain a ring structure on A, and that the projection on the ith factor is a ring homomorphism. Furthermore, A together with these projections clearly satisfies the required universal property. Note that the usual inclusion of A i on the ith factor is not a ring homomorphism because it does not map the unit element ei of A i on the unit element of A. Indeed, it maps ei on the element of A having ei as ith component, and 0 ( = O i ) as all other components. Let A be a ring. Elements x, y of A are said to be zero divisors if x =F 0, y =F 0, and xy = 0. Most of the rings without zero divisors which we con sider will be commutative. In view of this, we define a ring A to be entire if 1 =F 0, if A is commutative, and if there are no zero divisors in the ring. (Entire rings are also called integral domains. However, linguistically, I feel 1,
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the need for an adjective. " Integral " would do, except that in English, " integral " has been used for " integral over a ring " as in Chapter VII, § 1 . In French, as in English, two words exist with similar roots : " integral " and " entire ". The French have used both words. Why not do the same in English? There is a slight psychological impediment, in that it would have been better if the use of " integral " and " entire " were reversed to fit the longstanding French use. I don't know what to do about this.) Examples. The ring of integers Z is without zero divisors, and is there fore entire. If S is a set with more than 2 elements, and A is a ring with 1 =F 0, then the ring of mappings Map(S, A) has zero divisors. (Proof?)
Let m be a positive integer =F 1 . The ring Z/mZ has zero divisors if and only if m is not a prime number. (Proof left as an exercise.) The ring of n x n matrices over a field has zero divisors if n > 2. (Proof?) The next criterion is used very frequently.
Let A be an entire ring, and let a, b be nonzero elements of A. Then a, b generate the same ideal if and only if there exists a unit u of A such that b = au. Proof If such a unit exists we have Ab = Aua = Aa. Conversely, assume Aa = Ab. Then we can write a = be and b = ad with some elements c, d E A. Hence a = adc, whence a ( 1  de) = 0, and therefore de = 1 . Hence c is a unit.
§2. C O M M UTATIV E R I N G S Throughout this section, we let A denote a commutative ring. A prime ideal in A is an ideal p =F A such that A/p is entire. Equiva lently, we could say that it is an ideal p =F A such that, whenever x, y E A and xy E p, then x E p or y E p . A prime ideal is often called simply a prime. Let m be an ideal. We say that m is a maximal ideal if m =F A and if there is no ideal a =F A containing m and =F m.
Every maximal ideal is prime. Proof Let m be maximal and let x, y E A be such that xy E m. Suppose x ¢ m. Then m + Ax is an ideal properly containing m, hence equal to A. Hence we can write l = u + ax with u E m and a E A. Multiplying by y we find
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y = yu + axy , whence y E m and m is therefore prime. A. Then a is contained in some maximal ideal m. Proof. The set of ideals containing a and =F A is inductively ordered by ascending inclusion. Indeed, if {b i } is a totally ordered set of such ideals, then 1 ¢ b i for any i, and hence 1 does not lie in the ideal b = U b i , which dominates all bi. If m is a maximal element in our set, then m =F A and m is a maximal ideal, as desired. The ideal {0} is a prime ideal of A if and only if A is entire. Let a be an ideal
=F
(Proof obvious.) We defined a field K to be a commutative ring such that 1 =I= 0 , and such that the multiplicative monoid of nonzero elements of K is a group (i . e . such that whenever x E K and x =t= 0 then there exists an inverse for x) . We note that the only ideals of a field K are K and the zero ideal .
If m is a maximal ideal of A, then A/m is a field. Proof If x E A, we denote by x its residue class mod m. Since m =F A we note that A/m has a unit element =F 0. Any nonzero element of A/m can be written as x for some x E A, x ¢ m. To find its inverse, note that m + Ax is an ideal of A =F m and hence equal to A. Hence we can write = u + yx with u E m and y E A. This means that yx = 1 (i.e. = 1 ) and hence that x has an inverse, as desired. Conversely, we leave it as an exercise to the reader to prove that : 1
If m is an ideal of A such that A/m is a field, then m is maximal. +
Let f : A A' be a homomorphism of commutative rings. Let p' be a prime ideal of A ' , and let p = f  1 (p'). Then p is prime . To prove this, let x, y E A, and xy E p. Suppose x ¢ p. Then f(x) ¢ p ' . But f(x)f(y) = f(xy) E p ' . Hence f(y) E p ' , as desired. As an exercise, prove that if f is surjective, and if m' is maximal in A', then f  1 (m') is maximal in A. Example. Let Z be the ring of integers . Since an ideal is also an additive subgroup of Z , every ideal =t= {0} is principal , of the form nZ for some integer n > 0 (uniquely determined by the ideal) . Let p be a prime ideal =I= {0 }, p = nZ . Then n must be a prime number, as follows essentially directly from the definition of a prime ideal . Conversely , if p is a prime number, then pZ is a prime ideal (trivial exercise) . Furthermore , pZ is a maximal ideal . Indeed , suppose pZ contained in some ideal nZ . Then p = nm for some integer m , whence n = p or n = 1 , thereby proving pZ maximal .
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If n is an integer, the factor ring Z/nZ is called the ring of integers modulo n. We also denote ZjnZ = Z(n). If n is a prime number p, then the ring of integers modulo p is in fact a field, denoted by FP . In particular, the multiplicative group of FP is called the group of nonzero integers modulo p. From the elementary properties of groups, we get a standard fact of elementary number theory : If x is an integer =I= 0 (mod p), then x P 1 = 1 (mod p). (For simplicity, it is customary to write mod p instead of mod pZ, and similarly to write mod n instead of mod nZ for any integer n.) Similarly, given an integer n > 1 , the units in the ring Z/nZ consist of those residue classes mod nZ which are represented by integers m =F 0 and prime to n. The order of the group of units in Z/nZ is called by definition lfJ(n) (where lfJ is known as the Euler phifunction) . Consequently, if x is an integer prime to n, then x " 2 we can find elements ai E a 1 and bi E a i such that
a.' + b.' = 1 '
n 0 The product ( ai + bi ) is equal to i =2
1,
.
l> = 2.
and lies in
i.e. in a 1 + a 2 · · a n . Hence •
n n at + ai = A . i=2 By the theorem for n = 2, we can find an element y 1 E A such that Y1
= 1
Y1 = 0
(mod at ),
(mod Ji a;) . &=2
We find similarly elements y2 , • • • , Yn such that and Then x = x 1 y 1 + · · · + Xn Yn satisfies our requirements.
for i =F j.
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In the same vein as above, we observe that if a 1 , ring A such that a 1 + · · · + an = A ' and if v 1 , , vn are positive integers, then •
•
•
•
•
95
, an are ideals of a
•
a� 1 + · · · + a�" = A.
The proof is trivial, and is left as an exercise. Corollary 2.2. Let a 1 , i i= j.
Let
•
•
•
, an be ideals of A. Assume that a i + ai = A for
n
f: A __.. n A/a i = (A/a 1 ) x · · · x (A/a n ) i =l be the map of A into the product induced by the canonical map of A onto n A/a i for each factor. Then the kernel of f is n a i , and f is surjective, i =t thus giving an isomorphism Proof That the kernel of f is what we said it is, is obvious. The surjectivity follows from the theorem. The theorem and its corollary are frequently applied to the ring of integers Z and to distinct prime ideals (p 1 ) , (Pn ). These satisfy the hypothesis of the theorem since they are maximal. Similarly, one could take integers m 1 , , mn which are relatively prime in pairs, and apply the theorem to the principal ideals (m 1 ) = m 1 Z, . . . , (mn ) = mn Z. This is the ultraclassical case of the Chinese remainder theorem. In particular, let m be an integer > 1 , and let ,
•
•
•
•
•
•
be a factorization of m ringisomorphism :
m = n P ir•· i into primes, with exponents ri > 1 . Then we have a Z/mZ
�
0 Z/p�t z. i
If A is a ring, we denote as usual by A * the multiplicative group of invertible elements of A. We leave the following assertions as exercises :
The preceding ringisomorphism of Z/mZ onto the product induces a group isomorphism (Z/mZ) * � 0 (Z/p�i Z) * . i In view of our isomorphism, we have
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If p is a prime num ber and r an integer > 1 , then cp(p r ) = (p 1)p r  1 . _
One proves this last formula by induction. If r = 1 , then Z/pZ is a field, and the multiplicative group of that field has order p 1 . Let r be > 1 , and consider the canonical ringhomomorphism Z/p r+ t Z __.. Z/p r z, 
arising from the inclusion of ideals (p r+ t ) c (p r ). We get an induced group homomorphism A.: (Z/p r + t Z) * __.. (Z/p r Z) * , which is surjective because any integer a which represents an element of Z/p r z and is prime to p will represent an element of (Z/p r +t Z) * . Let a be an integer representing an element of (Z/p r+ t Z) * , such that A.( a) = 1. Then a = 1 (mod p r z), and hence we can write a = 1 + xp r (mod p r+ t Z)
for some x E Z. Letting x = 0, 1 , . . . , p  1 gives rise to p distinct elements of (Z/p r+ t Z) * , all of which are in the kernel of A.. Furthermore, the element x above can be selected to be one of these p integers because every integer is congruent to one of these p integers modulo (p). Hence the kernel of A. has order p, and our formula is proved. Note that the kernel of A. is isomorphic to Z/pZ. (Proof?) Application : The ring of endomorphisms of a cyclic group. One of the
first examples of a ring is the ring of endomorphisms of an abelian group. In the case of a cyclic group, we have the following complete determination. Theorem 2.3. Let A be a cyclic group of order n. For each k E Z let fk : A � A be the endomorphism x � kx (writing A additively) . Then k � fk induc es a ring isomorphism Z/nZ = End(A) , and a group isomorphism (Z/nZ)* = Aut( A ) .
Proof Recall that the additive group structure on End(A) is simply addition of mappings, and the multiplication is composition of mappings. The fact that k �+ fk is a ringhomomorphism is then a restatement of the formulas 1a = a
'
(k + k' ) a = ka + k'a,
and
(kk ' ) a = k(k' a)
for k, k' E Z and a E A. If a is a generator of A, then ka = 0 if and only if k = 0 mod n, so Z/nZ is embedded in End(A). On the other hand, let f : A __.. A be an endomorphism. Again for a generator a, we have f( a) = ka
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for some k, whence f = fk since every x E A is of the form ma for some m E Z, and f(x) = f(ma) = mf(a) = mka = kma = kx. This proves the isomorphism Z/nZ � End(A). Furthermore, if k E (Z/nZ) * then there exists k' such that kk' = 1 mod n, so A has the inverse fk, and fk is an automorphism. Conversely, given any automorphism f with inverse g, we know from the first part of the proof that f = fk , g = gk ' for some k, k' , and 1 mod n , so k, k ' E (Z/nZ) * . This proves the f o g = id means that kk' isomorphism (Z/nZ) * = Aut(A). =
Note that if A is written as a multiplicative group C, then the map fk is given by x �+ x k . For instance, let Jln be the group of nth roots of unity in C. Then all automorphisms of Jln are given by with k E (Z/nZ) * .
§3.
PO LY N O M I A LS A N D G R O U P R I N G S
Although all readers will have met polynomial functions, this section lays the ground work for polynomials in general. One needs polynomials over arbitrary rings in many contexts. For one thing, there are polynomials over a finite field which cannot be identified with polynomial functions in that field. One needs polynomials with integer coefficients, and one needs to reduce these polynomials mod p for primes p. One needs polynomials over arbitrary commutative rings, both in algebraic geometry and in analysis, for instance the ring of polynomial differential operators. We also have seen the example of a ring B = A [S] generated by a set of elements over a ring A. We now give a systematic account of the basic definitions of polynomials over a commutative ring A. We want to give a meaning to an expression such as
a0 + a 1 X + · · · + an x n, where a i E A and X is a " variable ". There are several devices for doing so, and we pick one of them. (I picked another in my Undergraduate Algebra.) Consider an infinite cyclic group generated by an element X. We let S be the subset consisting of powers x r with r > 0. Then S is a monoid. We define the set of polynomials A [X] to be the set of functions S __.. A which are equal to 0 except for a finite number of elements of S. For each element a E A we denote by ax n the function which has the value a on x n and the value 0 for all other elements of S. Then it is immediate that a polynomial can be written uniquely as a finite sum
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a o X o + . . . + an X n for some integer n E N and ai E A. Such a polynomial is denoted by f( X ). The elements a i E A are called the coefficients of f We define the product according to the convolution rule. Thus, given polynomials n g(X) = L bjXi and f( X) L ai x i i =O j=O m
=
we define the product to be
f(X)g(X) =
�t: c+�k )
a bi X k . ;
It is immediately verified that this product is associative and distributive. We shall give the details of associativity in the m ore general context of a monoid ring below. Observe that there is a unit element, namely 1 X 0 • There is also an embedding given by A ___. A [X] One usually does not distinguish a from its image in A [X], and one writes a instead of aX 0 • Note that for c E A we have then cf(x) = L ca i X i . Observe that by our definition, we have an equality of polynomials L ai x i = L bi X i if and only if a i = bi for all i. Let A be a subring of a commutative ring B. Let x E B. If f E A [X] is a polynomial, we may then define the associated polynomial function by letting
fs(x) = f(x) = a 0 + a 1 x +
···
+ an x n .
Given an element b E B, directly from the definition of multiplication of polynomials, we find :
The association is a ring homomorphism of A [X] into B. This homomorphism is called the evaluation homomorphism, and is also said to be obtained by substituting b for X in the polynomial. (Cf. Proposition 3. 1 below.) Let x E B. We now see that the subring A [x] of B generated by x over A is the ring of all polynomial values f(x), for f E A [X]. If the evaluation map fH f(x) gives an isomorphism of A [X] with A [x], then we say that x is
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transcendental over A, or that x is a variable over A. In particular, X is a
variable over A.
Example. Let rx = fi. Then the set of all real numbers of the form a + brx, with a, b E Z, is a subring of the real numbers, generated by fi. Note that rx is not transcendental over Z, because the polynomial X 2  2 lies
in the kernel of the evaluation map ! �+ f( fi). On the other hand, it can be shown that e = 2 . 7 1 8 . . . and n are transcendental over Q. See Appendix 1 . Example.
Let p be a prime number and let K = Z/pZ. Then K is a field. Let f(X) = X P  X E K [X]. Then f is not the zero polynomial. But fK is the zero function. Indeed, fK(O) = 0. If x E K, x =F 0, then since the multiplicative group of K has order p  1 , it follows that x p  l = 1 , whence x P = x, so f(x) = 0 . Thus a nonzero polynomial gives rise to the zero function on K. There is another homomorphism of the polynomial ring having to do with the coefficients. Let cp : A + B be a homomorphism of commutative rings. Then there is an associated homomorphism of the polynomial rings A [X] + B[X], such that The verification that this mapping is a homomorphism is immediate, and further details will be given below in Proposition 3.2, in a more general context. We call ! �+ cpf the reduction map. Examples.
In some applications the map cp may be an isomorphism. For instance, if f(X) has complex coefficients, then its complex conju gate f(X) = L a i X i is obtained by applying complex conjugation to its coefficients. Let p be a prime ideal of A. Let cp: A + A ' be the canonical homo morphism of A onto Ajp. If f(X) is a polynomial in A [X], then cpf will sometimes be called the reduction of / modulo p. For example, taking A = Z and p = (p) where p is a prime number, we can speak of the polynomial 3X4  X + 2 as a polynomial mod 5, viewing the coefficients 3,  1 , 2 as integers mod 5, i.e. elements of Z/5Z. We may now combine the evaluation map and the reduction map to generalize the evaluation map. Let q>: A � B be a homomorphism of commutative rings. Let x E B . There is a unique homomorphism extending
A [X] + B
such that
and for this homomorphism, L a i X i H L cp(a i )x i .
X H X,
'P
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The homomorphism of the above statement may be viewed as the composite
A [X]
+
B[X]
evx
B
where the first map applies cp to the coefficients of a polynomial, and the second map is the evaluation at x previously discussed. Example.
In Chapter IX, §2 and §3, we shall discuss such a situation in several variables, when (lfJf) (x) = 0, in which case x is called a zero of the polynomial f n
When writing a polynomial f(X) = L ai X i, if an =F 0 then we define n i= 1 to be the degree of f. Thus the degree of f is the smallest integer n such that ar = 0 for r > n. If f = 0 (i.e. f is the zero polynomial), then by con vention we define the degree of f to be oo . We agree to the convention that  00 +  00 =  00 , oo < n  oo + n = oo, '
for all n E Z, and no other operation with  oo is defined. A polynomial of degree 1 is also called a linear polynomial. If f =F 0 and degf = n, then we call an the leading coefficient of f We call a 0 its constant term. Let be a polynomial in A [X], of degree m, and assume g =F 0. Then
f(X)g(X) = a0 b0 + · · · + an bmx m + n .
Therefore :
If we assume that at least one of the leading coefficients an or bm is not a divisor of 0 in A, then deg (fg) = deg f + deg g and the leading coefficient of fg is an bm . This holds in particular when an or bm is a unit in A, or when A is entire. Consequently, when A is entire, A [X] is also entire. If f or g = 0, then we still have deg( fg) = deg f + deg g if we agree that  oo + m = oo for any integer m. One verifies trivially that for any polynomial f, g E A [X] we have deg(f + g) < max(deg f, deg g), again agreeing that oo < m for every integer m.
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1 01
Po l y n o m i a l s i n seve ra l va r i a b l es
We now go to polynomials in several variables. Let A be a subring of a commutative ring B. Let x 1 , • • • , xn E B. For each ntuple of integers (v 1 , . . . , vn ) = (v) E N n, we use vector notation, letting (x) = (x 1 , . . . , xn ), and M( v) (x) = x ; 1 • • • x ;". The set of such elements forms a monoid under multiplication. Let A [x] = A [x 1 , . . . , xn ] be the subring of B generated by x 1 , . . . , xn over A. Then every element of A [x] can be written as a finite sum
L a( v) M( v) (x)
with a( v) E A. Using the construction of polynomials in one variable repeatedly, we may form the ring A [ X 1 , . . . , Xn ] = A [X 1 ] [X2 ] · · · [Xn ], selecting Xn to be a variable over A [X 1 , . . . , Xn _ 1 ]. Then every element f of A [X 1 , . . . , Xn ] = A [X] has a unique expression as a finite sum
Therefore by induction we can write f uniquely as a sum
f=
L a v 1 . ' ' Vn x ; 1 . . . x:�11 ) x:n ( � Vn0 V 1 , . . . , Vn 1
= L a( v) M( v) (X) = L a( v) x; l . . . x:n
with elements a E A, which are called the coefficients of f. The products M (X) = x ; 1 • • • X;"
will be called primitive monomials. Elements of A [X] are called polynomials (in n variables). We call a its coefficients. Just as in the onevariable case, we have an evaluation map. Given (x) = (x 1 , . . . , xn ) and f as above, we define f(x) = L a ( v) M( v) (x) = L a X ; 1 • • • x;". Then the evaluation map
such that
f � f(x)
is a ringhomomorphism. It may be viewed as the composite of the suc cessive evaluation maps in one variable Xi � x i for i = n, . . . , 1, because A [X] c B [X]. Just as for one variable, if f(X) E A [X] is a polynomial in n variables, then we obtain a function
1 02
I I,
RINGS
by
§3
(x) � f(x).
We say that f(x) is obtained by substituting (x) for (X) in f, or by specializing (X) to (x). As for one variable, if K is a finite field, and f E K [ X ] one may have f =F 0 but fK = 0. Cf. Chapter IV, Theorem 1 .4 and its corollaries. Next let cp: A � B be a homomorphism of commutative rings. Then we have the reduction map (generalized in Proposition 3.2 below) We can also compose the evaluation and reduction. An element (x) E B n is called a zero of f if (lfJf )(x) = 0. Such zeros will be studied in Chapter IX. Go back to A as a subring of B. Elements x 1 , , xn E B are called algebraically independent over A if the evaluation map •
•
•
f � f(x) is injective . Equivalently, we could say that if f E A [X] is a polynomial and f(x) = 0, then f = 0; in other words , there are no nontrivial polynomial _relations among x 1 , , Xn over A . •
•
•
Example. It is not known if e and
n
are algebraically independent over the rationals. It is not even known if e + n is rational. We now come to the notion of degree for several variables. By the degree of a primitive monomial we shall mean the integer I v i = v 1 + · · · A polynomial
+
vn (which is > 0). (a E A)
will be called a monomial (not necessarily primitive). If f(X) is a polynomial in A [X] written as v 1 • • • X V" a f(X) = � X 1 n ' � ( v)
then either f = 0, in which case we say that its degree is oo, or f =F 0, and then we define the degree of f to be the maximum of the degrees of the monomials M (X) such that a =F 0. (Such monomials are said to occur in the polynomial.) We note that the degree of f is 0 if and only if ° f(X) = a0 X P · · · Xn for some a0 E A, a 0 =F 0. We also write this polynomial simply f( X) = a0 , i.e. writing 1 instead of x 1o . . . xno ' in other words, we identify the polynomial with the constant a 0 •
II,
§3
PO LYNO M IALS AN D G RO U P R I N G S
1 03
Note that a polynomial f(X 1 , , Xn ) in n variables can be viewed as a polynomial in Xn with coefficients in A [X 1 , , Xn  1 ] (if n > 2). Indeed, we can write dn f( X) = L jj (X 1 ' . . . ' xn  1 )X1 , j= O •
•
•
•
•
•
where jj is an element of A [X 1 ' . . . ' xn  1 ]. By the degree of I in Xn we shall mean its degree when viewed as a polynomial in Xn with coefficients in A [X 1 , . . . , Xn _ 1 ]. One sees easily that if this degree is d, then d is the largest integer occurring as an exponent of Xn in a monomial with a< v> =F 0. Similarly, we define the degree of f in each variable Xi ( i = 1, , n). The degree of f in each variable is of course usually different from its degree (which is sometimes called the total degree if there is need to prevent ambiguity). For instance, . . .
has total degree 4, and has degree 3 in X 1 and 2 in X2 As a matter of notation, we shall often abbreviate " degree " by " deg. " For each integer d > 0, given a polynomial f, let J be the sum of all monomials occurring in f and having degree d. Then •
Suppose f =F 0. We say that f is homogeneous of degree d if f = f ; thus f can be written in the form
V 1 + · · · + Vn
with
=
d
if a( v) =F 0.
We shall leave it as an exercise to prove that a nonzero polynomial f in n variables over A is homogeneous of degree d if and only if, for every set of n + 1 algebraically independent elements u, t 1 , , tn over A we have f(ut 1 , , utn ) = u d_[(t 1 , tn ) . . .
• . .
. • . ,
·
We note that if f, g are homogeneous of degree d, e respectively, and fg =F 0, then fg is homogeneous of degree d + e. If d = e and f + g =F 0, then f + g is homogeneous of degree d. Remark. In view of the isomorphism A [X 1 ,
. . . , Xn J
�
A [t 1 ,
. . . , tn ]
between the polynomial ring in n variables and a ring generated over A by n
1 04
II,
RINGS
§3
algebraically independent elements, we can apply all the terminology we have defined for polynomials, to elements of A [t 1 , . . . , tn ]. Thus we can speak of the degree of an element in A [t], and the rules for the degree of a product or sum hold. In fact, we shall also call elements of A [t] polynomials in (t). Algebraically independent elements will also be called variables (or indepen dent variables), and any distinction which we make between A [X] and A [t] is more psychological than mathematical. Suppose next that A is entire. By what we know of polynomials in one variable and induction, it follows that A [X 1 , , Xn ] is entire. In particular, suppose f has degree d and g has degree e. Write f = j + terms of lower degree, •
•
•
g = g <e> + terms of lower degree. Then fg = j g <e> + terms of lower degree, and if fg =F 0 then j g <e> =F 0. Thus we find : deg(fg) = deg f + deg g, deg(f + g) < max(deg f, d e.g g). We are now finished with the basic terminology of polynomials. We end this section by indicating how the construction of polynomials is actually a special case of another construction which is used in other contexts. Inter ested readers can skip immediately to Chapter IV, giving further important properties of polynomials. See also Exercise 33 of Chapter XIII for har monic polynomials. The g ro u p ri ng or m o n o i d ri ng
Let A be a commutative ring. Let G be a monoid, written multiplica tively. Let A [G] be the set of all maps �: G __.. A such that �(x) = 0 for almost all x E G. We define addition in A [G] to be the ordinary addition of mappings into an abelian (additive) group. If �' p E A [G], we define their product �P by the rule ( �p ) (z) = L �(x)p(y). xy=z
The sum is taken over all pairs (x, y) with x, y E G such that xy = z . This sum is actually finite, because there is only a finite number of pairs of elements (x, y) E G x G such that �(x)p(y) =F 0. We also see that (�p) (t) = 0 for almost all t, and thus belongs to our set A [ G]. The axioms for a ring are trivially verified. We shall carry out the proof of associativity as an example. Let �, p, y E A [ G]. Then
I I,
§3
PO LYN O M IALS AN D G RO U P R I N G S
1 05
( (�p)y) (z) = L (�p ) (x)y(y)
xy=z
�x z L�x a(u)P(v)] y(y) a(u)p(v) y(y)J  x� z L�x =
= L �(u) P (v)y(y),
(u,v,y) uvy=z
this last sum being taken over all triples ( u v, y) whose product is z . This last sum is now symmetric, and if we had computed ( a (f3y))(z) , we would have found this sum also. This proves associativity. The unit element of A [G] is the function J such that J (e) = 1 and J (x) = 0 for all x E G, x =F e. It is trivial to verify that � = J� = �J for all � E A [G]. We shall now adopt a notation which will make the structure of A [G] clearer. Let a E A and x E G. We denote by a · x (and sometimes also by ax) the function whose value at x is a, and whose value at y is 0 if y =F x. Then an element � E A [G] can be written as a sum � = L �(x) · x. xe G Ind eed , if {ax } x e G is a set of elements of A almost all of which are 0, and we set then for any y E G we have p(y) = ay (directly from the definitions). This also shows that a given element � admits a unique expression as a sum L ax · x. With our present notation, multiplication can be written
(xLe G ax · x) (yLe G by · Y) = x,yL ax by · xy
and addition can be written L ax · x + L bx · x = L (ax + bx ) · x, xeG xe G xeG which looks the way we want it to look. Note that the unit element of A [G] is simply 1 · e. We shall now see that we can embed both A and G in a natural way in A [G]. Let cp0 : G __.. A [G] be the map given by cp0 (x) = 1 · x. It is immediately verified that cp0 is a multiplicative monoidhomomorphism, and is in fact injective, i.e. an em bedding. Let f0 : A __.. A [G] be the map given by f0 (a) = a · e.
1 06
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§3
It is immediately verified that fo is a ringhomomorphism, and is also an embedding. Thus we view A as a subring of A [G]. One calls A [G] the monoid ring or monoid algebra of G over A, or the group algebra if G is a group. Examples.
When G is a finite group and A = k is a field, then the group ring k[G] will be studied in Chapter XVIII. Polynomial rings are special cases of the above construction. In n vari ables, consider a multiplicative free abelian group of rank n. Let X 1 , . . . , Xn be generators. Let G be the multiplicative subset consisting of elements x;• · · · x;" with vi > 0 for all i. Then G is a monoid, and the reader can verify at once that A [G] is j ust A [X 1 , , Xn ]. As a matter of notation we usually omit the dot in writing an element of the ring A [G], so we write simply L ax x for such an element. More generally, let I = {i} be an infinite family of indices, and let S be the free abelian group with free generators Xi, written multiplicatively. Then we can form the polynomial ring A [X] by taking the monoid to consist of products M< v> (X) = 0 Xiv i, iE I •
•
•
where of course all but a finite number of exponents v i are equal to 0. If A is a subring of the commutative ring B, and S is a subset of B, then we shall also use the following notation. Let v: S __.. N be a mapping which is 0 except for a finite number of elements of S. We write M( v) (S) = n X v( x) . xeS Thus we get polynomials in infinitely many variables. One interesting exam ple of the use of such polynomials will occur in Artin's proof of the existence of the algebraic closure of a field, cf. Chapter V, Theorem 2.5. We now consider the evaluation and reduction homomorphisms in the present context of monoids. Proposition 3.1 . Let cp: G __.. G' be a homomorphism of monoids. Then r
there exists a unique homomorphism h: A [G] __.. A [G'] such that h(x) = cp(x) for all x E G and h(a) = a for all a E A . Proof In fact, let rx = L ax x E A [ G]. Define h(rx) = L ax cp(x).
Then h is immediately verified to be a homomorphism of abelian groups, and h(x) = cp(x). Let P = L byY · Then
h(rxp) =
� c� % ax by)
0, equal to 0 for almost all p E P such that a = u n p v ( p) . peP
Furthermore, the unit u and the integers v(p) are uniquely determined by a. We call v(p) the order of a at p, also written ordP a . If A is a factorial ring, then an irreducible element p generates a prime ideal (p). Thus in a factorial ring, an irreducible element will also be called a prime element, or simply a prime. We observe that one can define the notion of least common multiple (l.c.m.) of a finite number of nonzero elements of A in the usual manner : If a 1 . . . an E A are such elements, we define a l.c.m. for these elements to be any c E A such that for all primes p of A we have ordP c = max ordP a i. '
'
i
This element c is well defined up to a unit. If a, b e A are nonzero elements, we say that a, b are relaively prime if the g.c.d. of a an d b is a u ni t . Example. The ring of integers Z is factorial. Its group of units consists of 1 and 1 . It is natural to take as representative prime element the positive prime element (what is called a prime number) p from the two possible choices p and p. Similarly, we shall show later that the ring of polynomials in one variable over a field is factorial, and one selects represen tatives for the prime elements to be the irreducible polynon1ials with leading coefficient 1 . Examples. It will be proved in Chapter I V that if R is a factorial ring, then the polynomial ring R [X 1 , , Xn ] in n variables is factorial. In partic ular, if k is a field, then the polynomial ring k [X 1 , , Xn ] is factorial. Note that k[X 1 ] is a principal ring, but for n > 2, the ring k[X 1 , Xn ] is not principal. In Exercise 5 you will prove that the localization of a factorial ring is factorial. In Chapter IV, §9 we shall prove that the power series ring k [ [X 1 , . . . , Xn ] ] is factorial. This result is a special case of the more general statement that a regular local ring is factorial, but we do not define regular local rings in this book. You can look them up in books on commutative 

•
.
.
•
•
•
•
•
•
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algebra. I recommend : H. MATSUMURA, Commutative Algebra, second edition, Benj aminCummings, New York, 1 980 H. MATSUMURA, Commutative Rings, Cambridge University Press, Cambridge, UK, 1 986
Examples from algebraic and complex geometry. Roughly speaking, reg
ular local rings arise in the following context of algebraic or complex geom etry. Consider the ring of regular functions in the neighborhood of some point on a complex or algebraic manifold. This ring is regular. A typical example is the ring of convergent power series in a neighborhood of 0 in en . In Chapter IV, we shall prove some results on power series which give some algebraic background for those analytic theories, and which are used in proving the factoriality of rings of power series, convergent or not. Conversely to the above examples, singularities in geometric theories may give rise to examples of nonfactoriality. We give examples using notions which are sufficiently basic so that readers should have encountered them in more elementary courses. Examples of nonfactorial rings. Let k be a field, and let x be a variable over k. Let R = k[x 2 , x 3 ]. Then R is not factorial (proof?). The ring R may be viewed as the ring of regular functions on the curve y 2 = x 3 , which has a singularity at the origin, as you can see by drawing its real graph. Let R be the set of all numbers of the form a + b�, where a, b E Z. Then the only units of R are + 1 , and the elements 3, 2 + �, 2  F5 are irreducible elements, giving rise to a nonunique factorization 3 2 = (2 + �)(2  �). (Do Exercise 10.) Here the nonfactoriality is not due to singularities but due to a nontrivial ideal class group of R , which is a Dedekind ring. For a definition see the exercises of Chapter III, or go straight to my book Algebraic Number Theory, for instance. As Trotter once pointed out (Math. Monthly, April 1 988), the relation sin 2 x = (1 + cos x) (l  cos x) may be viewed as a nonunique factorization in the ring of trigonometric polynomials R [ sin x, cos x], generated over R by the functions sin x and cos x. This ring is a subring of the ring of all functions, or of all differenti able functions. See Exercise 1 1 . EX E R C1 S ES We let A denote a commutative ring. 1 . Suppose that 1
:/;
0 in A. Let S be a multiplicative subset of A not containing 0.
Let p be a maximal element in the set of ideals of A whose intersection with S is empty. Show that p is prime.
115
EX E R C I S E S
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2. Let f : A A' be a surjective homomorphism of rings, and assume that A is local, A' :/; 0. Show that A' is local. 3. Let p be a prime ideal of A. Show that A ., has a unique maximal ideal, consisting of all elements ajs with a e p and s ¢ p . 4. Let A be a principal ring and S a multiplicative subset with 0 ¢ S. Show that principal.
s • A is
5. Let A be a factorial ring and S a multiplicative subset with 0 ¢ S. Show that s • A is factorial, and that the prime elements of s • A are those primes p of A such that ( p) n S is empty. 6. Le! A be a factorial ring and p a prime element. Show that the local ring A
is principal.
7. Let A be a principal nng and a 1 , . . . , a, nonzero elements of A.
(a 1 , . . . , a,) = (d). (i = 1 , . . . , n).
Let
Show that d is a greatest common divisor for the a i
8. Let p be a prime number, and let A be the ring Z/p r z (r = integer > 1 ). Let G be the group of units in A, i.e. the group of integers prime to p, modulo pr. Show that G is cyclic, except in the case when
p = 2,
r > 3, in which case it is of type (2, 2r  2 ) . [Hint : In the general case, show that G is the product of a cyclic group generated by 1 + p, and a cyclic group of order p  1 . In the exceptional case, show that G is the product of the group { + 1 } with the cyclic group generated by the residue class of 5 mod 2r.] 9. Let i be the complex number
.J=l.
hence factorial. What are the units ?
Show that the ring Z [i] is principal, and
10. Let D be an integer > 1 , and let R be the set of all element a + b� with a, b e z. (a) Show that R is a ring. (b) Using the fact that complex conjugation is an automorphism of C, show that complex conjugation induces an automorphism of R. (c) Show that if D > 2 then the only units in R are + 1 . (d) S how that 3, 2 +
R, 2  R are irreducible elements in Z [RJ.
1 1 . Let R be the ring of trigonometric polynomials as defined in the text. Show that R consists of all functions f on R which have an expression of the form , f(x ) = a0 + L (am cos mx + bm sin mx), m =l where a0 , am , bm are real numbers. Define the trigonometric degree degtr( f ) to be the maximum of the integers r, s such that ar , bs :/; 0. Prove that
Deduce from this that R has no divisors of 0, and also deduce that the functions sin x and 1  cos x are irreducible elements in that ring.
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1 2. Let P be the set of positive integers and R the set of functions defined on P with values in a commutative ring K. Define the sum in R to be the ordinary addition of functions, and defi ne the convolution product by the formula (f • g) (m)
= L f(x) g (y), xy=m
where the sum is taken over all pairs (x, y) of positive integers such that xy = m. (a) Show that R is a commutative ring, whose unit element is the function � such that � ( 1 ) = 1 and �(x) = 0 if x :/; 1 . (b) A function f is said to be multiplicative if f(mn ) = f(m)f(n ) whenever m, n are relatively prime. If J, g are multiplicative, show that f • g is multiplicative. (c) Let J1. be the Mobius function such that JJ.( 1 ) = 1 , JJ.(p 1 · · · p,) = (  1 )' if p 1 , . . . , p, are distinct primes, and JJ. (m) = 0 if m is divisible by p 2 for some prime p. Show that J1. • c:p1 = �' where c:p1 denotes the constant function having value 1 . [Hint : Show first that J1. is multiplicative, and then prove the assertion for prime powers.] The Mobius inversion formula of elementary number theory is then nothing else but the relation J1. • c:p1 • f = f.
Dedekind rings Prove the following statements about a Dedekind ring o. To simplify terminology, by an ideal we shall mean nonzero ideal unless otherwise specified. We let K denote the quotient field of o.
1 3. Every ideal is finitely generated. [Hint : Given an ideal a, let b be the fractional ideal such that ab = o. Write 1 = L a;b; with a; e a and b; e b. Show that a = (a 1 , . . . , a,.).]
14. Every ideal has a factorization as a product of prime ideals, uniquely determined up to permutation.
1 5. Suppose o has only one prime ideal p . Let t e p and t ¢ p 2 . Then p = (t) is principal.
1 6. Let o be any Dedekind ring. Let p be a prime ideal. Let o ., be the local ring at
p . Then o ., is Dedekind and has only one prime ideal.
1 7. As for the integers, we say that a l b (a divides b) if there exists an ideal
c
such that
b = a c. Prove : (a) a l b if and only if b c a. (b) Let a, b be ideals. Then a + b is their greatest common divisor. In particular, a, b are relatively prime if and only if a + b = o.
1 8. Every prime ideal p is maximal. (Remember, p
:/;
0 by convention.) In particular,
if p 1 , p,. are distinct primes, then the Chinese remainder theorem applies to . rl r t h etr powers p 1 . . . , p,.". U se t h Is " to prove : •
•
.
,
,
1 9. Let a, b be ideals. Show that there exists an element c e K (the quotient field of o) such that ca is an ideal relatively prime to b. In particular, every ideal class in Pic(o) contains representative ideals prime to a given ideal. For a continuation, see Exercise 7 of Chapter VII.
CHA PT E R
Ill
M odul es
Although this chapter is logically selfcontained and prepares for future topics , in practice readers will have had some acquaintance with vector spaces over a field . We generalize this notion here to modules over rings . It is a standard fact (to be reproved) that a vector space has a basis , but for modules this is not always the case . Sometimes they do; most often they do not. We shall look into cases where they do . For examples of modules and their relations to those which have a basis , the reader should look at the comments made at the end of §4 .
§1 .
BAS I C D E FI N ITI O N S
Let A be a ring. A left module over A , or a left A module M is an abelian group, usually written additively, together with an operation of A on M (viewing A as a multiplicative monoid by RI 2), such that, for all a , b E A and x, y E M we have
(a + b)x = ax + bx and a(x + y) = ax
+ ay.
We leave it as an exercise to prove that a(  x) =  (ax) and that Ox = 0. By definition of an operation, we have 1 x = x. In a similar way, one defines a right Amodule. We shall deal only with left Amodules, unless otherwise specified, and hence call these simply Amodules, or even modules if the reference is clear.
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MODULES
Let M be an Amodule. By a submodule N of M we mean an additive sub group such that AN c N. Then N is a module (with the operation induced by that of A on M). Examples
We note that A is a module over itself. Any commutative group is a Zmodule. An additive group consisting of 0 alone is a module over any ring. Any left ideal of A is a module over A . Let J be a twosided ideal of A . Then the factor ring A/J i s actually a module over A . If a E A and a + J is a coset of J in A , then one defines the operation to be a (x + J) = ax + J . The reader can verify at once that this defines a module structure on A/1. More general , if M is a module and N a submodule , we shall define the factor module below . Thus if L is a left ideal of A , then A/L is also a module . For more examples in this vein , see §4 . A module over a field is called a vector space . Even starting with vector spaces , one is led to consider modules over rings . Indeed , let V be a vector space over the field K. The reader no doubt already knows about linear maps (which will be recalled below systematically) . Let R be the ring of all linear maps of V into itself. Then V is a module over R . Similarly , if V = K n denotes the vector space of (vertical) ntuples of elements of K, and R is the ring of n X n matrices with components in K, then V is a module over R. For more comments along these lines , see the examples at the end of §2. Let S be a nonempty set and M an Amodule . Then the set of maps Map(S , M ) is an Amodule. We have already noted previously that it is a com mutative group, and for f E Map(S, M ) , a E A we define af to be the map such that (aj)(s) = af(s) . The axioms for a module are then trivially verified . For further examples , see the end of this section . For the rest of this section, we deal with a fixed ring A , and hence may omit the prefix A  . Let A be an entire ring and let M be an Amodule . We define the torsion submodule Mtor to be the subset of elements x E M such that there exists a E A , a =f=. 0 such that ax = 0 . It is immediately verified that Mtor is a submodule . Its structure in an important case will be determined in § 7 . Let a be a left ideal, and M a module. We define aM to be the set o f all elements with a i E a and x i E M. It is obviously a submodule of M. If a, b are left ideals, then we have associativity, namely a(bM)
=
( ab )M.
Ill, §1
BASIC DEFIN ITIONS
119
We also have some obvious distributivities, like (a + b)M = aM + bM. If N , N ' are submodules of M, then a(N + N ' ) = aN + aN ' . Let M be an Amodule, and N a submodule. We shall define a module structure on the factor group M/N (for the additive group structure). Let x + N be a coset of N in M, and let a E A . We define a(x + N) to be the coset ax + N. It is trivial to verify that this is well defined (i.e. if y is in the same coset as x, then ay is in the same coset as ax), and that this is an opera tion of A on M/ N satisfying the required condition, making M/ N into a module, called the factor module of M by N. By a modulehomomorphism one means a map
f : M __.. M ' of one module into another (over the same ring A), which is an additive group homomorphism, and such that
f(ax)
=
af(x)
for all a E A and x E M. It is then clear that the collection of Amodules is a category, whose morphisms are the modulehomomorphisms usually also called homomorphisms for simplicity, if no confusion is possible. If we wish to refer to the ring A , w e also say that f is an A homomorphism, or also that it is an Alinear map. If M is a module, then the identity map is a homomorphism. For any module M ' , the map ( : M __.. M ' such that ((x) = 0 for all x E M is a homo morphism, called zero. In the next section , we shall discuss the homomorphisms of a module into itself, and as a result we shall g i v e further examples of modules which arise in practice . Here we continue to tabulate the translation of basic properties of groups to modules .
Let M be a module and N a submodule. We have the canonical additive grouphomomorphism f: M
__..
M/N
and one verifies trivially that it is a modulehomomorphism. Equally trivially, one verifies that f is universal in the category of homomorphisms of M whose kernel contains N.
If f : M __.. M' is a modulehomomorphism, then its kernel and image are submodules of M and M' respectively (trivial verification). Letf : M � M ' be a homomorphism. By the cokernel off we mean the factor mo du l e M '/ Im f = M ' /J(M) . One may also mean the canonical homomorphi sm
1 20
I l l , §1
MODULES
M ' � M '/f(M) rather than the module itself. The context should make clear which is meant. Thus the cokernel is a factor module of M ' . Canonical homomorphisms discussed in Chapter I , §3 apply to modules mutatis mutandis. For the convenience of the reader, we summarise these homomorphisms:
Let N , N ' be two submodules of a module M . Then N module, and we have an isomorphism N/(N n N')
If M
::J
M'
::J
�
(N
+
N')/N' .
(M/M")/(M'/M")
�
MjM' .
+
N' is also a sub
M" are modules, then
Iff : M __.. M' is a modulehomomorphism, and N' is a submodule of M', then f  1{N') is a submodule of M and we have a canonical injective homomorphism f : Mjf  1(N') __.. M'jN'.
Iff is surjective, then J is a moduleisomorphism. The proofs are obtained by verifying that all homomorphisms which ap peared when dealing with abelian groups are now Ahomomorphisms of modules. We leave the verification to the reader. As with groups, we observe that a modulehomomorphism which is bijective is a moduleisomorphism. Here again, the proof is the same as for groups, adding only the observation that the inverse map, which we know is a group isomorphism, actually is a moduleisomorphism. Again, we leave the verifica tion to the reader. As with abelian groups, we define a sequence of modulehomomorphisms M' !. M � M" to be exact if lm f = Ker g. We have an exact sequence associated with a submodule N of a module M, namely 0 __.. N __.. M __.. M/N __.. 0, the map of N into M being the inclusion, and the subsequent map being the canonical map. The notion of exactness is due to EilenbergSteenrod. If a homomorphism u : N __.. M is such that
O __. N � M is exact, then we also say that u is a monomorphism or an embedding. Dually, if is exact, we say that u is an epimorphism .
BASIC DEFINITIONS
Ill, §1
1 21
Algebras
There are some things in mathematics which satisfy all the axioms of a ring except for the existence of a unit element . We gave the example of L 1 (R) in Chapter II , § 1 . There are also some things which do not satisfy associativity , but satisfy distributivity . For instance let R be a ring , and for x, y E R define the bracket product [x, y] = xy  yx. Then this bracket product is not associative in most cases when R is not com mutative , but it satisfies the distributive law. Examples.
A typical example is the ring ofdifferential operators with coo coefficients , operating on the ring of coo functions on an open set in R n . The bracket product [D t , D2 ] = D t o D2  D2 o D t of two differential operators is again a differential operator. In the theory of Lie groups , the tangent space at the origin also has such a bracket product . Such considerations lead us to define a more general notion than a ring . Let A be a commutative ring . Let E, F be modules . By a bilinear map g: E X E � F we mean a map such that given x E E, the map y � g(x, y) is Alinear, and given y E E, the map x � g(x, y) is Alinear. By an Aalgebra we mean a module together with a bilinear map g : E x E � E. We view such a map as a law of composition on E. But in this book, unless otherwise specified , we shall assume that our algebras are associative and have a unit element . Aside from the examples already mentioned , we note that the group ring A [G] (or monoid ring when G is a monoid) is an Aalgebra , also called the group (or monoid ) algebra . Actually the group algebra can be viewed as a special case of the following situation . Let f : A � B be a ringhomomorphism such that f(A) is contained in the center of B , i . e . , f(a) commutes with every element of B for every a E A . Then we may view B as an Amodule , defining the operation of A on B by the map (a , b) � f(a)b for all a E A and b E B . The axioms for a module are trivially satisfied , and the multiplicative law of composition B x B � B is clearly bilinear (i . e . , Abilinear) . In this book , unless otherwise specified , by an algebra over A , we shall always mean a ringhomomorphism as above . We say that the algebra is finitely gen erated if B is finitely generated as a ring over f(A ) . Several examples o f modules over a polynomial algebra or a group algebra will be given in the next section , where we also establish the language of representations .
1 22
MODULES
§2.
TH E G R O U P O F H O M O M O R P H I S M S
I l l , §2
Let A be a ring, and let X, X' be Amodules. We denote by HomA(X', X) the set of Ahomomorphisms of X' into X. Then HomA(X' , X) is an abelian group, the law of addition being that of addition for mappings into an abelian group. If A is commutative then we can make HomA(X', X) into an Amodule, by defining aj for a E A and jE HomA(X', X) to be the map such that
(af)(x) = af(x). The verification that the axioms for an Amodule are satisfied is trivial. However, if A is not commutative, then we view Hom A( X', X) simply as an abelian group: We also view HomA as a functor. It is actually a functor of two variables, contravariant in the first and covariant in the second. Indeed, let Y be an Amodule, and let X' !. X be an Ahomomorphism. Then we get an induced homomorphism HomA(f, Y) : HomA(X, Y) __.. HomA(X', Y) (reversing the arrow !) given by
g � g of This is illustrated by the following sequence of maps : X' !. X � Y. The fact that HomA(f, Y) is a homomorphism is simply a rephrasing of the property (g 1 + g 2 ) o f = g 1 o f + g 2 o f, which is trivially verified. If f = id, then composition with f acts as an identity mapping on g, i.e. g o id = g. If we have a sequence of Ahomomorphisms X' __.. X __.. X", then we get an induced sequence
Proposition 2.1 .
A sequence
X' � X __.. X" __.. 0
is exact if and on/y if the sequence HomA(X', Y) � HomA{X, Y) � HomA(X", Y) � 0
is exact for all Y.
I l l , §2
TH E G ROUP OF HOMOMORPHISMS
1 23
Proof This is an important fact, whose proof is easy. For instance, suppose the first sequence is exact. If g : X" + Y is an A homomorphism, its image in HomA(X, Y) is obtained by composing g with the surjective map of X on X". If this composition is 0, it follows that g = 0 because X + X" is surjective. As another example, consider a homomorphism g : X + Y such that the composition
I \
is 0. Then g vanishes on the image of A.. Hence we can factor g through the factor module, X/Im A.
X
g
y
Since X + X" is surjective, we have an isomorphism X/Im A. ++ X". Hence we can factor g through X", thereby showing that the kernel of HomA(X', Y) � HomA{X, Y) is contained in the image of HomA(X, Y) � HomA(X", Y). The other conditions needed to verify exactness are left to the reader. So is the converse. We have a similar situation with respect to the second variable, but then the functor is covariant. Thus if X is fixed, and we have a sequence of A homomorphisms Y' + Y + Y", then we get an induced sequence
Proposition 2.2.
A sequence
0 + Y' + Y + Y", is exact if and only if 0 + HomA{X, Y') + HomA(X, Y) + HomA(X, Y")
is exact for all X.
1 24
I l l , §2
MODU LES
The verification will be left to the reader. It follows at once from the defini tions. We note that to say that 0 � Y' � Y is exact means that Y' is embedded in Y, i.e. is isomorphic to a submodule of Y. A homomorphism into Y' can be viewed as a homomorphism into Y if we have Y' c: Y. This corresponds to the injection Let Mod( A ) and Mod( B) be the categories of modules over rings A and B, and let F : Mod(A ) + Mod(B) be a functor. One says that F is exact if F transforms exact sequences into exact sequences. We see that the Hom functor in either variable need not be exact if the other variable is kept fixed. In a later section, we define conditions under which exactness is preserved. Endomorphisms .
Let M be an Amodule . From the relations
and its analogue on the right, namel y
and the fact that there is an identity for composition, namely idM , we conclude that HomA(M, M ) is a ring, the multiplication being defined as composition of mappings. If n is an integer > 1, we can write f n to mean the iteration of f with itself n times, and define f 0 to be id. According to the general definition of endomorphisms in a category , we also write EndA (M) instead of HomA(M, M) , and we call EndA(M) the ring of endomorphisms. Since an Amodule M is an abelian group, we see that Homz(M, M) ( = set of grouphomomorphisms of M into itself) is a ring, and that we could have defined an operation of A on M to be a ringhomomorphism A � Homz(M, M). Let A be commutative. Then M is a module over EndA(M ) . If R is a subring of EndA(M) then M is a fortiori a module over R . More generally , let R be a ring and let p : R � EndA(M) be a ring homomorphism. Then p is called a representation of R on M. This occurs especially if A = K is a field . The linear algebra of representations of a ring will be discussed in Part III , in several contexts , mostly finitedimensional . Infinitedimensional examples occur in anal ysis , but then the representation theory mixes algebra with analysis , and thus goes beyond the level of this course . Let K be a field and let V be a vector space over K. Let D : V � V be an endomorphism (Klinear map) . For every polynomial ; P(X) E K[X] , P(X) = L a;X with a; E K, we can define Example.
TH E G ROUP OF HOMOMORPH ISMS
I l l ' §2
1 25
P(D) = L a;D ; : V � V as an er:tdomorphism of V. The association P(X) � P(D ) gives a representation p : K[X] � EndK(V) , which makes V into a K[X]module. It will be shown in Chapter IV that K[X] is a principal ring . In §7 we shall give a general structure theorem for modules over principal rings , which will be applied to the above example in the context of linear algebra for finitedimensional vector spaces in Chapter XIV , § 3 . Readers acquainted with basic linear algebra from an undergraduate course may wish to read Chapter XIV already at this point. Examples for infinitedimensional vector spaces occur in analysis . For instance , let V be the vector space of complexvalued coo functions on R . Let D = d/ dt be the derivative (if t is the variable) . Then D : V � V is a linear map, and C[X] has the representation p : C[X] � Endc(V) given by P � P(D ) . A similar situation exists in several variables , when we let V be the vector space of Coo functions in n variables on an open set of R n . Then we let D; = a/ at; be the partial derivative with respect to the i th variable ( i = 1 , . . . , n) . We obtain a representation such that p(X; ) = D; . Example . Let H be a Hilbert space and let A be a bounded hermitian oper
ator on A . Then one considers the homomorphism R[X] � R[A] C End(H) , from the polynomial ring into the algebra of endomorphisms of H, and one extends this homomorphism to the algebra of continuous functions on the spec trum of A . Cf. my Real and Fun ctional Analysis, Springer Verlag, 1 993 .
h
Representations form a category as follows . We define a morphism of a representation p : R � End A (M) into a representation p' : R � EndA (M' ) , or in other words a homomorphism of one representation of R to another , to be an Amodule homomorphism h : M � M ' such that the following diagram is commutative for every a E R:
j
M p ( a)
j h ....
+
M 
M' p ' ( a)
j [h l
M'
In the case when h is an isomorphism, then we may replace the above diagram by the commutative diagram EndA(M)
Ry
� EndA(M')
1 26
I l l , §2
MODU LES
where the symbol [h] denotes conjugation by h , i . e . for f E EndA(M ) we have [ h]f = h f h 1 • Representations: from a monoid to the monoid algebra. Let G be a monoid . By a representation of G on an Amodule M, we mean a homomor phism p : G � EndA(M) of G into the multiplicative monoid of EndA(M) . Then we may extend p to a homomorphism of the monoid algebra 0
0
A [G] � EndA(M) , by letting
It is immediately verified that this extension of p to A [G] is a ring homomorphism, coinciding with the given p on elements of G . Examples: modules over a group ring.
The next examples will follow a certain pattern associated with groups of automorphisms . Quite generally, sup pose we have some category of objects , and to each object K there is associated an abelian group F(K) , functorially with respect to isomorphisms . This means that if u: K � K' is an isomorphism , then there is an associated isomorphism F(u) : F(K' ) � F(K' ) such that F(id) = id and F( uT) = F(u) F(T) . Then the group of automorphisms Aut(K) of an object operates on F(K) ; that is , we have a natural homomorphism o
Aut(K) � Aut(F(K) ) given by u � F( u) . Let G = Aut(K) . Then F(K) (written additively) can be made into a module over the group ring Z[G] as above . Given an element a = L auu E Z[G] , with au E Z , and an element x E F(K) , we define ax = L auF(u)x. The conditions defining a module are trivially satisfied . We list several concrete cases from mathematics at large , so there are no holds barred on the terminology . Let K be a number field (i . e . a finite extension of the rational numbers) . Let G be its group of automorphisms . Associated with K we have the following objects: the ring of algebraic integers o K ; the group of units ok ; the group of ideal classes C(K) ; the group of roots of unity �(K) . Then G operates on each of those objects , and one problem is to determine the structure of these objects as Z[G] modules . Already for cyclotomic fields this
I l l , §3
DIR ECT PRODUCTS AND SUMS OF MODULES
1 27
determination gives rise to substantial theories and to a number of unsolved problems . Suppose that K is a Galois extension of k with Galois group G (see Chapter VI) . Then we may view K itself as a module over the group ring k[ G] . In Chapter VI , § 1 3 we shall prove that K is isomorphic to k[G] as module over k[G] itself. In topology , one considers a space X0 and a finite covering X. Then Aut(X/X0) operates on the homology of X, so this homology is a module over the group r1 ng . With more structure , suppose that X is a projective nonsingular variety , say over the complex numbers . Then to X we can associate: the group of divisor classes (Picard group) Pic(X); in a given dimension, the group of cycle classes or Chow group CHP(X) ; the ordinary homology of X; the sheaf cohomology in general . If X is defined over a field K finitely generated over the rationals , we can associate a fancier cohomology defined algebraically by Grothendieck, and func torial with respect to the operation of Galois groups . Then again all these objects can be viewed as modules over the group ring of automorphism groups , and major problems of mathematics consist in deter mining their structure . I direct the reader here to two surveys , which contain extensive bibliographies . [CCFT 9 1 ]
P. CASSOUNOGUES , T . CHINBURG , A . FROHLICH , M . J . TAYLOR , L functions and Galois modules , in Lfunctions and Arithmetic J . Coates and M . J . Taylor (eds . ) , Proceedings of the Durham Symposium July 1 989 , London Math, Soc . Lecture Note Series 153 , Cambridge University Press
( 1 99 1 ) ,
[La 82]
§3 .
pp .
75 1 39
S. LANG , Units and class groups in number theory and algebraic geometry ,
Bull. AMS
Vol . 6 No . 3
( 1 982) ,
pp .
2533 1 6
D I R ECT P R O D U CTS A N D S U M S O F M O D U LE S
Let A be a ring . Let {M;};E 1 be a family of modules . We defined their direct product as abelian groups in Chapter I , §9. Given an element (x;);E J of the direct product, and a E A , we define a (x; ) = ( ax; ) . In other words, we multiply by an element a componentwise . Then the direct product IlM; is an Amodule . The reader will verify at once that it is also a direct product in the category of Amodules .
1 28
I l l , §3
MODULES
Similarly , let M = EB Mi ie I be their direct sum as abelian groups. We define on M a structure of Amodule : If ( x Ji e i is an element of M, i.e. a family of elements X ; E M; such that X ; = 0 for almost all i, and if a E A, then we define
a(x ;)ie i = (ax ; ) i e i ' that is we define multiplication by a componentwise. It is trivially verified that this is an operation of A on M which makes M into an Amodule. If one refers back to the proof given for the existence of direct sums in the category of abelian groups, one sees immediately that this proof now extends in the same way to show that M is a direct sum of the family {M i } i e J as Amodules. (For instance, the map A.i : Mi __. M such that A.1{x ) has jth component equal to x and ith component equal to 0 for i 1= j is now seen to be an Ahomomorphism.) This direct sum is a coproduct in the category of A modules . Indeed , the reader can verify at once that given a family of Ahomomorphisms {/; : M; � N} , the map f defined as in the proof for abelian groups is also an A isomorphism and has the required properties . See Proposition 7 . 1 of Chapter I . When I is a finite set, there is a useful criterion for a module t o be a direct product. Proposition
i=
1,
.
3. 1 . Let M be an A module and n an integer > 1. f"'or each
. . , n let CfJ i : M __.. M be an A homomorphism such that n
L CfJ i = id and
o
cpi = 0
i= j
. i= 1 Then cpf = cp ifor all i. Let Mi = cp ; {M ), and let cp : M __.. 0 M; be such that C{J;
if i
cp (x) = {cp 1{x ), . . . , CfJn (x)).
Then cp is an Aisomorphism of M onto the direct product 0 M; . Proof For each j, we have
n
id = cpj o L cp i = (/J j o (/J j = cpf, i= 1 thereby proving the first assertion. It is clear that cp is an Ahomomorphism. Let x be in its kernel. Since (/J j
=
(/Jj
o
n
x = id (x) = L cp ;( x ) i= l
I l l , §3
D I RECT PRODUCTS AND SUMS O F MODULES
1 29
we conclude that x = 0, so lfJ is injective. Given elements Y i E M i for each i = 1, . . . , n, let x = y 1 + · · · + Yn · We obviously have qJ1{yi ) = 0 if i =F j. Hence
for each j = 1 , . . . , n. This proves that of our proposition.
lfJ
is surjective, and concludes the proof
We observe that when I is a finite set, the direct sum and the direct product are equal. Just as with abelian groups, we use the symbol (f) to denote direct sum. Let M be a module over a ring A and let S be a subset of M. By a linear combination of elements of S (with coefficients in A) one means a sum
where {ax } is a set of elements of A, almost all of which are equal to 0. These elements ax are called the coefficients of the linear combination . Let N be the set of all linear combinations of elements of S. Then N is a submodule of M, for if
L ax x and
xeS
L bx x
xeS
are two linear combinations, then their sum is equal to
and if c E A, then
and these elements are again linear combinations of elements of S. We shall call N the submodule generated by S, and we call S a set of generators for N. We sometimes write N = A (S). If S consists of one element x, the module generated by x is also written Ax, or simply (x), and sometimes we say that (x) is a principal module.
A module M is said to be finitely generated, or of finite type, or finite over A, if it has a finite number of generators. A subset S of a module M is said to be linearly independent (over A) if when ever we have a linear combination
1 30
I l l , §3
MODULES
which is equal to 0, then ax two linear combinations
=
0 for all x E S. If S is linearly independent and if
L ax x and L bxx are equal, then ax = bx for all x E S. Indeed, subtracting one from the other yields L (ax  bx)x = 0, whence ax  bx = 0 for all x. If S is linearly indepen dent we shall also say that its elements are linearly independent. Similarly, a family {xi}i e i of elements of M is said to be linearly independent if whenever we have a linear combination � �
i
EI
a·X· ' l
=
0'
then a i = 0 for all i. A subset S (resp. a family {x i }) is called linearly dependent if it is not linearly independent, i.e. if there exists a relation L ax x xeS
=
0 resp.
" a l·X·' i �
=
0
EI
with not all ax (resp. a i) = 0. Warning. Let x be a single element of M which is linearly independent. Then the family {xi} i 1 , . . . such that x i = x for all i is linearly dependent if n > 1, but the set consisting of x itself is linearly inde pendent. Let M be an Amodule, and let {Mi} i e i be a family of submodules. Since we have inclusionhomomorphisms =
,
n
we have an induced homomorphism which is such that for any family of elements (x i ) i e I , all but a finite number of which are 0, we have
A. . ((x i))
=
L xi . iEI
If A. * is an isomorphism, then we say that the family { M J i e I is a direct sum decomposition of M. This is obviously equivalent to saying that every element of M has a unique expression as a sum
with x i E Mb and almost all x i in this case.
=
0. By abuse of notation, we also write
I l l , §3
DIRECT PRODUCTS AND SUMS OF MODULES
1 31
If the family { M i } is such that every element of M has some expression as a sum L x i (not necessarily unique), then we write M = L Mi . In any case, if {M;} is an arbitrary family of submodules, the image of the homomorphism A. * above is a submodule of M, which will be denoted by L M i . If M is a module and N, N' are two submodules such that N + N' = M and N n N' = 0, then we have a moduleisomorphism
M
�
N (f) N',
just as with abelian groups, and similarly with a finite number of submodules. We note, of course, that our discussion of abelian groups is a special case of our discussion of modules, simply by viewing abelian groups as modules over Z. However, it seems usually desirable (albeit inefficient) to develop first some statements for abelian groups, and then point out that they are valid (obviously) for modules in general. Let M, M', N be modules. Then we have an isomorphism of abelian groups
and similarly
The first one is obtained as follows. Iff : M (f) M' __.. N is a homomorphism, then f induces a homomorphismf1 : M __.. N and a homomorphismf2 : M' __.. N by composing f with the injections of M and M' into their direct sum re spectively :
M __.. M (f) {0} M' __.. {0} (f) M'
c c
M (f) M' !. N, M (f) M' !. N .
We leave it to the reader to verify that the association
gives an isomorphism as in the first box. The isomorphism in the second box is obtained in a similar way. Given homomorphisms ft : N � M and
1 32
I l l , §3
MODULES
we have a homomorphism f : N + M x M' defined by It is trivial to verify that the association
gives an isomorphism as in the second box. Of course , the direct sum and direct product of two modules are isomorphic , but we distinguished them in the notation for the sake of functoriality , and to fit the infinite case , see Exercise 22 . 3.2. Let 0 + M' .!. M !!.. M" + 0 be an exact sequence of modules. The following conditions are equivalent : Propo�ition
1 . There exists a homomorphism
: M" + M such that g o cp = id. 2. There exists a homomorphism 1/J : M M ' such that 1/1 of = id. cp
+
If these conditions are satisfied, then we have isomorphisms : M = Ker g (f) lm cp,
M = Im f ffi Ker 1/J, M
�
M ' (f) M" .
Proof. Let us write the homomorphisms on the right : M � M" + 0. qJ
Let x E M. Then
x  qJ(g(x)) is in the kernel of g, and hence M = Ker g This sum is direct, for if
+
Im qJ .
x=y+z with y E Ker g and z E lm qJ , z = qJ{ w) with w E M" , and applying g yields g(x) = w. Thus w is uniquely determined by x, and therefore z is uniquely determined by x. Hence so is y, thereby proving the sum is direct. The arguments concerning the other side of the sequence are similar and will be left as exercises, as well as the equivalence between our conditions. When these conditions are satisfied , the exact sequence of Proposition 3 . 2 is said to split . One also says that fjJ splits f and cp splits g.
I l l , §3
D I RECT PRODUCTS AN D SUMS OF MODULES
1 33
Abelian categories
Much in the theory of modules over a ring is arrowtheoretic. In fact, one needs only the notion of kernel and cokernel (factor modules). One can axi omatize the special notion of a category in which many of the arguments are valid, especially the arguments used in this chapter. Thus we give this axi omatization now, although for concreteness, at the beginning of the chapter, we continue to use the language of modules. Readers should strike their own balance when they want to slide into the more general framework. Consider first a category a such that Mor(E, F) is an abelian group for each pair of objects E, F of a, satisfying the following two conditions : AB 1 .
The law of composition of morphisms is bilinear, and there exists a zero object 0, i.e. such that Mor(O, E) and Mor(E, 0) have precisely one element for each object E. AB 2. Finite products and finite coproducts exist in the category.
Then we say that a is an additive category. Given a morphism E .!.. F in a, we define a kernel off to be a morphism E ' + E such that for all objects X in the category, the following sequence is exact : 0 + Mor(X, E') + Mor(X, E) + Mor(X, F). l
We define a cokernel for f to be a morphism F + F" such that for all objects X in the category, the following sequence is exact : 0 � Mor(F'', X) � Mor(F, X) � Mor(E, X) .
It is immediately verified that kernels and cokernels are universal in a suitable category, and hence uniquely determined up to a unique isomorphism if they exist. AB 3. Kernels and cokernels exist. AB 4.
If j : E + F is a morphism whose kernel is 0, then j is the kernel of its co kernel. If f : E + F is a morphism whose co kernel is 0, then f is the cokernel of its kernel. A morphism whose kernel and cokernel are 0 is an isomorphism.
A category a satisfying the above four axioms is· called an abelian category. In an abelian caegory , the group of morphisms is usually denoted by Hom, so for two objects E, F we write Mor(E, F)
=
Hom{E, F).
The morphisms are usually called homomorphisms. Given an exact sequence 0 + M' + M,
1 34
I l l , §3
MODULES
we say that M ' is a subobject of M, or that the homomorphism of M ' into M is a monomorphism . Dually, in an exact sequence
M + M" + 0, we say that M" is a quotient object of M, or that the homomorphism of M to M" is an epimorphism, instead of saying that it is surjective as in the category of modules. Although it is convenient to think of modules and abelian groups to construct proofs, usually such proofs will involve only arrowtheoretic argu ments , and will therefore apply to any abelian category . However, all the abelian categories we shall meet in this book will have elements , and the kernels and co kernels will be defined in a natural fashion , close to those for modules , so readers may restrict their attention to these concrete cases . Examples of abelian categories . Of course , modules over a ring form an
abelian category , the most common one. Finitely generated modules over a Noetherian ring form an abelian category , to be studied in Chapter X . Let k be a field . We consider pairs (V, A) consisting of a finitedimensional vector space V over k, and an endomorphism A : V � V. By a homomorphism (morphism) of such pairs f : (V, A) � (W, B ) we mean a khomomorphism f : V � W such that the following diagram is commutative :
v
f
w
It is routinely verified that such pairs and the above defined morphisms form an abelian category . Its elements will be studied in Chapter XIV . Let k be a field and let G be a group. Let Modk (G) be the category of finite dimensional vector spaces V over k, with an operation of G on V, i . e . a homo morphism G � Autk (V) . A homomorphism (morphism) in that category is a k homomorphism f : V � W such that f(ax) = af(x) for all x E V and a E G . It is immediate that Mod k ( G) is an abelian category . This category will be studied especially in Chapter XVIII . In Chapter XX , § 1 we shall consider the category of complexes of modules over a ring . This category of complexes is an abelian category . In topology and differential geometry , the category of vector bundles over a topological space is an abelian category . Sheaves of abelian groups over a topological space form an abelian category , which will be defined in Chapter XX , §6.
I l l , §4
§4.
1 35
FREE MODULES
F R E E M O D U L ES
Let M be a module over a ring A and let S be a subset of M. We shall say that S is a basis of M if S is not empty, if S generates M, and if S is linearly independent. If S is a basis of M, then in particular M =F {0} if A =F {0} and every element of M has a unique expression as a linear combination of elements of S. Similarly, let {x;} ;e i be a nonempty family of elements of M. We say that it is a basis of M if it is linearly independent and generates M. If A is a ring, then as a module over itself, A admits a basis, consisting of the unit element 1 . Let I be a nonempty set, and for each i E I, let A; = A, viewed as an A module. Let
Then F admits a basis, which consists of the elements e; of F whose ith com ponent is the unit element of A;, and having all other components equal to 0. By a free module we shall mean a module which admits a basis, or the zero module. Theorem 4. 1 . Let A be a ring and M a module over A. Let I be a nonempty e
set, and let {x;}; e 1 be a basis of M. Let N be an Amodule, and let {y i } i i be a family of elements of N. Then there exists a unique homomorphism f : M __.. N such that f(x i ) = Y ; for all i. e
Proof Let x be an element of M. There exists a unique family {a i } i i of elements of A such that We define It is then clear that f is a homomorphism satisfying our requirements, and that it is the unique such, because we must have
e
Corollary 4.2.
Let the notation be as in the theorem, and assume that {y i } i is a basis of N. Then the homomorphism f is an isomorphism, i.e. a module isomorphism. Proof By symmetry, there exists a unique homomorphism g : N __. M
1
1 36
I l l , §4
MODULES
=
X ; for all i, and fo g and g of are the respective identity map
Corollary 4.3.
Two modules having bases whose cardinalities are equal are
such that g(y;) pings. .
isomorphic.
Proof. Clear. We shall leave the proofs of the following statements as exercises. Let M be a free module over A, with basis {x i } i e J , so that
Let a be a two sided ideal of A. Then aM is a submodule of M. Each submodule of Ax;. We have an isomorphism (of Amodules)
MjaM
�
ax;
is a
EB Ax;/ax; . ie I
Furthermore, each Ax;/ax; is isomorphic to Aja, as Amodule.
Suppose in addition that A is commutative. Then Aja is a ring. r·urthermore MjaM is a free module over Aja, and each Ax;/ax; is free over Aja. lfx; is the image of X ; under the canonical homomorphism then the single element i; is a basis of Ax;/ax; over A/a. All of these statements should be easily verified by the reader. Now let A be an arbitrary commutative ring . A module M is called principal if there exists an element x E M such that M = Ax. The map
a � ax (for a E A) is an Amodule homomorphism of A onto M, whose kernel is a left ideal a, and inducing an isomorphism of Amodules A/ a
�
M.
Let M be a finitely generated module , with generators { v i , . . . , vn } . Let F be a free module with basis { e i , . . . , en } . Then there is a unique surjective homomorphism / : F + M such that f(e; ) = V; . The kernel off is a submodule M I · Under certain conditions , M I is finitely generated (cf. Chapter X , § 1 on Noetherian rings) , and the process can be continued . The systematic study of this process will be carried out in the chapters on resolutions of modules and homology .
I l l , §4
FREE MODULES
1 37
Of course , even if M is not finitely generated , one can carry out a similar construction, by using an arbitrary indexing set. Indeed , let { v;} (i E / ) be a family of generators . For each i , let F; be free with basis consisting of a single element e; , so F; A . Let F be the direct sum of the modules F; ( i E I ) , as in Proposi tion 3 . 1 . Then we obtain a surjective homomorphism f : F � M such that f(e; ) = V; . Thus every module is a factor module of a free module. Just as we did for abelian groups in Chapter I, §7 , we can also define the free module over a ring A generated by a nonempty set S. We let A (S) be the set of functions cp : S + A such that cp(x) = 0 for almost all x E S. If a E A and x E S, we denote by ax the map cp such that cp(x) = a and cp(y) = 0 for y =f=. x. Then as for abelian groups , given cp E A (S) there exist elements a; E A and X; E S such that cp = a l x l + . . . + a n xn . =
It is immediately verified that the family of functions { «Sx } (x E S) such that 8x (x) = 1 and 8x ( Y) = 0 for y =f=. x form a basis for A(S) . In other words , the ex pression of cp as 2: a;x; above is unique . This construction can be applied when S is a group or a monoid G , and gives rise to the group algebra as in Chapter II , § 5 . Projective modules
There exists another important type of module closely related to free modules , which we now discuss . Let A be a ring and P a module. The following properties are equivalent, and define what it means for P to be a projective module . P 1.
Given a homomorphism f: P __.. M" and surjective homomorphism g : M __.. M", there exists a homomorphism h : P __.. M making the following diagram commutative.
M + M" g
+
0
P 2.
Every exact sequence 0 __.. M' __.. M" __.. P __.. 0 splits.
P 3.
There exists a module M such that P (f) M is free, or in words, P is a direct summand of a free module.
P 4. The functor M �+ HomA(P, M) is exact.
We prove the equivalence of the four conditions.
1 38
I l l , §4
MODULES
Assume P 1. Given the exact sequence of P 2, we consider the map f = id in the diagram
M"
+
P
.
0
Then h gives the desired splitting of the sequence. Assume P 2. Then represent P as a quotient of a free module (cf. Exercise 1 ) F __.. P __.. 0, and apply P 2 to this sequence to get the desired splitting, which represents F as a direct sum of P and some module. Assume P 3. Since HomA(X (f) Y, M) = HomA(X, M) (f) HomA{ Y, M), and since M �+ HomA(F, M) is an exact functor if F is free, it follows that HomA(P, M) is exact when P is a direct summand of a free module, which proves P 4.
Assume P 4 . The proof of P 1 will be left as an exercise.
Examples.
It will be proved in the next section that a vector space over a field is always free , i . e . has a basis . Under certain circumstances , it is a theorem that projective modules are free. In §7 we shall prove that a finitely generated projective module over a principal ring is free . In Chapter X , Theorem 4 . 4 we shall prove that such a module over a local ring is free ; in Chapter XVI , Theo rem 3 . 8 we shall prove that a finite flat module over a local ring is free; and in Chapter XXI , Theorem 3 . 7 , we shall prove the QuillenSuslin theorem that if A = k[X 1 , Xn ] is the polynomial ring over a field k, then every finite pro jective module over A is free . Projective modules give rise to the Grothendieck group. Let A be a ring . Isomorphism classes of finite projective modules form a monoid . Indeed , if P is finite projective, let [P] denote its isomorphism class . We define •
•
•
,
[P] + [Q]
=
[P (f) Q] .
This sum is independent of the choice of representatives P , Q in their class . The conditions defining a monoid are immediately verified . The corresponding Groth endieck group is denoted by K(A) . We can impose a further equivalence relation that P is equivalent to P ' if there exist finite free modules F and F ' such that P EB F is isomorphic to P' EB F '. Under this equivalence relation we obtain another group denoted by K0 (A) . If A is a Dedekind ring (Chapter II , § I and Exercises 1 3 1 9) it can be shown that this group is isomorphic in a natural way with the group of ideal classes Pic( A) (defined in Chapter II , § 1 ) . See Exercises 1 1 , 1 2 , 1 3 . It is also a
I l l , §5
VECTOR SPAC ES
1 39
problem to determine K0(A) for as many rings as possible , as explicitly as pos sible . Algebraic number theory is concerned with K0(A) when A is the ring of algebraic integers of a number field . The QuillenSuslin theorem shows if A is the polynomial ring as above , then K0(A) is trivial . Of course one can carry out a similar construction with all finite modules . Let [M] denote the isomorphism class of a finite module M. We define the sum to be the direct sum . Then the isomorphism classes of modules over the ring form a monoid , and we can associate to this monoid its Grothendieck group . This construction is applied especially when the ring is commutative . There are many variations on this theme . See for instance the book by Bass , Algebraic Ktheory , Benjamin , 1 968 . There is a variation of the definition of Grothendieck group as follows . Let F be the free abelian group generated by isomorphism classes of finite modules over a ring R , or of modules of bounded cardinality so that we deal with sets . In this free abelian group we let f be the subgroup generated by all elements [M]  [M' ]  [M"] for which there exists an exact sequence 0 � M' � M � M" � 0 . The factor group F /f is called the Grothendieck group K(R) . We shall meet this group again in §8 , and in Chapter XX , §3 . Note that we may form a similar Grothendieck group with any family of modules such that M is in the family if and only if M' and M" are in the family . Taking for th e family finite projective modules , one sees easily that the two possible definitions of the Grothendieck group coincide in that case .
§5 .
V E CTO R S PAC ES
A module over a field is called a vector space. Theorem 5. 1 . Let V be a vector space over a field K, and assume that
V =F {0}. Let r be a set of generators of V over K and let S be a subset of r which is linearly independent. Then there exists a basis CB of V such that s c CB c r. Proof Let l: be the set whose elements are subsets T of r which contain S and are linearly independent. Then l: is not empty (it contains S), and we contend that l: is inductively ordered. Indeed, if { �} is a totally ordered subset
1 40
I l l , §5
MODULES
of l: (by ascending inclusion), then U � is again linearly independent and con tains S. By Zorn ' s lemma, let CB be a maximal element of l:. Then CB is linearly independent. Let W be the subspace of V generated by CB. If W =F V, there exists some element x E r such that x ¢ W. Then CB u {x} is linearly inde pendent, for given a linear combination
we must have b
=
0, otherwise we get x
=

L b  1 ay y E W. ye 1 . We shall prove that any other basis must also have m elements. For this it will suffice to prove : If w 1 , , wn are elements of V which are linearly independent over K, then n < m (for we can then use symmetry). We proceed by induction. There exist elements c 1 , , em of K such that •
•
•
•
•
•
•
•
•
(1) and some c; , say c 1 , is not equal to 0. Then v 1 lies in the space generated by w 1 , v 2 , , vm over K, and this space must therefore be equal to V itself. Furthermore, w 1 , v 2 , , vm are linearly independent, for suppose b h . . . , bm are elements of K such that •
•
•
•
•
•
,
If b 1 =F 0, divide by b 1 and express w 1 as a linear combination of v 2 , vm . Subtracting from ( 1 ) would yield a relation of linear dependence among the vb which is impossible. Hence b 1 = 0, and again we must have all b; = 0 because the v; are linearly independent. •
•
•
I l l , §5
VECTOR SPAC ES
1 41
Suppose inductively that after a suitab le renumbering of the v i , we have found w b . . . , wr (r < n) such that is a basis of V. We express wr + 1 as a linear combination (2) with c i E K. The coefficients of the V; in this relation cannot all be 0 ; otherwise there would be a linear dependence among the wi . Say cr+ 1 =F 0. Using an argument similar to that used above, we can replace vr + 1 by wr + 1 and still have a basis of V. This means that we can repeat the procedure until r = n, and therefore that n < m, thereby proving our theorem. We shall leave the general case of an infinite basis as an exercise to the reader. [Hint : Use the fact that a finite number of elements in one basis is contained in the space generated by a finite number of elements in another basis.] If a vector space V admits one basis with a finite number of elements, say m, then we shall say that V is finite dimensional and that m is its dimension. In view of Theorem 5.2, we see that m is the number of elements in any basis of V. If V = {0}, then we define its dimension to be 0, and say that V is 0dimensional. We abbreviate dimension " by " dim " or dimK " if the reference to K is needed for clarity. When dealing with vector spaces over a field, we use the words subspace and factor space instead of submodule and factor module. 4'
Theorem 5.3.
44
Let V be a vector space over afield K, and let W be a subspace.
Then
dim K V = dimK W + dim K VjW.
Iff: V + U is a homomorphism of vector spaces over K, then dim V = dim Ker f + dim Im f
Proof. The first statement is a special case of the second, taking for f the canonical map. Let { u ; L e i be a basis of l m f, and let { wi } i e J be a basis of Ker f. Let { v ; } ; be a family of elements of V such that 1· ( v ;) = u; for each i E J. We contend that e
1
is a basis for V. This will obviously prove our assertion.
1 42
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MODULES
Let x be an element of V. Then t here exist elements {aJ i e i of K almost all of which are 0 such that f(x ) = L a i u i . ie I Hence f(x  L a i v;) = f(x )  L a i f(vi) = 0. Thus is in the kernel off, and there exist elements { bi} i e 1 of K almost all of which are 0 such that ' X  i..J a·I V I· = � � b}· W)· ' From this we see t hat x = L a i vi + L bi wi , and that {vi , wi } generates V. It remains to be shown that the family {vi , wi } is linearly independent. Suppose that there exist elements c; , di such that ' 0=� � C·I V I· + i..J dJ· W)· '
Applying /yields whence all c i = 0. From this we conclude at once that all di = 0, and hence that our family {v; , wi } is a basis for V over K, as was to be shown. Corollary 5.4.
Let V be a vector space and W a subspace. Then dim W < dim V.
If V
is finite dimensional and dim W
=
dim V then W
=
V.
Proof Clear.
§6. T H E D U A L S P AC E A N D D U A L M O D U L E be a free module over a commutative ring A . We view A as a free module of rank 1 over itself. By the dual module E v of E we shall mean the module Hom(£, A). Its elements will be called functionals . Thus a functional on E is an Alinear map f : E � A . If x E E and f E E v , we sometimes denote f(x) by (x, f) . Keeping x fixed , we see that the symbol (x, f) as a function of f E E v is Alinear in its second argument, and hence that x induces a linear map on E v, which is 0 if and only if x = 0 . Hence we get an injection E � E vv which is not always a surjection . Let E
1 43
THE DUAL SPAC E AND DUAL MODULE
I l l , §6
Let {x;};E 1 be a basis of E. For each i E I let fi be the unique functional such that fi(xi ) = 5ij (in other words , 1 if i = j and 0 if i =t= j ) . Such a linear map exists by general properties of bases (Theorem 4. 1 ) . Theorem 6. 1 . Let E be a finite free module over the commutative ring A ,
of finite dimension n . Then E v is also free, and dim E v = n . If {x i , . . . , xn } is a basis for E, and fi is the functional such that fi(xi ) = 5;i ' then {fi , . . . , fn } is a basis for E v . Proof. Let f E E v and let a; = f(x; ) (i = 1 , . . . , n) . We have f(c l x i + · · · + cn xn ) = c i f(x i ) + · · · + cnf(xn ) . Hence f = a I fi + · · · + anfn , and we see that the fi generat e E v . Furthermore , they are linearly independent , for if b I fI + · · · + bn J�"n = O with b; E K, then evaluating the lefthand side on X; yields b iJ· 1·(X·) i l = 0' whence b; = 0 for all i . This proves our theorem . Given a basis {x;} (i = 1 , . . . , n) as in the theorem , we call the basis {{;} the dual basis . In terms of these bases , W,e can express an element A of E with coordinates (a I , . . . , an ) , and an element B of E v with coordinates (b 1 , , bn ) , such that •
•
•
Then in terms of these coordinates , we see that (A , B) = a I b 1 + · · · + an b n = A · B is the usual dot product of ntuples . When E is free finite dimensional, then the map E � E v v which to each x E V associates the functional f � (x, f) on E v is an isomorphism of E onto E v v . Proof. Note that since {fi , . . . , fn } is a basis for E v , it follows from the definitions that {x I ' . . . ' xn } is the dual basis in E ' so E = E vv . Corollary 6.2.
Theorem 6.3. Let U, V, W be finite free modules over the commutative ring
A , and let
,\
cp
o� w� v� u� o
be an exact sequence of A homomorphisms. Then the induced sequence
0
�
HomA(U, A) � HomA(V, A) � HomA(W, A) � 0
1 44
I l l , §6
MODU LES
.
z .e. is also exact. Proof This is a conse q uence of P2, because a free module is projective. We now consider properties which have specifically to do with vector spaces , because we are going to take factor spaces . So we assume that we deal with vector spaces over a field K. Let V, V ' be two vector spaces, and suppose given a mapping v
X
V ' __.. K
denoted by (x, x') �+ (x, x' ) for x E V and x' E V'. We call the mapping bilinear if for each x E V the function x' 1+ (x, x') is linear, and similarly for each x' E V ' the function x �+ (x, x') is linear. An element x E V is said to be orthogonal (or perpendicular) to a subset S' of V' if (x, x' ) = 0 for all x' E S'. We make a similar definition in the opposite direction. It is clear that the set of x E V orthogonal to S' is a sub space of V. We define the kernel of the bilinear map on the left to be the subspace of V which is orthogonal to V ' , and similarly for the kernel on the right. Given a bilinear map as above ,
v
X
V ' __.. K,
let W ' be its kernel on the right and let W be its kernel on the left. Let x' be an element of V ' . Then x' gives rise to a functional on V, by the rule x �+ (x, x' ), and this functional obviously depen ds only on the coset of x' modulo W ' ; in other words, if x'1 = x� (mod W ' ), then the functionals x �+ (x, x'1 ) and x 1+ (x, x� ) are equal. Hence we get a homomorphism V'
�
yv
whose kernel is precisely W ' by definition , whence an injective homomorphism 0 � V '/W ' � v v. Since all the functionals arising from elements of V ' vanish on W, we can view them as functionals on V/W, i . e . as elements of (V/W)v . So we actually get an injective homomorphism 0 � V '/W ' � (V/W) v . One could give a name to the homomorphism g : V' � y v
I l l , §6
THE DUAL S PAC E AN D DUAL MODULE
1 45
such that
(x, x ' )
=
(x, g(x ' ))
for all x E V and x ' E V ' . However, it will usually be possible to describe it by an arrow and call it the induced map, or the natural map. Giving a name to it would tend to make the terminology heavier than necessary. X
V' � K be a bilinear map , let W, W' be its kernels on the left and right respectively, and assume that V' /W' is finite dimensional. Then the induced homomorphism V' /W' � (V/W)v is an isomorphism . Theorem 6.4. Let V
Proof. By symmetry , we have an induced homomorphism V/W � (V'/W' )v which is injective . Since dim(V'/W')v
=
dim V'/W'
it follows that V/W is finite dimensional . From the above injective homomor phism and the other, namely 0 � V'/W' � (V/W)v , we get the inequalities dim V/W < dim V '/W' and dim V '/W ' < dim V/W, whence an equality of dimensions. Hence our homomorphisms are surjective and inverse to each other, thereby proving the theorem. Remark 1 . Theorem 6.4 is the analogue for vector spaces of the duality Theorem 9. 2 of Chapter I . Remark 2.
Let A be a commutative ring and let E be an Amodule . Then we may form two types of dual : E" Ev
= =
Hom(£, Q/Z) , viewing E as an abelian group; HomA(E, A ) , viewing E as an Amodule .
Both are called dual , and they usually are applied in different contexts . For instance , E v will be considered in Chapter XIII , while E" will be considered in the theory of injective modules , Chapter XX , §4 . For an example of dual module E v see Exercise 1 1 . I f by any chance the two duals arise together and there is need to distinguish between them , then we may call E" the Pontrjagin dual .
1 46
I l l , §7
MODULES
Indeed , in the theory of topological groups G , the group of continuous homo morphisms of G into R/Z is the classical Pontrjagin dual , and is classically denoted by G " , so I find the preservation of that terminology appropriate . Instead of R/Z one may take other natural groups isomorphic to R/Z . The most common such group is the group of complex numbers of absolute value 1 , which we denote by S 1 The isomorphism with R/Z is given by the map •
Remark 3. A bilinear map V x V � K for which V' = V is called a bilinear form . We say that the form is nonsingular if the corresponding maps V' � V v and V � (V' )v
are isomorphisms . Bilinear maps and bilinear forms will be studied at greater length in Chapter XV . See also Exercise 33 of Chapter XIII for a nice example .
§7 .
M O D U L E S OV E R P R I N C I P A L R I N G S
Throughout this section, we assume that R is a principal entire ring. All modules are over R, and homomorphisms are Rhomomorphisms, unless otherwise specified. The theorems will generalize those proved in Chapter I for abelian groups. We shall also point out how the proofs of Chapter I can be adj usted with sub stitutions of terminology so as to yield proofs in the present case. Let F be a free module over R, with a basis {xJ i e i · Then the cardinality of I is uniquely determined , and is called the dimension of F. We recall that this is proved , say by taking a prime element p in R , and observing that F /pF is a vector space over the field R/pR , whose dimension is precisely the cardinality of / . We may therefore speak of the dimension of a free module over R . Theorem 7 . 1 . Let F be a free module, and M a sub module. Then M is free,
and its dimension is less than or equal to the dimension of F.
Proof For simplicity, we give the proof when F has a finite basis {x i }, i = 1, . . . , n. Let M r be the intersection of M with (x 1 , , xr), the module generated by x b . . . , x r . Then M 1 = M n (x 1 ) is a submodule of (x 1 ), and is therefore of type (a 1 x 1 ) with some a 1 E R. Hence M 1 is either 0 or free, of di mension 1 . Assume inductively that Mr is free of dimension < r. Let a be the set consisting of all elements a E R such that there exists an element x E M .
which can be written
•
•
MODU LES OVE R PRINC I PAL R I N G S
I l l , §7
1 47
with b; E R. Then a is obviously an ideal, and is principal, generated say by an element ar+ 1 . If ar + 1 = 0, then M r + 1 = Mr and we are done with the inductive step. If ar + 1 =1= 0, let w E M r + 1 be such that the coefficient of w with respect to Xr + 1 is ar + 1 • If X E M + 1 then the coefficient of X with respect to X r + 1 is divisible by a r + 1 ' and hence there exists c E R such t hat X cw lies in M r · Hence r

On the other hand, it is clear that Mr n (w) is 0, and hence th at this sum is direct, thereby proving our t heorem. (For the infinite case, see Appendix 2, §2.) Corollary 7 .2. Let E be a finitely generated module and E' a submodule.
Then E ' is finitely generated. P ro of We can represent E as a factor m odule of a free module F with a
, vn are generators of E, we take a free finite number of generators : If v 1 , module F with basis {x 1 , , xn } and map X; on V; . Th e inverse image of E' in F is a submodule, which is free, and finitely generated, by the theorem. Hence E ' is finitely generated. The assertion also follows using simple properties of Noetherian rings and modules. .
.
•
•
.
.
If one wants to translate the proofs of Chapter I, then one makes t he following definitions. A free 1 dimensio,nal module over R is called infinite cyclic. An infinite cyclic module is isomorphic to R, viewed as module over itself. Thus every nonzero submodule of an infinite cyclic module is infinite cyclic . The proof given in Chapter I for the analogue of Theorem 7 . 1 applies without further change . Let E be a module. We say that E is a torsion module if given x E £, there exists a E R, a =1= 0, such that ax = 0. The generalization of finite abelian group is finitely generated torsion module. An element x of E is called a torsion element if there exists a E R, a =F 0, such that ax = 0. Let E be a module . We denote by Etor the submodule consisting of all torsion elements of E, and call it the torsion submodule of E. If Etor = 0, we say that E is torsion free . Theorem 7.3. Let E be finitely generated. Then
E/Etor is free. There exists
a free sub module F of E such that E is a direct sum
E
=
Etor EB F.
The dimension of such a submodule F is uniquely determined. Proof.
We first prove that E / Etor is torsion free . If x E E, let i denote its residue class mod Etor · Let b E R , b =I= 0 be such that bi = 0 . Then bx E E10p and hence there exists c E R , c =I= 0 , such that cbx = 0. Hence x E Etor and i = 0, thereby proving that E/Etor is torsion free . It is also finitely generated .
1 48
I l l , §7
MODULES
Assume now that M is a torsion free module which is finitely generated . Let { v 1 , , vn } be a maximal set of elements of M among a given finite set of generators {y 1 , , Ym } such that { v 1 , , vn } is linearly independent . If y is one of the generators , there exist elements a, b 1 , , bn E R not all 0 , such that •
•
•
•
•
•
•
•
•
•
ay
+
b 1 V1 + · · · + bn Vn
•
=
•
0.
Then a 1= 0 (otherwise we contradict the linear independence of v 1 , , vn ). Hence ay lies in ( v b . . . , vn ). Thus for each j = 1 , . . . , m we can find ai E R, ai 1= 0, such that ai yi lies in ( v b . . . , vn ) . Let a = a 1 · · · am be the product. Then aM is contained in ( v 1 , , vn ), and a 1= 0. The map •
•
•
.
.
•
x �+ ax is an injective homomorphism , whose image is contained in a free module. This image is isomorphic to M, and we conclude from Theorem 7 . 1 that M is free , as desired . To get the submodule F we need a lemma . Lemma 7 .4. Let E, E ' be modules, and assume that E' is free. Let f : E __.. E '
be a surjective homomorphism. Then there exists a free submodule F of E such that the restriction off to F induces an isomorphism ofF with E ' , and such that E = F (f) K er f Proof Let {xa ie i be a basis of E' . For each i, let x i be an element of E such that f(x i ) = x� . Let F be the submodule of E generated by all the elements x i , i E I. Then one sees at once that the family of elements {x i }i e i is linearly inde pendent, and therefore that F is free. Given x E E, there exist elements ai E R such that
Then x  L a i x i lies in the kernel off, and therefore E = K er f + F. It is clear that Ker f n F = 0, and hence that the sum is direct, thereby proving the lem ma. We apply the lemma to the homomorphism E � E/Eto r in Theorem 7 . 3 to get our decomposition E = Eto r EB F. The dimension of F is uniquely determined , because F is isomorphic to E/ Eto r for any decomposition of E into a direct sum as stated in the theorem . The dimension of the free module F in Theorem 7 . 3 is called the rank of E. In order to get the structure theorem for finitely generated modules over R, one can proceed exactly as for abelian groups. We shall describe the dictionary which allows us to transport the proofs essentially withou t change. Let E be a module over R. Let x E E. The map a 1+ ax is a homo morphism of R onto the submodule generated by x, and the kernel is an ideal, which is principal, generated by an element m E R. We say that m is a period of x. We
I l l , §7
MODULES OVER PRINC I PAL RINGS
1 49
note that m is determined up to multiplication by a unit (if m =F 0). An element c E R, c =F 0, is said to be an exponent for E (resp. for x) if c E = 0 (resp. ex = 0). Let p be a prime element. We denote by E{p) the submodule of E consisting of all elements x having an exponent which is a power pr (r > 1). A psubmodule of E is a submodule contained in E(p) . We select once and for all a system of representatives for the prime elements of R (modulo units). For instance, if R is a polynomial ring in one variable over a field, we take as representatives the irreducible polynomials with leading coefficient 1 . Let m E R, m =F 0. We denote by Em the kernel of the map x 1+ mx. It consists of all elements of E having exponent m. A module E is said to be cyclic if it is isomorphic to Rj(a ) for some element a E R. Without loss of generality if a =F 0, one may assume that a is a product of primes in our system of representatives, and then we could say that a is the order of the module. Le t r 1 , , rs be integers > 1 . A pmodule E i s said to be of type •
•
•
if it is isomorphic to the product of cyclic modules Rj(pr i) (i = 1, . . , s). If p is fixed, then one could say that the module is of type (r b . . . , r5) (relative to p ). All the proofs of Chapter I , §8 now go over without change. Whenever w e argue on the size of a positive inte g e r m , w e have a similar argument on th e number of prime factors appearing in its prime factorization . If we deal with a prime power p r , w e can view the order as bei ng determined by r. The read er can now check that the proofs of Chapter I , §8 are applicable . However, we shall de v el op the theory once again without assuming any knowledge of Chapter I , § 8 . Thus our treatment is selfcontained . .
Theorem 7 . 5 . Let E be a .finitely generated torsion module =F 0. Then E is
the direct sum
E = EB E(p), p
taken over all primes p such that E(p) =F 0. Each E(p) can be written as a direct sum with 1
l)for some prime element p. Let x 1 E E be an element of period pr. Let E = E/{x 1 ). Let , Ym be independent elements of E. Then for each i there exists a repre .Y 1 , sentative Y i E E of Y i , such that the period of Yi is the same as the period of Y i · The elements x b Y b . . . , Ym are independent. Proof Let y E E have period pn for some n > 1. Let y be a representative of y in E. Then pny E {x 1 ), and hence Lemma 7 6 .
.
•
.
.
c E R, p � c, for some s < r. If s = r, w e see that y has the same period as y. If s < r, then p5CX 1 has period pr  s, and hence y has period pn + r  s . We must have
n + r  s < r, because pr is an exponent for E. Thus we obtain n < s, and we see that is a representative for y, whose period is pn . Let Yi be a representative for Y ; having the same period. We prove that x b y 1 , . . . , Ym are independent. Suppose that a, a 1 , , a m E R are elements such that .
•
.
MODULES OVE R PRINCIPAL R I N G S
I l l , §7
1 51
Then By hypothesis, we must have a i Yi = 0 for each i. If pr i is the period of yi , then pr i divides a i . We then conclude that a i yi = 0 for each i, and hence finally that ax 1 = 0, thereby proving the desired independence. To get the direct sum decomposition of E(p), we first note that E( p) is finitely generated. We may assume without loss of generality that E = E( p ). Let x 1 be an element of E whose period pr• is such that r 1 is maximal. Let E = Ej(x 1 ) . We contend that dim EP as vector space over RjpR is strictly less than dim EP . Indeed, if y 1 , • • • , Ym are linearly independent elements of EP over R/pR , then Lemma 7 .6 implies that dim EP > m + 1 because we can always find an element of {x 1 ) having period p, independent of Y b . . . , Y m · Hence dim E P < dim EP . We can prove the direct sum decomposition by induction. If E =F 0, there exist elements x 2 , . . • , X5 having periods pr 2 , • • • , prs respectively, such that r2 > · · · > rs  By Lemma 7 . 6 , there exist representatives x2 , . . . , xr in E such that xi has period pr i and x 1 , • . . , x r are independent. Since pr 1 is such that r 1 is maximal, we have r 1 > r 2 , and our decomposition is achieved. The uniqueness will be a consequence of a more general uniqueness theorem, which we state next. Theorem 7. 7.
Let E be a finitely generated torsion module, E is isomorphic to a direct sum of nonzero factors
E
=F 0. Then
where q 1 , , q r are nonzero elements of R, and q 1 l q 2 1 · · · l qr . The sequence of ideals (q 1 ), , (q r) is uniquely determined by the above conditions. • • .
• • .
Proof. Using Theorem 7 . 5 , decompose E into a direct sum of psubmodules , say E(p 1 ) (f) · · · (f) E(p 1 ), and then decompose each E(pJ into a direct sum of cyclic submodules of periods p'ii i . We visualize these symbolically as described by the following diagram :
E( p1 ) :
r 1 1 < r1 2 < · · ·
A horizontal row describes the type of the module with respect to the prime at the left. The exponents r ii are arranged in increasing order for each fixed i = 1 , . . . , I. We let q b . . . , q r correspond to the columns of the matrix of exponents, in other words q 1  pr11 1 pr22 1 . . . pr, l l , q 2  pr11 2pr222 . . . pr, 12 , _
_
1 52
I l l , §7
MODULES
The direct sum of the cyclic modules represented by the first column is then isomorphic to R/(q 1 ), because, as with abelian groups, the direct sum of cyclic modules whose periods are relatively prime is also cyclic. We have a similar remark for each column, and we observe that our proof actually orders the qi by increasing divisibility, as was to be shown. Now for uniqueness. Let p be any prime, and suppose that E = R/(pb) for some b E R, b =1= 0. Then EP is the submodule bR/(pb), as follows at once from unique factorization in R. But the kernel of the composite map
R __.. bR
__..
bR/(pb)
is precisely (p). Thus we have an isomorphism
R/(p)
�
bR/(pb).
Let now E be expressed as in the theorem, as a direct sum of r terms. An element
is in EP if and only if pv; = 0 for all i. Hence EP is the direct sum of the kernel of multiplication by p in each term. But EP is a vector space over R/(p ) and its dimension is therefore equal to the number of terms R/( q;) such that p divides q; . Suppose that p is a prime dividing q 1 , and hence q; for each i = 1, . . . , r. Let E have a direct sum decomposition into d terms satisfying the conditions of the theorem, say ,
E = R/( q '1 ) ffi · · · ffi R/( q�) . Then p must divide at least r of the elements qj , whence r < s. By symmetry, r = s, and p divides qj for all j. Consider the module p E . By a preceding remark, if we write q; = pb; , then
and b 1 I · · · I br . Some of the b; may be units, but those which are not units determine their principal ideal uniquely, by induction. Hence if
but (b i + 1 ) =F ( 1 ) , then the sequence of ideals
MODU LES OVE R PRINC I PAL RINGS
I l l , §7
1 53
is uniquely determined . This proves our uniqueness statement, and concludes the proof of Theorem 7 . 7 . , (qr ) are called the invariants of E. The ideals (q 1 ) , For one of the main applications of Theorem 7 . 7 to linear algebra, see Chapter XV , §2 . The next theorem is included for completeness . It is called the elementary divisors theorem . •
•
•
Theorem 7 .8.
Let F be a free module over R, and let M be a .finitely generated submodule =I= 0. Then there exists a basis CB of F, elements e I , . . . , em in this basis, and nonzero elements a 1 , . . . , am E R such that : (i) The elements a i e 1 , , am em form a basis of M over R . (ii) We have a; I a; + 1 for i = 1 , . . . , m  1 . •
•
•
The sequence of ideals (a 1 ), conditions.
•
•
•
, (am) is uniquely determined by the preceding
Proof. Write a finite set of generators for M as linear combination of a finite number of elements in a basis for F. These elements generate a free submodule of finite rank , and thus it suffices to prove the theorem when F has finite rank , which we now assume . We let n = rank(F) . The uniqueness is a corollary of Theorem 7 . 7 . Suppose we have a basis as in the theorem . Say a 1 , , a5 are units , and so can be taken to be = 1 , and as +j = qi with q 1 l q 2 l . . . I qr nonunits . Observe that F IM = F is a finitely generated module over R, having the direct sum expression •
F IM = F =
•
•
r
EB1 (RI qi R)ei EB free module of rank n
j=
 (r + s)
where a bar denotes the class of an element of F mod M. Thus the direct sum over j = 1 , . . . , r is the torsion submodule of F, whence the elements q i , . . . , qr are uniquely determined by Theorem 7 . 7 . We have r + s = m, so the rank of F IM is n  m, which determines m uniquely . Then s = m  r is uniquely determined as the number of units among a 1 , , am . This proves the uniqueness part of the theorem. Next we prove existence . Let A be a functional on F, in other words , an element of HomR(F, R) . We let JA = A(M) . Then JA is an ideal of R . Select A 1 such that A I (M) is maximal in the set of ideals {JA } , that is to say , there is no properly larger ideal in the set {JA } . Let A. 1 (M) = {a 1 ). Then a 1 =F 0, because there exists a nonzero element of M, and expressing this element in terms of some basis for F over R, with some nonzero coordinate , we take the projection on this coordinate to get a func tional whose value on M is not 0. Let x 1 E M be such that A. 1 (x 1 ) = a1 • For any functional g we m ust have g(x 1 ) E {a 1 ) [immediate from the maximality of •
•
•
1 54
I l l , §7
MODULES
A t (M)].
Writing x t in terms of any basis of F, we see that its coefficients must all be divisible by a t · (If some coefficient is not divisible by a 1 , p roj ect on this coefficient to get an impossible functional.) Therefore we can write X t = a t e t with some element e 1 E F. Next we prove that F is a direct sum F = Re t Et> Ker A t · Since A t (e t ) = 1 , it is clear that Re t n Ker A t = 0. Furthermore, given x E F we note that x  A t (x )e t is in the kernel of A t . Hence F is th e sum of the in dicated submodules, and therefore the direct sum. We note that Ker A 1 is free , being a submodule of a free module (Theorem 7 . 1 ) . W e let F 1 = Ker A 1 and M 1 = M n Ker A 1 We see at once that M = Rx 1 EB M 1 • Thus M 1 is a submodule of F t and its dimension is one less than the dimension of M. From the maximality condition on A t (M) , it follows at once that for any functional A on F 1 , the image A (M ) will be contained in At (M) ( b ec ause otherwise , •
a suitable linear combination of functionals would yield an ideal larger than (a 1 )) . We can the refore c omp l ete the existence proof by induction . In Theorem 7 . 8 , we call the ideals (a 1 ) , , (am) the invariants of M in F. For another characterization of these invariants , see Chapter XIII , Proposition •
•
•
4 . 20 . Example. First, see examples of situations similar to those of Theorem 7 . 8 i n Exercises 5 , 7 , and 8 , and for Dedekind ri ng s in Exercise 1 3 . Example. Another way to obtain a module M as in Theorem 7 . 8 is as a module of relations . Let W be a finitely generated module over R , with genera tors w 1 , , wn . By a relation among { w 1 , , wn } w e mean an element n (a1 , . . . , an ) E R such that L a; W; = 0 . The set of such relations is a sub n module of R , to which Theorem 7 . 8 may be applied . •
•
•
•
•
•
It is also possible to formulate a proof of Theorem 7 . 8 by considering M as a submodule of R n , and applying the method of row and column operations to get a desired basis . In this context, we make some further comments which may serve to illustrate Theorem 7 . 8 . We assume that the reader is acquainted with matrices over a ring. B y row operations we mean: i n terch an ging two rows; adding a multiple of one row to another; mu ltipl y i n g a row by a unit in the ring. We define column operations s i mi lar l y . These row an d co lu mn operat ions correspond to m ult ip li ca t ion with t h e socalled elementary matrices in the ring. Theorem 7.9. Assume that the elementary matrices in R generate GL,iR).
b e a nonzero ma trix with components in R. Then with a finite n umber of row and column O)Jerations, it is possi b le to bring the matrix to the form Let (xu)
EULERPO I NCAR E MAPS
I l l ' §8
1 55
0 0 0 0 0
... 0
0 0 0
with a 1 · · · am � 0 and a 1 l a2 l · · · I am .
We leave the proof for the reader. Either Theorem 7.9 can be viewed as equivalent to Theorem 7 .8, or a direct proof may be given. In any case, Theorem 7 . 9 can be used in the following context. Consider a system of linear equations
with coefficients in R . Let F be the submodule of R n generated by the vectors X = (x 1 , , xn ) which are solutions of this system. By Theorem 7 . 1 , we know that F is free of dimension < n . Theorem 7 . 9 can be viewed as providing a normalized basis for F in line with Theorem 7 . 8 . •
•
•
Further example. As pointed out by Paul Cohen , the row and column method can be applied to modules over a power series ring o [ [X] ] , where o is a complete discrete valuation ring . Cf. Theorem 3 . 1 of Chapter 5 in my Cyclo tomic Fields I and II (Springer Verlag , 1 990) . For instance , one could pick o it self to be a power series ring k[ [T]] in one variable over a field k, but in the theory of cyclotomic fields in the above reference , o is taken to be the ring of padic integers. On the other hand, George B ergman has drawn my attention to P. M. Cohn' s ' 'On the structure of GL;. of a ring, ' ' IHES Pub!. Math. No. 30 ( 1 966), giving examples of principal rings where one cannot use row and column operations in Theorem 7 . 9.
§8.
E U LERPO I N CARE MAPS
The present section may be viewed as providing an example and application of the JordanHolder theorem for modules . But as pointed out in the examples and references below , it also provides an introduction for further theories . Again let A be a ring. We continue to consider Amodules. Let r be an abelian group, written additively. Let cp be a rule which to certain modules associates an element of r, subj ect to the following condition:
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I l l , §8
MODULES
JfO __.. M' __.. M __.. M" __.. 0 is exact, then qJ(M) is defined if and only if qJ(M') and qJ(M") are defined, and in that case, we have lfJ( M) = lfJ( M') + lfJ( M"). Furthermore qJ(O) is defined and equal to 0. Such a rule lfJ will be called an EulerPoincare mapping on the category of Amodules. If M' is isomorphic to M, then from the exact sequence 0 __.. M' __.. M __.. 0 __.. 0 we conclude that qJ{M') is defined if qJ{M) is defined, and that qJ{M') = qJ(M). Thus if qJ{M) is defined for a module M, lfJ is defined on every submodule and factor module of M. In particular, if we have an exact sequence of modules M' __.. M __.. M" and if qJ{M') and qJ{M") are defined, then so is qJ{M), as one sees at once by considering the kernel and image of our two maps, and using the definition. Examples.
We could let A = Z, and let qJ be defined for all finite abelian groups, and be equal to the order of the group. The value of qJ is in the multi plicative group of positive rational numbers. As another example, we consider the category of vector spaces over a field k. We let lfJ be defined for finite dimensional spaces, and be equal to the dimension. The values of cp are then in the additive group of integers . In Chapter XV we shall see that the characteristic polynomial may be con sidered as an EulerPoincare map . Observe that the natural map of a finite module into its image in the Groth endieck group defined at the end of §4 is a universal EulerPoincare mapping . We shall develop a more extensive theory of this mapping in Chapter XX , §3 . If M is a module (over a ring A), then a sequence of submodules
is also called a finite filtration, and we call r the length of the filtration. A module M is said to be sim ple if it does not contain any submodule other than 0 and M itself, and if M =F 0. A filtration is said to be simple if each Md Mi + 1 is simple. The JordanHolder theorem asserts that two simple filtrations of a module are equivalent. A module M is said to be of finite length if it is 0 or if it admits a simple (finite) filtration. By the JordanHolder theorem, the length of such a simple filtration is the uniquely determined, and is called the length of the module . In the language of Euler characteristics, the JordanHolder theorem can be re formulated as follows :
TH E SNAKE LEMMA
I l l ' §9
1 57
Theorem 8. 1 . Let qJ be a rule which to each simple module associates an
element of a commutative group r, and such that if M qJ{M)
=
�
M ' then
qJ{M' ).
Then qJ has a unique extension to an EulerPoincare mapping defined on all modules offinite length. Proof Given a simple filtration we define
qJ{M)
r
=
1
L qJ(Mi /M i + 1 ) .
i= 1
The JordanHolder theorem shows immediately that this is welldefined, and that this extension of qJ is an Eule.rPoincare map. In particular, we see that the length function is the EulerPoincare map taking its values in the additive group of integers, and having the value 1 for any simple module.
§9. T H E S N A K E L E M M A This section gives a very general lemma, which will be used many times , so we extract it here . The reader may skip it until it is encountered , but already we give some exercises which show how it is applied: the five lemma in Exercise 1 5 and also Exercise 26 . Other substantial applications in this book will occur in Chapter XVI , §3 in connection with the tensor product, and in Chapter XX in connection with complexes, resolutions, and derived functors . We begin with routine comments . Consider a commutative diagram of homo morphisms of modules .
N N ' + h Then f induces a homomorphism Ker d ' Indeed , suppose d 'x '
= 0.
�
Then df(x ' )
Ker d.
= 0
because df(x ' )
=
hd '(x ' )
= 0.
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I l l , §9
MODULES
Similarly , h induces a homomorphism Coker d '
�
Coker d
in a natural way as follows . Let y ' E N' represent an element of N ' / d ' M ' . Then hy ' mod dM does not depend on the choice of y ' representing the given element , because if y " = y ' + d 'x ', then =
hy'' = hy ' + h d 'x ' = hy ' + dfx'
hy ' mod dM.
Thus we get a map
h * : N ' /d ' M ' = Coker d '
�
N/dM = Coker d,
which is immediately verified to be a homomorphism. In practice , given a commutative diagram as above , one sometimes writes f instead of h , so one writes f for the horizontal maps both above and below the diagram. This simplifies the notation , and is not so incorrect: we may view M', N ' as the two components of a direct sum , and similarly for M, N. Then f is merely a homomorphism defined on the direct sum M ' EB N ' into M EB N. The snake lemma concerns a commutative and exact diagram called a snake diagram:
)
l l 
M' _...;;_!+ M d
'
d
M" + 0
9 +
d
0 + N' + N f
g
"
...... N"
Let z" E Ker d" . We can construct elements of N ' as follows . Since g is surjective , there exists an element z E M such that gz = z". We now move vertically down by d, and take dz. The commutativity d"g = gd shows that gdz = 0 whence dz is in the kernel of g in N. By exactness , there exists an element z ' E N' such that fz ' = dz. In brief, we write Z
' = f  l o d o g  1 Z".
Of course, z' is not well defined because of the choices made when taking inverse images. However, the snake lemma will state exactly what goes on. Lemma 9. 1 . (Snake Lemma). b:
Given a snake diagram as above, the map
Ker d"
__..
Coker d '
given by bz " = f  1 o d o g  1 z " is well defined, and we have an exact sequence Ker d '
__..
Ker d __.. Ker d"
d
__..
Coker d '
where the maps besides b are the natural ones.
__..
Coker d __.. Coker d"
111, §1 0
D I R ECT AN D INVE RSE LIMITS
1 59
Proof. It is a routine verification that the class of z' mod lm d' is in dependent of the choices made when taking inverse images, whence defining the ma p b. The proof of the exactness of the sequence is then routine, and consists in chasing around diagrams. It should be carried out in full detail by the reader who wishes to acquire a feeling for this type of triviality. As an example, we shall prove that Ker 5 C Im g * where g* is the induced map on kernels. Suppose the image of z " is 0 in Coker d' . By definition, there exists u ' e M' such that z ' = d ' u ' . Then dz = fz ' = fd' u ' = dju ' by commutativity. Hence
d(z  fu ' ) = 0, and z  fu' is in the kernel of d. But g(z  fu ' ) = gz = z". This means that z " is in the image of g* ' as desired. All the remaining cases of exactness will be left to the reader.
l
l
l
The original snake diagram may be completed by writing in the kernels and cokernels as follows (whence the name of the lemma) : Ker d'
l
......
M'
0
.... ,
§1 0.
Ker d"
l
M"
M '
N'
Coker d
Ker d
Coker d
l
0
N"
N I
......
.....
Coker d
II
D I R E C T A N D I N V E R S E L I M I TS
We return to limits , which we considered for groups in Chapter I . We now consider lint its in other categories (rings , modules) , and we point out that limits satisfy a universal property , in line with Chapter I , § 1 1 . Let I = { i } be a directed system of indices , defined in Chapter I , § 1 0 . Let C1 be a category , and {A,} a family of objects in CI . For each pair i , j such that
1 60
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MODU LES
i < j assume given a morphism
i fJ. •· A I· + A J·
such that, whenever i < j < k, we have f t o f� = fl and f� = id.
Such a family will be called a directed family of morphisms. A direct limit for the family { f� } is a universal object in the following category e . Ob(e) consists of pairs (A, ( f i)) where A E Ob(CI) and ( f i) is a family of morphisms f i : A i + A, i E J, such that for all i < j the following diagram is commutative : A l·
��
A J·
'\ f A
(Universal of course means universally repelling.) Thus if (A, ( f i)) is the direct limit, and if (B, (g i)) is any object in the above category, then there exists a unique morphism lfJ : A + B which makes the following diagram commutative :
For simplicity, one usually writes A = � lim A ,. ,
i
omitting the f � from the notation. Theorem 10. 1 . Direct limits exist in the category of abelian groups, or more
generally in the category of modules over a ring.
Proof Let {Mi} be a directed system of modules over a ring. Let M be their direct sum. Let N be the submodule generated by all elements Xij =
( . . . , 0, X, 0, . . . , jj (X), 0, . . . )
111, §1 0
D I R ECT AND I NVERSE LIMITS
1 61
where, for a given pair of indices (i, j) with j > i, x ii has component x in M; , f�(x) in M i , and component 0 elsewhere. Then we leave to the reader the veri fication that the factor module M/N is a direct limit, where the maps of M; into MjN are the natural ones arising from the composite homomorphism
Let X be a topological space , and let x E X. The open neigh borhoods of x form a directed system , by inclusion . Indeed , given two open neighborhoods U and V, then U n V is also an open neighborhood contained in both U and V. In sheaf theory, one assigns to each. U an abelian group A ( U) and for each pair U ::J V a homomorphism h� : A( U) � A(V) such that if U ::J V ::J W then hw h� = h((, . Then the family of such homomorphisms is a directed family . The direct limit Example.
a
fun A(U) u
is called the stalk at the point x. We shall give the formal definition of a sheaf of abelian groups in Chapter XX , §6. For further reading , I recommend at least two references. First, the selfcontained short version of Chapter II in Hartshorne ' s Algebraic Geometry , Springer Verlag, 1 977 . (Do all the exercises of that section, concerning sheaves . ) The section is only five pages long . Second , I recommend the treatment in Gunning's Introduction to Holomorphic Functions of Several Variables, Wadsworth and Brooks/Cole, 1 990. We now reverse the arrows to define inverse limits. We are again given a directed set I and a family of objects A ; . Ifj > i we are now given a morphism
f� · A . + A · l •
J
l
satisfying the relations
if j > i and i > k. As in the direct case, we can define a category of objects (A, /;) with /; : A + A; such that for all i, j the following diagram is com mutative :
A universal object in this category is called an inverse limit of the system (A; ,f �).
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As before, we often say that A = lim A; i
is the inverse limit, omitting the f� from the notation. Theorem 10.2. Inverse limits exist in the category of groups, in the category
of modules over a ring, and also in the category of rings.
Proof. Let { G;} be a directed family of groups , for instance, and let f be their inverse limit as defined in Chapter I , § 1 0. Let p; : r � G; be the projection (defined as the restriction from the projection of the direct product , since r is a subgroup of TI G; ). It is routine to verify that these data give an inverse limit in the category of groups. The same construction also applies to the category of rings and modules. Example. Letp be a prime number. For n > m we have a canonical surjective ring homomorphism fJ:, : Z/p n z � Z/pmz . The projective limit is called the ring of padic integers, and i s denoted by ZP . For a consideration of this ring as a complete discrete valuation ring, see Exercise 1 7 and Chapter XII . Let k be a field . The power series ring k[ [T] ] in one variable may be viewed as the projective J imit of the factor polynomial rings k[T] /(T n ) , where for n > m we have the canonical ring homomorphism A similar remark applies to power series in several variables . More generally , let R be a commutative ring and let J be a proper ideal . If
n > m we have the canonical ring homomorphism JJ:, : R/J n � R/Jm .
Let R1 = lim R/J n be the projective limit. Then R has a natural homomorphism into R 1. If R is a Noetherian local ring , then by Krull ' s theorem (Theorem 5 . 6 of Chapter X) , one knows that nJ n = {0 }, and so the natural homorphism of R in its completion is an embedding. This construction is applied especially when J is the maximal ideal . It gives an algebraic version of the notion of holomorphic functions for the following reason . Let R be a commutative ring and J a proper ideal . Define a ]Cauchy se quence {xn } to be a sequence of elements of R satisfying the following condition . Given a positive integer k there exists N such that for all n , m > N we have xn  x E Jk . Define a null sequence to be a sequence for which given k there m exists N such that for all n > N we have xn E Jk. Define addition and multipli
DI RECT AND INVE RSE LIMITS
I l l , §1 0
1 63
cation of sequences termwise . Then the Cauchy sequences form a ring e , the null sequences form an ideal .N', and the factor ring e /.N' is called t he Jadic completion of R . Prove these statements as an exercise , and also prove that there is a natural isomorphism n e/.N' R/J . fun n Thus the inverse limit li!!! R/J is also called the Jadic completion . See Chapter XII for the completion in the context of absolute values on fields . =
Examples. In certain situations one wants to determine whether there exist solutions of a system of a polynomial equationf(X 1 , , Xn ) = 0 with coefficients in a power series ring k[T] , say in one variable . One method is to consider the ring mod (TN) , in which case this equation amounts to a finite number of equations in the coefficients . A solution of f(X) = 0 is then viewed as an inverse limit of truncated solutions . For an early example of this method see [La 52] , and for an extension to several variables [Ar 68] . .
[La 52] [Ar 68]
•
•
S. L ANG , On quasi algebraic closure , Ann of Math . SS ( 1 952) , pp . 373 390 M . ARTIN , On the solutions of analytic equations , Invent. Math . S ( 1 968) , pp .
27729 1
See also Chapter XII , § 7 . In Iwasawa theory , one considers a sequence of Galois cyclic extensions Kn n over a number field k of degree p with p prime , and with Kn C Kn + I · Let Gn be the Galois group of Kn over k. Then one takes the inverse limit of the group n rings (Z/p Z) [Gn] , following Iwasawa and Serre . Cf. my Cyclotomic Fields , Chapter 5 . In such towers of fields , one can also consider the projective limits of the modules mentioned as examples at the end of § 1 . Specifically , consider n the group of p  t h roots of unity fJ,p n , and let Kn = Q( fJ,pn + l ) , with K0 = Q(fJ,p ) . We let Tp ( fJ, ) = li!!! fJ,p n under the homomorphisms fJ,p n + I � fJ,pn given by ( � (P. Then Tp ( fJ, ) becomes a module for the projective limits of the group rings . Similarly , one can consider inverse limits for each one of the modules given in the examples at the end of § 1 . (See Exercise 1 8 . ) Th e determination of the structure of these inverse limits leads to fundamental problems in number theory and algebraic geometry . After such examples from real life after basic algebra, we return to some general considerations about inverse limits .
Let (Ai , f{) = (Ai) and (Bi , g{) = (Bi) be two inverse systems of abelian groups indexed by the same indexing set. A homomorphism (A i ) + (Bi) is the obvious thing, namely a family of homomorphisms
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for each i which commute with the maps of the inverse systems :
A sequence is said to be exact if the corresponding sequence of groups is exact for each i. Let ( A n) be an inverse system of sets, indexed for simplicity by the positive integers, with connecting maps
We say that this system satisfies the MittagLefller condition ML if for each n, the decreasing sequence u m , n(A m ) (m > n) stabilizes, i.e. is constant for m sufficiently large . This condition is satisfied when u m , n is surjective for all m, n. We note that trivially, the inverse limit functor is left exact, in the sense that given an exact sequence
then is exact . Proposition 10.3. Assume that (A n ) satisfies ML . Given an exact sequence
of inverse systems, then is exact. The only point is to prove the surjectivity on the right. Let (en) be an element of the inverse limit. Then each inverse image g  1 (cn ) is a coset of A n , so in bijection with A n . These inverse images form an inverse system, and the ML condition on (A n ) implies ML on (g  1 {cn ) ). Let Sn be the stable subset Proof
sn = () u!, n(g  1 ( c m )) . m�n
I l l , Ex
EXERCISES
1 65
Then the connecting maps in the inverse system (Sn ) are surjective, and so there
is an element (bn) in the inverse limit. It is immediate that g maps this element on the given (e n), thereby concluding the proof of the Proposition. Proposition 10.4. Let (Cn ) be an inverse system of abelian groups satisfying ML , and let (um , n ) be the system of connecting maps. Then we have an exact
sequence
Proof. For each positive integer N we have an exact sequence with a finite prod uct N
0 +
lim
l �n �N
en +
en n n =l
N
_... ..,.� 1 u
en + 0 . n n =l
The map u is the natural one, whose effect on a vector is {0 , . . . ' 0 ,
em ,
0,
. . . ' 0) 1+ (0, . . . ' 0, u m , m  l em , 0, . . . ' 0).
One sees immediately that the sequence is exact. The infinite products are in verse limits taken over N. The hypothesis implies at once that ML is satisfied for the inverse limit on the left, and we can therefore apply Proposition 1 0 . 3 to conclude the p roo f .
EX E R C I S ES 1 . Let V be a vector space over a field K, and let U, W be subspaces. Show that dim U + dim W = dim( U + W) + dim( U () W).
2. Generalize the dimension statement of Theorem 5.2 to free modules over a commutative
ring. [Hint : Recall how an analogous statement was proved for free abelian groups, and use a maximal ideal instead of a prime number.]
3 . Let R be an entire ring containing a field k as a subring . Suppose that R is a finite dimensional vector space over k under the ring multiplication . Show that R is a field . 4 . Direct sums.
(a) Prove in detail that the conditions given in Proposition 3 . 2 for a sequence to split are equivalent . Show that a sequence 0 7 M' _£, M � M" 7 0 splits if and only if there exists a submodule N of M such that M is equal to the direct sum I m f EB N, and that if this is the case , then N is isomorphic to M". Complete all the details of the proof of Proposition 3 . 2.
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(b) Let E and E;( i = I , . . . , m) be modules over a ring . Let cp; : E; � E and t/1; : E � E; be homomorphisms having the following properties: Ill '1' 1
•
0 {()
•
'1' 1
= id '
m
I (/J;
i= 1
if i :F j, o
t/1 ; = id.
Show that the map x � ( t/1 1 K, . . . , t/Jm x) is an isomorphism of E onto t he direct product of the E; (i = 1 , . . . , m), and that the map
is an isomorphism of t his direct product onto E. Conversely , if E is equal to a direct product (or direct sum) of submodules E; (i = I , . . . , m) , if we let cp; be the inclusion of E; in E, and t/1; the projection of E on E; , then these maps satisfy the abovementioned properties .
5 . Let A be an additive subgroup of Euclidean space R ", and assume that in every bounded
region of space, there is only a finite number of elements of A. Show that A is a free abelian group on < n generators. [Hint : Induction on the maximal number of linearly independent elements of A over R. Let v b . . . , vm be a maximal set of such elements, and let A0 be the subgroup of A contained in the Rspace generated by vb . . . , vm  t · By induction, one may assume that any element of A 0 is a linear integral combination of vb . . . , vm  t · Let S be the subset of elements v e A of the form v = a 1 v 1 + · · · + am vm with real coefficients a; sat isfying
0 < a; < 1
if i = 1, . . . , m

1
0 < am < 1. If v� is an element of S with the smallest am :/= 0, show that {v 1 , . . . , vm  h v�} is a basis of A over Z.]
Note . The above exercise is applied in algebraic number theory to show that the group of units in the ring of integers of a number field modulo torsion is isomorphic to a lattice in a Euclidean space . See Exercise 4 of Chapter VII . 6.
(ArtinTate) . Let G be a finite group operating on a finite set S . For w 1 w by [ w] , so that we have the direct sum
E
S , denote
·
Z (S ) =
L Z[w] .
WE S
Define an action of G on Z(S) by defining u[w] = [uw] (for w E S) , and extending u to Z(S) by linearity . Let M be a subgroup of Z(S) of rank # [S] . S how that M has a Zba sis { y w}wes such that uy w = Yaw for all w E S . (Cf. my Algebraic Number Theory , Chapter IX, §4, Theorem 1 . )
7 . Let M be a finitely generated abelian group . By a seminorm on M we mean a real valued function v � I v I satisfying the following properties:
EXERC ISES
I l l , Ex
1 67
I v I > 0 for all v E M ; l nv l = l n l l v l for n E Z; I v + w I I v I + I w I for all v, w E M.
· Let d be the exponent of M /M 1 • Then dM has a finite index in M 1 • Let ni ,j be the smallest positive integer such that there exist integers nj, 1 , , nj , j  l •
•
•
satisfy ing
Without loss of generality we may assume 0 W 1 , • • • , W form the desired basis . ] r
m . Let ZP be its inverse limit. Show that ZP maps sur jectively on each Z/p n z; that ZP has no divisors of 0, and has a unique maximal ideal generated by p . Show that ZP is factorial , with only one prime , namely p itself.
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(b) Next consider all ideals of Z as forming a directed system , by divisibility . Prove that
Z ' n p P
!!!!! Z / ( a ) = (a)
where the limit is taken over all ideals (a) , and the product is taken over all pnmes p .
18. (a) Let {An } be an inversely directed sequence of commutative rings , and let {Mn } be an inversely directed sequence of modules , Mn being a module over A n such that the following diagram is commutative :
The vertical maps are the homomorphisms of the directed sequence , and the horizontal maps give the operation of the ring on the module . Show that !!!!! Mn is a module over lim A n . (b) Let M be a pdivisible group . Show that Tp (A) is a module over ZP . (c) Let M , N be pdivisible groups . Show that Tp (M EB N) = Tp (M) EB Tp (N) , as modules over zp .
Direct limits 1 9 . Let (A ; ,/�) be a directed family of modules. Let ak E A k for some k, and suppose that the image of ak in the d irect limit A is 0. Show t hat there exists some index j > k such that f�(ak) = 0. In other words� whether some element in some group A; vanishes tn t he direct limit can already be seen within the original data. One way to see this is to use the construction of Theorem 1 0. 1 .
20. Let
I, J be two directed sets, and give t he product I
x
J the obvious ordering that
(i, j) < (i', j') if i < i' and j < j'. Let Aii be a family of abelian groups, with homo morphisms indexed by I x J, and forming a directed family. Show that t he direct limits lim limA ii and lim lim A ii
i
j
j
i
exist and are isomorphic in a natural way. State and prove t he same result for inverse limits.
2 1 . Let (M� , f�), (M ; , g�) be directed systems of modules over a ring. By a
homomorphism
one means a family of homomorphisms ui : M; + Mi for each i which commute with t he f� , g�. Suppose we are given an exact sequence
0 + (MD � (M i) � (M�') + 0 of directed systems, meaning that for each i, the sequence
0 + M� + M . + M�' + 0 l
l
l
I l l , Ex
EXERC ISES
1 71
is exact. Show that t he duect limit preserves exactness, that is
I s exact. 22 . (a) Let {M;} be a famtly of modules over a nng. Fo r any module N show that Hom( ffi M; , N)
=
0 Hom(M; , N)
(b) Show that Hom{N,
0 M;)
=
0 Hom(N, M;).
23 . Let {M;} be a directed family of modules over a ring. For any module N show t hat ltm Hom(N, M;)
=
Hom(N, lim M;)
24 . Show that any mod ule is a direct limit of finitely generated submodules. A module M is called
finitely presented
if there is an exact sequence
w here F0 , F 1 are free with finite bases. The image ofF 1 in F 0 is said to be t he submodule of relations, among t he free basis elements of F0 . 25 . Show that any module is a direct limit of finitely presented modules (not necessarily submodules). In other words, given M, t here exists a directed system { M ; , Jj} with M; finitely presented for all i such that
[Hint : Any finitely generated submod ule is such a direct limit, since an infinitely generated module of relations can be viewed as a limit of finitely generated modules of relations. Make this precise to get a proof.] 26 . Let E be a module over a ring. Let {M;} be a directed family of mod ules. If E is finitely generated, show that t he natural homomorp hism lim Hom(£, M;) � Hom(£, lim M;) I S I nJective. If E is finitely presented, show that this homomorphism is an isomorphism. Hint : First prove the statements when E is free with finite basis. Then, say E is finitely presented by an exact sequence F 1 � F0 � E � 0. Consider t he diagram :
l
l
l
0
.
lim Hom(£, M;) . lim Hom(F0 , M;) 4 lim Hom(F 1 , M;)
0
...
Hom(£, lim M;) ... Hom(F 0 , lim M;) ... H om(F 1 , lim M;)
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Graded Algebras Let A be an algebra over a field k. By a filtration of A we mean a sequence of k vector spaces A; ( i = 0, 1 , . . . ) such that
Ao
c
A1
c
A2
c
and
···
U A; = A ,
A; +j for all i, j > 0. In particular, A is an A o algebr a . We then call A a fil tered algebra. Let R be an algebra. We say that R is graded if R is a di re ct sum R ffi R; of subspaces s uc h that R;Rj c R;+j for all i, j > 0.
and A ; Aj
c
==
27. Let A be a filtered algebra. Define R; for i > 0 by R; = A ;/ A; _ 1 . By definition,
A  1 == {0} . Le t R ffi R; , and R; == gr; (A ) . Define a natural product on R m aking R into a graded algebra, denoted by gr(A) , and called the associated graded algebra. ==
28. Let A , B be filtered algeb ras, A
==
U A ; and B U B;. Let L: A ==
linear m ap preserving the filtration, that is L( A;) L(c)L(a) for c e Ao and a e A; for all i. (a) Show that L induces an (Ao , B0 ) l inear ma p
c
B be an (Ao , Bo ) B; for all i, and L( ca) == +

gr ; (L) : gr; (A)
+
gr; (B)
for all i.
(b) Suppose that gr; (L) is an isomorphism for all i. Show that L is an (Ao , Bo ) isomorphism.
29. Su ppose k has characteristic 0. Let n be the set of all strictly upper triangular ma trices of a given size n x n over k. (a) For a given matrix X e n, let Dt (X) , . . . , Dn (X) be its diagonals, so D1 = Dt (X) is the main diagonal, and is 0 by the definition of n. Let n; be the subset of n consisting of those matrices whose diagonals D1 , , Dni are 0. Thus no = {0}, n1 consists of all matrices whose components are 0 except possibly for Xnn ; n2 consists of all matrices whose components are 0 except possibly those in the last two diagonals; and so forth. Show that each n; is an algebra, and its elements are nilpotent (in fact the ( i + 1 )th power of its elements is 0). (b) Let U be the set of elements I + X with X e n. Show that U is a multi plicative group. (c) Let exp be the exponential series defined as usual. Show that exp defines a polynomial function on n (all but a finite number of terms are 0 when eval uated on a nilpotent matrix), and establishes a bijection •
exp: n + U. Show that the inverse is given by the standard log series.
•
•
C H A PT E R
IV
Po l yn o m i a ls
This chapter provides a continuation of Chapter II, §3. We prove stan dard properties of polynomials. Most readers will be acquainted with some of these properties, especially at the beginning for polynomials in one vari able. However, one of our purposes is to show that some of these properties also hold over a commutative ring when properly formulated. The Gauss lemma and the reduction criterion for irreducibility will show the importance of working over rings. Chapter IX will give examples of the importance of working over the integers Z themselves to get universal relations. It happens that certain statements of algebra are universally true. To prove them, one proves them first for elements of a polynomial ring over Z, and then one obtains the statement in arbitrary fields (or commutative rings as the case may be) by specialization. The CayleyHamilton theorem of Chapter XV, for instance, can be proved in that way. The last section on power series shows that the basic properties of polynomial rings can be formulated so as to hold for power series rings. I conclude this section with several examples showing the importance of power series in various parts of mathematics.
§1 .
BAS I C P R O P E RTI ES FO R PO LY N O M IALS I N O N E VA R IA B L E
We start with the Euclidean algorithm. Theorem 1.1.
Let A be a commutative ring, let f, g E A [X] be poly nomials in one variable, of degrees > 0, and assume that the leading 1 73
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coefficient of g is a unit in A. Then there exist unique polynomials q, r E A [X] such that f = gq + r and deg r < deg g. Proof. Write
f( X) = an x n + · · · + a 0 , 4 b4X + · · · + b0 , = X ) g(
where n = deg f, d = deg g so that an , b4 =F 0 and b4 is a unit in A. We use induction on n. If n = 0, and deg g > deg f, we let q = 0, r = f. If deg g = deg f = 0, then we let r = 0 and q = an bi 1 • Assume the theorem proved for polynomials of degree < n (with n > 0). We may assume deg g < deg f (otherwise, take q = 0 and r = f). Then
where f1 ( X) has degree < n. By induction, we can find q 1 , r such that 1 f( X) = an bi x n  dg ( X ) + q 1 ( X ) g ( X) + r( X)
and deg r < deg g. Then we let to conclude the proof of existence for q, r. As for uniqueness, suppose that
with deg r 1 < deg g and deg r2 < deg g. Subtracting yields Since the leading coefficient of g is assumed to be a unit, we have
Since deg ( r2  r 1 ) < deg g, this relation can hold only if q 1 q 1 = q 2 , and hence finally r 1 = r2 as was to be shown.

q 2 = 0, i.e.
Theorem 1.2. Let k be a field. Then the polynomial ring in one variable
k [X] is principal.
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1 75
Proof Let a be an ideal of k[X], and assume a =F 0. Let g be an element of a of smallest degree > 0. Let f be any element of a such that f =F 0. By the Euclidean algorithm we can find q, r E k[X] such that f = qg + r and deg r < deg g. But r = f q g, whence r is in a. Since g had minimal degree > 0 it follows that r = 0, hence that a consists of all polynomials qg (with q E k [X] ). This proves our theorem. By Theorem 5.2 of Chapter II we get : 
Corollary 1.3.
The ring k [X] is factorial.
If k is a field then every nonzero element of k is a unit in k, and one sees immediately that the units of k [X] are simply the units o f k. (No polyno mial of degree > 1 can be a unit because of the addition formula for the degree of a product.) A polynomial f(X) E k [X] is called irreducible if it has degree > 1 , and if one cannot write f(X) as a product
f(X) = g(X)h(X) h E k[X] , and both g , h � k. Elements of k are usually called constant polynomials, so we can also say that in such a factorization , one of g o r h must be constant. A poly n om ia l is called monic if it has leading coefficient 1 . Let A be a commutative ring and f(X) a polynomial in A [X]. Let A be a subring of B. An element b E B is called a root or a zero o f f in B if f(b) = 0. Similarly, if (X) is an ntuple of variables, an ntuple (b) is called a zero of f if f(b) = 0. w i th
g,
Theorem 1.4. Let k be a field and f a polynomial in one variable X in k [X], of degree n > 0. Then f has at most n roots in k, and if a is a root
off in k, then X  a divides f(X).
Proof Suppose f(a) = 0. Find q, r such that f(X) = q(X) (X  a) + r(X) and deg r < 1 . Then
0=
f(a) = r(a).
Since r = 0 or r is a nonzero constant, we must have r = 0, whence X  a divides f(X). If a 1 , , am are distinct roots of f in k, then inductively we see that the product •
•
•
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divides f(X), whence m < n, thereby proving the theorem. The next corollaries give applications of Theorem 1 .4 to pol ynomial functions. Corollary 1.5. Let k be a field and T an infinite subset of k. Let
f(X) E k [X]
be a polynomial in one variable. If f(a) = 0 for all a E f = 0, i.e. f induces the zero function.
T,
then
Let k be a field, and let S1 , , Sn be infinite subsets of k. Let f(X , Xn ) be a polynomial in n variables over k. If f(a 1 , , an ) = 0 for all ai E Si (i = 1, . . . , n), then f = 0. Corollary 1.6. 1,
•
•
•
•
•
•
•
•
•
Proof By induction. We have just seen the result is true for one variable. Let n > 2, and write f(X 1 ' . . . ' Xn )
L _h (X1 ' . . . ' xn  1 )Xj
=
j
as a polynomial in X" with coefficients in k [X 1 , such that for some j we have !'i ( b1 ,
•
•
•
•
•
•
, Xn _1 ]. If there exists
, bn  1 ) 1= 0, then
is a nonzero polynomial in k [Xn J which takes on the value 0 for the infinite set of elements Sn . This is impossible. Hence fj induces the zero function on s1 X . . . X sn  1 for all j, and by induction we have fj = 0 for all j. Hence f = 0, as was to be shown. Corollary 1.7. Let k be an infinite field and f a polynomial in n variables over k. Iff induces the zero function on k , then f = 0.
We shall now consider the case of finite fields. Let k be a finite field with q elements. Let f(X1 , . . . , X") be a polynomial in n variables over k. Write If a 1= 0, we recall that the monomial M ( X) occurs in f Suppose this is the case, and that in this monomial M(X), some variable Xi occurs with an exponent vi > q . We can write
tt = integer > 0. 1
If we now replace Xti by xr + in this monomial, then we obtain a new polynomial which gives rise to the same function as f The degree of this new polynomial is at most equal to the degree of f
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1 77
Performing the above operation a finite number of times, for all the monomials occurring in f and all the variables X 1 , , Xn we obtain some polynomial f * giving rise to the same function as f, but whose degree in each variable is < q. •
Corollary 1.8. Let k be a finite field with
•
•
elements. Let f be a polynomial in n variables over k such that the degree of f in each variable is < q. Iff induces the zero function on k, then f = 0. q
Proof By induction. If n = 1 , then the degree of f is < q, and hence f cannot have q roots unless it is 0 . The inductive step is carried out just as we did for the proof of Corollary 1 .6 above. Let f be a polynomial in n variables over the finite field k . A polynomial g whose degree in each variable is < q will be said to be reduced. We have shown above that there exists a reduced polynomial f * which gives the same function as f on k< n> . Theorem 1 .8 now shows that this reduced polynomial is unique. Indeed, if g 1 , g 2 are reduced polynomials giving the same function, then g 1  g 2 is reduced and gives the zero function. Hence g 1  g 2 = 0 and g 1 = g2 . We shall give one more application of Theorem 1 .4. Let k be a field. By a multiplicative subgroup of k we shall mean a subgroup of the group k * (nonzero elements of k) . Theorem 1.9. Let k be a field and let U be a finite multiplicative sub
group of k . Then U is cyclic.
Proof Write U as a product of subgroups U(p) for each prime p, where U (p) is a pgroup. By Proposition 4.3(vi) of Chapter I, it will suffice to prove that U(p) is cyclic for each p. Let a be an element of U(p) of maximal period p r for some integer r. Then x P.. = 1 for every element x E U(p), and hence all elements of U(p) are roots of the polynomial X P..  1 . The cyclic group generated by a has p r elements. If this cyclic group is not equal to U(p), then our polynomial has more than p r roots, which is impossible. Hence a generates U(p), and our theorem is proved. Corollary 1.10. If k is a finite field, then k* is cyclic.
An element ' in a field k such that there exists an integer n > 1 such that ' n = 1 is called a root of unity, or more precisely an nth root of unity. Thus the set of nth roots of unity is the set of roots of the polynomial xn  1 . There are at most n such roots, and they obviously form a group, which is
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cyclic by Theorem 1 .9. We shall study roots of unity in greater detail later. A generator for the group of nth roots of unity is called a primitive nth root of unity. For example, in the complex numbers, e2n i/n is a primi tive nth root of unity, and the nth roots of unity are of type e2ni vfn with 1 < v < n. The group of roots of unity is denoted by p. The group of roots of unity in a field K is denoted by p(K). A field k is said to be algebraically closed if every polynomial in k [ X ] of degree > 1 has a root in k . In books on analysis, it is proved that the complex numbers are algebraically closed. In Chapter V we shall prove that a field k is always contained in some algebraically closed field. If k is algebraically closed then the irreducible polynomials in k [X] are the poly nomials of degree 1 . In such a case, the unique factorization of a polynomial f of degree > 0 can be written in the form r
f(X) = c n ( X  � i )m i i =1
with c E k, c =F 0 and distinct roots � 1 , , �r · We next develop a test when mi > 1 . Let A be a commutative ring. We define a map •
•
•
D : A [X] � A [X]
of the polynomial ring into itself. If f(X) = an x n + . . . + ao with a i E A , we define the derivative n L Df(X ) = / ' (X) = va"x v l = nan x n  1 + . . . + a 1 . =
v l One verifies easily that if f, g are polynomials in A [X], then (f + g) ' = f ' + g ' ,
(fg)' = f 'g + fg ' ,
and if a E A , then (af)' = af' .
Let K be a field and f off in K. We can write
a
nonzero polynomial in K[X] . Let a be
a
root
f(X ) = ( X  a)mg (X )
with some polynomial g ( X) relatively prime to X  a (and hence such that g (a) =F 0). We call m the multiplicity of a in f, and say that a is a multiple root if m > 1 .
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Proposition 1.1 1 .
Let K, f be as above. The element a of K is a multiple root off if and only if it is a root and f'(a) = 0. Proof Factoring f as above, we get f' (X) = (X  a) mg ' (X) + m(X  a)m  1 g(X). If m > 1 , then obviously f' (a) = 0. Conversely, if m = 1 then
f' (X) = (X  a)g'(X) + g(X), whence f'(a) = g(a) =F 0. Hence if f'(a) = 0 we must have m > 1, as desired. Proposition 1.12. Let f E K [X]. If K has characteristic 0, and f has degree > 1, then f' =F 0. Let K have characteristic p > 0 and f have
degree > 1 . Then f ' = 0 if and only if, in the expression for f(X) given by n
f(X) = L a v x v, v =l p divides each integer v such that a v =F 0. Proof If K has characteristic 0, then the derivative of a monomial a v x v such that v > 1 and av =F 0 is not zero since it is va v x v  t . If K has characteristic p > 0, then the derivative of such a monomial is 0 if and only if P I v, as contended. Let K have characteristic p > 0, and let f be written as above, and be such that f'(X) = 0. Then one can write d
with bll E K.
f(X) = L bllX Pil /l =l
Since the binomial coefficients
(�) are divisible by p for 1 < v < p  1 we
see that if K has characteristic p, then for a, b E K we have
Since obviously (ab)P = a Pb P, the map
is a homomorphism of K into itself, which has trivial kernel, hence is injective. Iterating, we conclude that for each integer r > 1 , the map x � x P,.
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is an endomorphism of K, called the Frobenius endomorphism. Inductively, if c1 , , en are elements of K, then (c l + . . . + cn Y' = c f + . . . + c: . •
•
•
Applying these remarks to polynomials, we see that for any element a E K we have
If c E K and the polynomial has one root a in K, then a�"" = c and X�""  c = (X  a)I"". Hence our polynomial has precisely one root, of multiplicity stance, (X  1 )P,. = X P'"  1.
§2.
p r.
For in
PO LYN O M IALS OV E R A FACTO R IA L R I N G
Let A be a factorial ring, and K its quotient field. Let a E K, a =F 0. We can write a as a quotient of elements in A, having no prime factor in common. If p is a prime element of A, then we can write where b E K, r is an integer, and p does not divide the numerator or denominator of b. Using the unique factorization in A, we see at once that r is uniquely determined by a, and we call r the order of a at p (and write r = ord a). If a = 0, we define its order at p to be oo. P If a, a' E K and aa' =F 0, then ordp (aa') = ordP a + ordP a'. This is obvious. Let f(X) E K [X] be a polynomial in one variable, written f(X) = a 0 + a 1 X + · · · + an x n. If f = 0, we define ordP f to be oo. If f =F 0, we define ordP f to be
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1 81
ordP f = min ordP a h the minimum being taken over all those i such that ai =F 0. If r = ordP f, we call up r a pcontent for f, if u is any unit of A. We define the content of f to be the product. n p ordpf, the product being taken over all p such that ordP f =F 0, or any multiple of this product by a unit of A. Thus the content is well defined up to multiplication by a unit of A. We abbreviate content by cont. If b E K, b =F 0, then cont(bf) = b cont(f). This is clear. Hence we can write f(X) = c · f1 (X) where c = cont(f), and f1 (X) has content 1 . In particular, all coefficients of f1 lie in A , and their g.c.d. is 1 . We define a polynomial with content 1 to be a primitive polynomial. Theorem 2. 1. (Gauss Lemma). Let A be a factorial ring, and let K be
its quotient field. Let f, g E K [X] be polynomials in one variable. Then cont(fg) = cont(f) cont(g).
Proof. Writing f = cf1 and g = dg 1 where c = cont(f) and d = cont(g), we see that it suffices to prove : If f, g have content 1, then fg also has content 1 , and for this, it suffices to prove that for each prime p, ordp (fg) = 0. Let f(X) = a x + · · · + a 0 , g(X) = bm X m + · · · + b0 , n
n
be polynomials of content 1 . Let p be a prime of A. It will suffice to prove that p does not divide all coefficients of fg. Let r be the largest integer such that 0 < r < n, ar =F 0, and p does not divide ar . Similarly, let bs be the coefficient of g farthest to the left, bs =F 0, such that p does not divide bs. Consider the coefficient of x r +s in f(X)g(X). This coefficient is equal to c = ar bs + ar + l bs  1 + . . . + ar 1 bs + l + . . . 
and p % ar bs. However, p divides every other nonzero term in this sum since in each term there will be some coefficient ai to the left of ar or some coefficient bi to the left of bs . Hence p does not divide c, and our lemma is proved.
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§2
We shall now give another proof for the key step in the above argument, namely the statement :
Iff, g e A [X] are primitive (i.e. have content 1) then fg is primitive. Proof. We have to prove that a given p rime p does not divide all the coefficients of fg . Consider reduction mod p, namely the canonical homo A/(p) = A. Denote the image of a polynomial by a bar, so morphism A g under the reduction homomorphi sm. Then f H f and g +
�+
fg By hypothesis, f
=F 0
g =F 0.
and
was to be shown.
Corollary 2.2. Let
K [X].
If
Since
A
is entire, it follows that fg
as
A [X] have a factorization f(X) = g(X)h(X) in ch = cont(h), and g = c9 g 1 , h = chh 1 , then f(X) = c9 chg 1 (X)h 1 (X),
and c9 ch is an element of A . then h e A [X] also. Proof
=F 0,
e
f( X)
c9 = cont(g),
= fg.
In
particular,
if f,
g e A [X] have content 1,
c9 ch e A. But cont (f) = c9 ch cont( g 1 h 1 ) = c9 ch,
The only thing to be proved is
whence our assertion follows.
Theorem 2.3. Let A be a factorial ring. Then the polynomial ring A [X]
in one var_iable is factorial. Its prime elements are the primes of A and poly nomials in A [X] which are irreducible in K[X ] and have content 1 .
Proof
Let f e
A [X],
f
=F 0.
Using the unique factorization in
and the preceding corollary, we can find a factorization
where
�[X].
c e A,
f(X) = c · p 1 (X) and
p1 ,
•
•
•
p,
,
···
K [X]
p, (X)
are polynomials in
A [X]
which are irreducible in
Extracting the�!"_ contents, we may assume without loss of generality
that the content of
P;
is
1
for each
i.
Then
c=
cont ( f ) by the Gauss lemma .
This gives us the existence of the factorization . It follows that each irreducible in
A [X ] .
Pi (X)
is
If we have another such factorization , say
f(X) = d · q 1 (X) · · · qs(X), then from the unique factorization in a permutation of the factors we have
K [X]
we conclude that
r = s,
and after
IV,
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C R IT E R I A FO R I R R E D U C I B I LITY
1 83
with elements ai E K. Since both Ph q i are assumed to have content 1, it follows that a i in fact lies in A and is a unit. This proves our theorem. Corollary 2.4. Let A be a factorial ring. Then the ring of polynomials in
n variables A [X , Xn J is factorial. Its units are precisely the units of A, and its prime elements are either primes of A or polynomials which are irreducible in K [X] and have content 1 . 1 ,
•
.
.
Proof. Induction. In view of Theorem 2.3, when we deal with polynomials over a factorial ring and having content 1 , it is not necessary to specify whether such polynomials are irreducible over A or over the quotient field K. The two notions are equivalent. Remark 1. The polynomial ring K [X 1 ,
, Xn J over a field K is not principal when n > 2. For instance, the ideal generated by X , Xn is not principal (trivial proof). •
•
•
1 ,
.
.
•
Remark 2.
It is usually not too easy to decide when a given polynomial (say in one variable) is irreducible. For instance, the polynomial X4 + 4 is reducible over the rational numbers, because
X4 + 4 = (X 2  2X + 2) (X 2 + 2X + 2). Later in this book we shall give a precise criterion when a polynomial x n  a is irreducible. Other criteria are given in the next section.
§3.
C R IT E R IA FO R I R R E D U C I B I LITY
The first criterion is : Theorem 3.1. (Eisenstein's Criterion). Let A be a factorial ring. Let K
be its quotient field. Let f(X) = an x n + · · · + a0 be a polynomial of degree n > 1 in A [X]. Let p be a prime of A , and assume : an =I= 0 (mod p),
a i = 0 (mod p) a 0 =/= 0 (mod p 2 ).
Then f(X) is irreducible in K [X].
for all i < n,
1 84
PO LYN O M IALS
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§3
Proof Extracting a g.c.d. for the coefficients of f, we may assume without loss of generality that the content of f is 1 . If there exists a factorization into factors of degree > 1 in K [X], then by the corollary of Gauss' lemma there exists a factorization in A [X], say f(X ) = g(X)h(X), g(X) = bd X d + · · · + b0 , h(X) = cm X "' + . . . + Co , with d, m > 1 and bd cm =F 0. Since b0 c 0 = a0 is divisible by p but not p 2 , it follows that one of b0 , c 0 is not divisible by p, say b0. Then p l c 0. Since c m bd = an is not divisible by p, it follows that p does not divide em . Let cr be the coefficient of h furthest to the right such that cr =I= 0 (mod p). Then Since p % b0 cr but p divides every other term in this sum, we conclude that p i ar , a contradiction which proves our theorem. Let a be a nonzero squarefree integer =F + 1. Then for any integer n > 1 , the polynomial x n  a is irreducible over Q. The polynomials 3X 5  1 5 and 2X 1 0  2 1 are irreducible over Q . Example.
There are some cases in which a polynomial does not satisfy Eisenstein's criterion, but a simple transform of it does. Example.
Let p be a prime number. Then the polynomial f(X) = x p  1 + · · · + 1
is irreducible over
Q.
Proof It will suffice to prove that the polynomial f(X + 1 ) is irreducible over Q. We note that the binomial coefficients
(p ) v
p! ' v ! (p  v) !
1 < v < p  1,
are divisible by p (because the numerator is divisible by p and the denomina tor is not, and the coefficient is an integer). We have (X + 1)P  1 X P + px p 1 + · · · + pX = f(X + 1 ) = (X + 1)  1 X from which one sees that f(X + 1 ) satisfies Eisenstein' s criterion . Example. Let E be a field and t an element of some field containing E such that t is transcendental over E. Let K b e the quotient field of E[ t] .
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C R IT E R I A FO R I R R E D U C I B I LITY
1 85
For any integer n > 1 the polynomial x n  t is irreducible in K [X]. This comes from the fact that the ring A = E [t] is factorial and that t is a prime in it. Theorem 3.2. (Reduction Criterion). Let A, B be entire rings, and let qJ : A + B
be a homomorphism. Let K, L be the quotient fields of A and B respec tively. Let f E A [X] be such that lfJf =F 0 and deg lfJf = deg f If lfJf is irreducible in L[X], then f does not have a factorization f(X) = g(X)h(X) with deg g, deg h > 1 . g, h e A [X] and Proof. Suppose f has such a factorization. Then lfJf = (({Jg) (lfJh). Since deg qJg < deg g and deg qJ h < deg h, our hypothesis implies that we must have equality in these degree relations. Hence from the irreducibility in L[X] we conclude that g or h is an element of A, as desired. In the preceding criterion, suppose that A is a local ring, i.e. a ring having a unique maximal ideal p, and that p is the kernel of lfJ · Then from the irreducibility of lfJf in L[X] we conclude the irreducibility of f in A [X]. Indeed, any element of A which does not lie in p must be a unit in A, so our last conclusion in the proof can be strengthened to the statement that g or h is a unit in A . One can also apply the criterion when A is factorial, and in that case deduce the irreducibility of f in K [X]. Example.
Let p be a prime number. It will be shown later that X P  X  1 is irreducible over the field Z/pZ. Hence X P  X  1 is irreduc ible over Q. Similarly, X 5  5X4  6X  1 is irreducible over Q. There is also a routine elementary school test whether a polynomial has a root or not. Proposition 3.3. (Integral Root Test). Let A be a factorial ring and K
its quotient field. Let
Let rx E K be a root of f, with rx = bjd expressed with b, d e A and b, d relatively prime. Then b l a0 and dian . In particular, if the leading coefficient an is 1, then a root rx must lie in A and divides a 0 •
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PO LYN O M I ALS
We leave the proof to the reader, who should be used to this one from way back. As an irreducibility test, the test is useful especially for a polynomial of degree 2 or 3, when reducibility is equivalent with the existence of a root in the given field.
§4.
H I L B E RT'S T H EO R E M
This section proves a basic theorem of Hilbert concerning the ideals of a polynomial ring. We define a commutative ring A to be Noetherian if every ideal is finitely generated. Theorem 4. 1.
Let A be a commutative Noetherian ring. Then the polyno mial ring A [X] is also Noetherian.
Proof. Let � be an ideal of A[X] . Let a; consist of 0 and the set of elements a E A appearing as leading coefficient in some polynomial a0 + a 1 X +
···
+ aX i
lying in � Then it is clear that a i is an ideal. (If a, b are in a i , then a + b is in a i as one sees by taking the sum and difference of the corresponding polynomials. If x E A, then xa E a i as one sees by multiplying the corre sponding polynomial by x.) Furthermore we have
in other words, our sequence of ideals { a i } is increasing. Indeed, to see this multiply the above polynomial by X to see that a E a i + t · By criterion (2) of Chapter X, § 1 , the sequence of ideals { a i } stops , say at a
Let
0
c a 1 c a 2 c · · · c ar = a r + 1 = · · ·
•
a o t , . . . , a o no be generators for a 0 , ............ .... ....... .
.
.
ar t ' . . . arn be generators for a r . ' ,. For each i = 0, . . . , r and j = 1 , . . . , ni let fu be a polynomial in �' of degree i, with leading coefficient aii. We contend that the polynomials /;,i are a set of generators for �Let f be a polynomial of degree d in � We shall prove that f is in the ideal generated by the /;,i , by induction on d. Say d > 0. If d > r, then we
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PA RTIAL F RACT I O N S
1 87
note that the leading coefficients of X d  rJr{' 1 '
•
•
· ' x d  rJrr n
r
generate ad . Hence there exist elements c 1 , , c,. polynomial f  c 1 x d rJ{' 1  . . .  cn,. x d rf, n,. •
•
r
e
A such that the
r
r
has degree < d, and this polynomial also lies in a linear combination
•
�
If d < r, we can subtract
to get a polynomial of degree < d, also lying in m. We note that the polynomial we have subtracted from f lies in the ideal generated by the hi · By induction, we can subtract a polynomial g in the ideal generated by the fii such that f  g = 0, thereby proving our theorem. We note that if cp : A __.. B is a surjective homomorphism of commutative rings and A is Noetherian, so is B. Indeed, let b be an ideal of B, so lfJ  1 (b) is an ideal of A . Then there is a finite number of generators (a h . . . , a ) for n 1 ( ) 1 cp  (b), and it follows since lfJ is surjective that b = lfJ lfJ  (b) is generated by cp(a 1 ), , cp(a,.), as desired. As an application, we obtain : •
•
•
Corollary 4.2. Let A be a Noetherian commutative ring,
let B = A [x 1 , , Xm ] be a commutative ring finitely generated over A . Then B is Noethe ria n. •
•
and
•
Proof Use Theorem 4. 1 and the preceding remark, representing B as a factor ring of a polynomial ring. Ideals in polynomial rings will be studied more deeply in Chapter IX. The theory of Noetherian rings and modules will be developed in Chapter X.
§5.
PA RTIA L F R ACTI O N S
In this section, we analyze the quotient field of a principal ring, using the factoriality of the ring. Theorem 5. 1. Let A be a principal entire ring, and let P be a set of
representatives for its irreducible elements. Let K be the quotient field of A, and let rx E K. For each p E P there exists an element rxP E A and an integer j(p) > 0, such that j(p) = 0 for almost all p E P, rxP and p i< P) are
1 88
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PO LY N O M IALS
relat ively prime, and rx
§5
rxP p":P p i
.
 � 
If we have another such expression
�
= L iP ' peP P > then j(p) = i(p) for all p, and rxP = PP mod p i
for all p. Proof We first prove existence, in a special case. Let a, b be rela tively prime nonzero elements of A. Then there exists x, y E A such that xa + yb = 1. Hence 1 X y = + ab b a . Hence any fraction cjab with c E A can be decomposed into a sum of two fractions (namely cxjb and cyja) whose denominators divide b and a respec tively. By induction, it now follows that any rx E K has an expression as stated in the theorem, except possibly for the fact that p may divide rxP . Canceling the greatest common divisor yields an expression satisfying all the desired conditions. As for uniqueness, suppose that rx has two expressions as stated in the theorem. Let q be a fixed prime in P. Then a

Pq rxq i q j(q)  q (q)

=

rxP PP i p)  j(p) . p p'tq p ( �
If j(q) = i(q) = 0, our conditions concerning q are satisfied. Suppose one of j(q) or i(q) > 0, say j(q), and say j(q) > i(q) . Let d be a least common multiple for all powers p i
and p i
such that p =F q . Multiply the above equation by dq i< q> . We get for some P E A. Furthermore, q does not divide d . If i(q) < j(q) then q divides rxq , which is impossible. Hence i(q) = j(q). We now see that q i divides rxq  pq , thereby proving the theorem. We apply Theorem 5. 1 to the polynomial ring k [X] over a field k. We let P be the set of irreducible polynomials, normalized so as to have leading coefficient equal to 1. Then P is a set of representatives for all the irreduc ible elements of k [X]. In the expression given for rx in Theorem 5. 1 , we can now divide rxP by p i< P>, i.e. use the Euclidean algorithm, if deg a.P > deg pi
. We denote the quotient field of k[X] by k(X), and call its elements rational
functions.
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1 89
Theorem 5.2. Let A = k[X] be the polynomial ring in one variable over a
field k. Let P be the set of irreducible polynomials in k [X] with leading coefficient 1. Then any element f of k(X) has a unique expression J;,(X) f(X) = L + g (X), ) ( P )i p e P P (X
where J;,, g are polynomials, J;, = 0 if j(p) = 0, J;, is relatively prime to p if j(p) > 0, and deg J;, < deg p i
if j(p) > 0. Proof. The existence follows at once from our previous remarks. The uniqueness follows from the fact that if we have two expressions, with elements J;, and cpP respectively, and polynomials g, h, then p i
divides J;,  cpP , whence J;,  cpP = 0, and therefore J;, = cpP , g = h. One can further decompose the term J;,/p i
by expanding J;, according to powers of p. One can in fact do something more general. Theorem 5.3.
Let k be a field and k [X] the polynomial ring in one variable. Let f, g E k [X], and assume deg g > 1 . Then there exist unique polynomials such that deg h < deg g and such that f = fo + ft g + · · · + iJg d .
Proof. We first prove existence. If deg g > deg f, then we take fo = f and h = 0 for i > 0. Suppose deg g < deg f We can find polynomials q, r with deg r < deg g such that f = qg + r, and since deg g > 1 we have deg q < deg f Inductively, there exist polyno mials h 0 , h 1 , . . . , hs such that q = h o + h 1 g + · · · + h 5 g 5, and hence f = r + h o g + . . . + hsg s +l ' thereby proving existence. As for uniqueness, let
f = fo + ft g + · · · + iJg d = CfJo + CfJ 1 g + · · · + lfJm g "' be two expressions satisfying the conditions of the theorem. Adding terms
1 90
IV,
PO LY N O M IA LS m
d. Subtracting, we get d · · · + ( fd  CfJd )g .
equal to 0 to either side, we may assume that 0 = ( fo  QJo ) +
§6
=
Hence g divides fo  cp0 , and since deg(/0  lfJo ) < deg g we see that fo = CfJo Inductively, take the smallest integer i such that h =F cpi (if such i exists). Dividing the above expression by g i we find that g divides h  cpi and hence that such i cannot exist. This proves uniqueness. ·
We shall call the expression for f in terms of g in Theorem 5.3 the gadic expansion of f If g(X) = X, then the gadic expansion is the usual expres sion of f as a polynomi al. In some sense, Theorem 5.2 redoes what was done in Theorem 8. 1 of Chapter I for Q/Z ; that is, express explicitly an element of K/A as a direct sum of its pcomponents. Remark.
§6. SY M M ETR I C PO LYN O M IALS Let A be a commutative ring and let t 1 , , tn be algebraically indepen dent elements over A. Let X be a variable over A [t 1 , , tn ]. We form the polynomial •
•
•
•
where each s i = s i (t 1 , . . . , tn ) is a polynomial in t 1 ,
.
.
•
•
•
, tn . Then for instance
and The polynomials s 1 , , sn are called the elementary symmetric polynomials of t 1 ' . . . ' tn . We leave it as an easy exercise to verify that s i is homogeneous of degree i in t 1 , , tn . Let a be a permutation of the integers ( 1 , . . . , n). Given a polynomial , tn ] , we define uf to be f(t ) E A[t] = A [t 1 , •
•
•
•
•
•
•
•
•
f( t I '
U
·
·
·
tn )
=
f ( tu( l ) '
·
·
·
'
tu( n) ) ·
If u, T are two permutations , then u Tf = u( Tj) and hence the symmetric group G on n letters operates on the polynomial ring A [ t] . A polynomial is called symmetric if uf = f for all u E G . It is clear that the set of symmetric polynomials is a subring of A [t], which contains the constant polynomials
IV, §6
SY M M ET R I C PO LY N O M IALS
1 91
(i.e. A itself) and also contains the elementary symmetric polynomials s 1 , • • • , sn . We shall see below that A [s 1 , • • • , sn J is the ring of symmetric polynomials. Let X 1 , • • • , Xn be variables. We define the weight of a monomial to be v 1 + 2v 2 + · · · + n vn . We define the weight of a polynomial g(X 1 , . . . , Xn ) to be the maximum of the weights of the monomials occurring . tn g. Theorem 6. 1.
Let f(t) E A [t 1 , • • • , tn J be symmetric of degree d. Then there exists a polynomial g(X 1 , • • • , Xn ) of weight < d such that
Proof By induction on n. The theorem is obvious if n = 1, because S1 = t1 . Assume the theorem proved for polynomials in n 1 variables. If we substitute tn = 0 in the expression for F(X), we find 1 (X  t 1 ) · · · (X  tn _ 1 )X = x n  (s 1 )0 X n  + · · · + (  1)n  1 (sn _ 1 )0 X, 
where (si) o is the expression obtained by substituting tn = 0 in s i . We see that (s 1 )0 , . • . , (sn _ 1 )0 are precisely the elementary symmetric polynomials in t 1 , . . . , tn  1 · We now carry out induction on d. If d = 0, our assertion is trivial. Assume d > 0, and assume our assertion proved for polynomials of degree Let f(t 1 , • • • , tn ) have degree d. There exists a polynomial < d. g 1 (X 1 , • • • , Xn  1 ) of weight < d such that We note that g 1 (s 1 , . . . , s n _ 1 ) has de gree has degree
> l u l 1 /2  e. The symbol >> means that the lefthand side is > the righthand side times a constant depending only on e. Again the proof is immediate from the abc conjecture. Actually, the hypothesis that u, v are relatively prime is not necessary ; the general case can be reduced to the relatively prime case by extracting common factors, and Hall stated his conjecture in this more general way. However, he also stated it without the epsilon in the exponent, and that does not work, as was realized later. As in the polynomial case, Hall's conjecture describes how small l u 3  v 2 1 can be, and the answer is not too small, as described by the righthand side. The Hall conjecture can also be interpreted as giving a bound for integral relatively prime solutions of v2 = u 3 + b with integral b. Then we find More generally, in line with conjectured inequalities from LangWaldschmidt [La 78], let us fix nonzero integers A, B and let u, v, k, m, n be variable, with u, v relatively prime and mv > m + n. Put Au m + Bv n = k. By the abc conjecture, one derives easily that m( m + n ( 1 + t:) n m (1) and lvl l u i . [Hint : Assume the contrary, and compare the degrees of the reduced polynomial belonging to 1
1
1

f(X)q  1
and ( 1  Xf  ) · · · ( 1  x:  ). The theorem is due to Chevalley.] (b) Refine the above results by proving that the number N of zeros of f in k is = 0 (mod p), arguing as follows. Let i be an integer > 1 . Show that
"' . {q 
�
xek
x'
=
0
1 = 1
q
if  1 divides i, otherwise.
Denote the preceding function of i by 1/J(i). Show that
21 4
IV, Ex
PO LYN O M I ALS
N
=
L
x e k("l
( 1  f(x)q  1 )
and for each ntuple {i 1 , . . . , i,.) of integers > 0 that
L X�1
x e k("l
•••
x!" = t/l(i 1 )
•··
t/l(i,.).
Show that both terms in the sum for N above yield 0 mod p. (The above argument is due to Warning.) (c) Extend Chevalley's theorem to r polynomials /1 , , /,. of degrees d 1 , . , dr respectively, in n variables. If they have no constant term and n > L d;, show that they have a nontrivial common zero. (d) Show that an arbitrary function f: kl2 0 f ' (t;). i=l
1 3. Polynomials will be taken over an algebraically closed field of characteristic 0. (a) Prove
Davenport's theorem.
Let f(t), g(t) be polynomials such that f3  g2 deg(/ 3  g 2 ) > 1 deg f + 1 .
:/;
0. Then
Or put another way, let h = f 3  g 2 and assume h :/; 0. Then deg f < 2 deg h  2. To do this, first assume f, g relatively prime and apply Mason's theorem. In general, proceed as follows. (b) Let A , B, f, g be polynomials such that Af, Bg are relatively prime :/; 0. Let h = Af3 + Bg 2 • Then deg f < deg A + deg B + 2 deg h  2. This follows directly from Mason's theorem. Then starting with f, g not necessarily relatively prime, start factoring out common factors until no longer possible, to effect the desired reduction. When I did it, I needed to do this step three times, so don't stop until you get it. (c) Generalize (b) to the case of f m  g" for arbitrary positive integer exponents m and n.
14. Prove that the generalized Szpiro conjecture implies the abc conjecture. 1 5. Prove that the abc conjecture implies the following conjecture : There are infinitely many primes p such that 2p t :f= 1 mod p2 • [Cf. the reference [Sil 88] and [La 90] at the end of §7.] 
1 6. Let w be a complex number, and let c = max( l , l w l ). Let F, G be nonzero polynomials in one variable with complex coefficients, of degrees d and d ' respec tively, such that I F I , I G l > 1 . Let R be their resultant. Then
I R I < c d + d ' [ I F( w) l + I G( w) I J I F i d ' I G i d (d + d')d + d ' .
(We denote by I F I the maximum of the absolute values of the coefficients of F.) 1 7. Let d be an integer > 3. Prove the existence of an irreducible polynomial of degree d over Q, having precisely d  2 real roots, and a pair of complex conj ugate roots. Use the following construction. Let b 1 , . . . , b  be distinct
d 2
21 6
IV, Ex
PO LYN O M IALS
integers, and let a be an integer > 0. Let
g(X) = (X 2 + a) (X  b 1 ) . . . (X  bd  1 ) = x d + cd  1 x d  1 + . . . + C o · Observe that c; e Z for all i. Let p be a prime number, and let p g,.(X) = g(X) + dil p so that g,. converges to g (i.e. the coefficients of g,. converge to the coefficients of g). (a) Prove that g,. has precisely d 2 real roots for n sufficiently large. (You may use a bit of calculus, or use whatever method you want.) (b) Prove that g,. is irreducible over Q. 
Integralvalued polynomials 1 8. Let P(X) e Q [X] be a polynomial in one variable with rational coefficients. It may happen that P(n) e Z for all sufficiently large integers n without necessarily P having integer coefficients. (a) Give an example of this. (b) Assume that P has the above property. c0, c 1 , . . . , cr such that
P(X) = c0 where
Prove that there are integers
( �) ( ) + c1
X
,
1
+ · · · + c.,
()
X = _!_ X (X  1 ) · · · (X  r + 1 ) r! r
is the binomial coefficient function. In particular, P(n) e Z for all n . Thus we rnay call P integral valued. (c) Let f : Z . Z be a function. Assume that there exists an integral valued polynomial Q such that the difference function �� defined by (�f) (n) = f(n)  f(n  1 ) is equal to Q(n) for all n sufficiently large. Show that there exists an integral valued polynomial P such that f(n) = P(n) for all n sufficiently large.
Exercises on symmetric functions 1 9. (a) Let X 1 , . . . , X,. be variables. Show that any homogeneous polynomial in Z [X1 , . . . , X,.] of degree > n(n  1 ) lies in the ideal generated by the elementary symmetric functions s 1 , . . . , s,.. (b) With the same notation show that Z [X 1 , . . . , X,.] is a free Z [s 1 , . . . , s,.] module with basis the monomials
with 0 < r; < n  i.
I V, Ex
EXE R CI S ES
(c) Let X 1 ,
, X, and Y1 ,
21 7
, Ym be two independent sets of variables. Let s 1 , . . . , s, be the elementary symmetric functions of X and s� , . . . , s� the elementary symmetric functions of Y (using vector vector notation). Show that Z[X, Y] is free over Z [s, s ' ] with basis x y, and the exponents (r), (q) •
•
•
•
•
•
satisfying inequalities as in (b) . (d) Let I be an ideal in Z [s, s ' ]. Let J be the ideal generated by I in Z [X, Y] . Show that
J n Z [ s, s ' ] = I.
20. Let A be a commutative ring. Let t be a variable. Let m , i and f(t) = L ai t g(t) = I bi t i i=O i=O be polynomials whose constant terms are a0 = b0 = 1. If f(t)g(t) = 1, show that there exists an integer N ( = (m + n) (m + n  1 ) ) such that any mono mial
ar1 1
·
·
·
a,r"
with L iri > N is equal to 0. [Hint : Replace the a's and b's by variables. Use Exercise 1 9 (b) to show that any monomial M(a) of weight > N lies in the ideal I generated by the elements
k ck = L a; bk  i i=O (letting a0 = b0 = 1 ). Note that ck is the kth elementary symmetric function of the m + n variables (X, Y).] [N o te : For some interesting contexts involving symmetric functions, see Cartier's talk at the Bourbaki Seminar, 1982 1983.]
lrings The following exercises start a train of thought which will be pursued in Exercise 33 of Chapter V ; Exercises 2224 of Chapter XVIII ; and Chapter XX, §3. These originated to a large extent in Hirzebruch's Riemann Roch theorem and its extension by Grothendieck who defined A.rings in general. Let K be a commutative ring. By loperations we mean a family of mappings
A_ i : K + K for each integer i > 0 satisfying the relations for all x e K :
0 A. (x) = 1 ,
and for all integers n > 0, and x, y e K,
, A."(x + y) = L A_i (x)A."  i (y). i=O
21 8
I V, Ex
PO LYN O M IALS
The reader will meet examples of such operations in the chapter on the alternat ing and symmetric products, but the formalism of such operations depends only on the above relations, and so can be developed here in the context of formal power series. Given a A.operation, in which case we also say that K is a lring, we define the power series 00
A.,(x) = L A_i(x)ti. i =O Prove the following statements.
2 1 . The map x ....+ A.,(x) is a homomorphism from the additive group of K into the multiplicative group of power series 1 + tK[[t]] whose constant term is equal to 1 . Conversely, any such homomorphism such that A.,(x) = 1 + x t + higher terms gives rise to A.operations.
22. Let s = at + higher terms be a power series in K [ [t] ] such that a is a unit in K. Show that there is a power series with
b e K. i
Show that any power series f(t) e K [ [t] ] can be written in the form h (s) for some other power series with coefficients in K. Given a A.operation on K, define the corresponding Grotbendieck power series
y,(x ) = A.,/( 1  r) (x) = A.s (x) where s = t/( 1  t). Then the map
X ....+ Yr (X ) is a homomorphism as before. We define yi(x) by the relation y,(x) = L y i(x)ti. Show that y satisfies the following properties.
23. (a) For every integer n > 0 we have
II
y " (x + y) = L yi(x )y "  i( y). .i =O (b) y,( 1 ) = 1/( 1  t). (c) y,(  1 ) = 1  t. 24. Assume that A_iu = 0 for i > 1. Show : (a) y,(u  1 ) = 1 + (u  1)t. 00
(b) Y r ( 1  u) = L ( 1  u)iti.
i =O
25. Bernoulli numbers. Define the Bernoulli numbers Bk as the coefficients tn the power senes
00
tk t = L Bk · F(t) = e  1 k= o k! t
I V, Ex
EX ER CIS ES
21 9
Of course, e' = L t" /n ! is the standard power series with rational coefficients 1/n !. Prove : (a) B0 = 1 , B 1 =  i, B2 = t . (b) F(  t) = t + F(t), and Bk = 0 if k is odd � 1 .
26. Be�noulli polynomials. Define the Bernoulli polynomials . senes expanston
F(t, X) =
te'x et  1
Ba:( X ) by the powe r
oo
k t = L Bk (X)  . k! k= o
It is clear that Bk = Bk (O), so the Bernoulli numbers are the constant terms of the Bernoulli polynomials. Prove : (a) B 0 (X) = 1 , B 1 (X) = X  !, B 2 (X) = X 2  X + t. (b) For each positive integer N,
(c) Bk (X) = X k  tkx k  1 + lower terms.
k kt (d) F ( t, X + 1 )  F( t, X) = texr = t L X , k. k 1 (e) Bk (X + 1 )  Bk (X) = kx  for k > 1 .
•
27. Let N be a positive integer and let f be a function on Z/NZ. Form the power senes
F1(t, X)
N1 te 1 ) of F, not all equal to 0, such that •
.
.
If rx =F 0, and rx is algebraic, then we can always find elements a i as above such that a0 =F 0 (factoring out a suitable power of rx). Let X be a variable over F. We can also say that rx is algebraic over F if the homomorphism F [X] � E 223
224
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ALG E B RAI C EXTE N S I O N S
§1
which is the identity on F and maps X on rx has a nonzero kernel. In that case the kernel is an ideal which is principal, generated by a single polyno mial p(X), which we may assume has leading coefficient 1 . We then have an isomorphism F[X]/( p(X) ) � F [rx], and since F [rx] is entire, it follows that p(X) is irreducible. Having normal ized p(X) so that its leading coefficient is 1, we see that p(X) is uniquely determined by rx and will be called THE irreducible polynomial of rx over F. We sometimes denote it by Irr(rx, F, X). An extension E of F is said to be algebraic if every element of E is algebraic over F. Proposition 1.1.
over F.
Let E be a finite extension of F. Then E is algebraic
Proof Let rx E E, rx =F 0. The powers of rx, 1 , rx, rx 2 , . . . , rx n,
cannot be linearly independent over F for all positive integers n, otherwise the dimension of E over F would be infinite. A linear relation between these powers shows that rx is algebraic over F. Note that the converse of Proposition 1 . 1 is not true ; there exist infinite algebraic extensions. We shall see later that the subfield of the complex numbers consisting of all algebraic numbers over Q is an infinite extension of Q. If E is an extension of F, we denote by [E : F] the dimension of E as vector space over F. It may be infinite. Proposition 1.2. Let k be a field and F
[ E : k]
=
c
E extension fields of k. Then
[ E : F] [F : k].
If { x i }i e i is a basis for F over k and { yi } i e J is a basis for E over F, then { x i yi } ( i, j) e ixJ is a basis for E over k. Proof Let z E E. By hypothesis there exist elements rxi E F, almost all rxi = 0, such that For each j E J there exist elements bii E k, almost all of which are equal to 0, such that
V,
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F I N IT E A N D ALG E B RA I C EXT E N S I O N S
225
and hence
bJl.. x . y . i j This shows that {x i yi } is a family of generators for E over k. We must show that it is linearly independent. Let { c ii } be a family of elements of k, almost all of which are 0, such that � � � � CIJ· ·X I· YJ· = 0 z = � i..J � i..J
j
i
1
r
•
Then for each j, C · ·X· = 0 i 1J I because the elements Yi are linearly independent over F. Finally c ii = 0 for each i because {x i } is a basis of F over k, thereby proving our propositi on. � f..J
The extension E of k is finite if and only if E is finite over F and F is finite over k. As with groups, we define a tower of fields to be a sequence Corollary 1.3.
F1 c F2 c · · · c Fn of extension fields . The tower is called finite if and only if each step is finite. Let k be a field, E an extension field, and rx E E. We denote by k(rx) the smallest subfield of E containing both k and rx. It consists of all quotients f(rx)jg(rx), where f, g are polynomials with coefficients in k and g(rx) =F 0. Proposition 1.4.
Let rx be algebraic over k. Then k(rx) = k[rx], and k(rx) is finite over k. The degree [k(rx) : k] is equal to the degree of Irr(rx, k, X). Proof Let p(X) = Irr(rx, k, X). Let f(X) E k [X] be such that f(rx) =F 0. Then p(X) does not divide f(X), and hence there exist polynomials g(X), h(X) E k[X] such that g(X)p(X) + h(X)f(X) = 1 . From this we get h(rx)f(rx) = 1 , and we see that f(rx) is invertible in k [rx]. Hence k [rx] is not only a ring but a field, and must therefore be equal to k(rx). Let d = deg p(X). The powers 1 , rx, . . . , rx d  1 are linearly independent over k, for otherwise suppose a0 + a 1 rx + · · · + ad _ 1 rx d  1 = 0
226
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with ai E k, not all ai = 0. Let g(X) = a 0 + · · · + a4 _ 1 x d  1 . Then g =F 0 and g(�) = 0. Hence p(X) divides g(X), contradiction. Finally, let f(a.) E k [�], where f(X) E k [X]. There exist polynomials q(X), r(X) E k [X] such that deg r < d and f(X) = q(X)p(X) + r(X). Then f(�) = r(�), and we see that 1, �, . . . , � d  1 generate k[�] as a vector space over k. This proves our proposition. Let E, F be extensions of a field k. If E and F are contained in some field L then we denote by EF the smallest subfield of L containing both E and F, and call it the compositum of E and F, in L. If E, F are not given as embedded in a common field L, then we cannot define the compositum. Let k be a subfield of E and let � 1 , . . . , �n be elements of E. We denote by k(� 1 ' . . . , �n ) the smallest subfield of E containing k and � 1 , . . . , �n · Its elements consist of all quotients f(� 1 ' · · · ' �n ) g(� 1 ' . . . , �n ) where f, g are polynomials in n variables with coefficients in k, and Indeed, the set of such quotients forms a field containing k and � 1 , Conversely, any field containing k and
�1 '
···'
•
•
•
, �n ·
�n
must contain these quotients. We observe that E is the union of all its subfields k(� 1 , , �n ) as (� 1 , . . . , �n ) ranges over finite subfamilies of elements of E. We could define the compositum of an arbitrary subfamily of subfields of a field L as the smallest subfield containing all fields in the family. We say that E is finitely generated over k if there is a finite family of elements � 1 , , �n of E such that E = k ( � 1 ' . . . ' �n ). We see that E is the compositum of all its finitely generated subfields over k. •
•
•
•
•
•
Proposition 1.5. Let E be a finite extension of k. Then E is finitely
generated.
Proof. Let { � 1 , . . . , �n } be a basis of E as vector space over k. Then certainly
E F, E V,
§1
EF
If
and
F
FI N IT E AN D ALG E B RA I C EXTE N S I O N S
EF /""F E""/
, rxn ) is finitely generated, and contained in L, then =
k(rx 1 ,
•
•
•
22 7
is an extension of k, both
is finitely generated over F. We often draw the following picture :
EF F F
k
E
E
F.
F.
Lines slanting up indicate an inclusion relation between fields. We also call the extension of the translation of to F, or also the lifting of to F. Let rx be algebraic over the field k. Let F be an extension of k, and assume k(rx), both contained in some field L. Then rx is algebraic over Indeed, the irreducible polynomial for rx over k has a fortiori coefficients in F, and gives a linear relation for the powers of rx over Suppose that we have a tower of fields :
E
E
each one generated from the preceding field by a single element. Assume that each rx i is algebraic over k, i = 1, . . . , n. As a special case of our preceding remark, we note that rx i +t is algebraic over k(rx 1 , , rx i ). Hence each step of the tower is algebraic. Proposition 1.6. Let
a field k, and assume rxi finite algebraic over k.
•
•
•
E
E
k (rx 1 , , rxn ) be a finitely generated extension of algebraic over k for each i = 1, . . . , n. Then is
=
•
•
•
E.
Proof From the above remarks, we know that can be obtained as the end of a tower each of whose steps is generated by one algebraic element, and is therefore finite by Proposition 1 .4. We conclude that is finite over k by Corollary 1 .3, and that it is algebraic by Proposition 1 . 1 . Let e be a certain class of extension fields F c We shall say that e is distinguished if it satisfies the following conditions : ( 1) Let k c c be a tower of fields. The extension k c is in e if and only if k c F is in e and c is in e . (2) If k c is in e , if is any extension of k, and F are both contained in some field, then c is in e . (3) If k c F and k c are in e and are subfields of a common field, then k c is in e .
F E F E E F F EF E F, E FE
E E,
228
E I F
EF F I "' E " 1
EF
/ � E F
ALG E B RAI C EXTE N S I O N S
V,
§1
The diagrams illustrating our properties are as follows :
I
EF
k (1)
�k/
k (2)
F EF. E/F F E E
(3)
E/F
These lattice diagrams of fields are extremely suggestive in handling exten sion fields. We observe that (3) follows formally from the first two conditions. Indeed, one views over k as a tower with steps k c c As a matter of notation, it is convenient to write instead of c to denote an extension. There can be no confusion with factor groups since we shall never use the notation to denote such a factor group when is an extension field of
E EF/F
F. ,
Proposition 1.7.
E/k EF F(a 1, a ,
F a1,EF/F, an E
The class of algebraic extensions is distinguished, and so is the class of finite extensions.
Prooj: Consider first the class of finite extensions. We have already proved condition ( 1). As for (2), assume that is finite, and let be any such extension of k. By Proposition 1 .5 there exist elements e that = k(a 1 , , n ) and hence is finitely = an )· Then generated by algebraic elements. Using Proposition 1 .6 we conclude that is finite. Consider next the class of algebraic extensions, and let •
F,
•
•
F E E E F. E. a n F0 k( ... , a0) . F0 a n, a F0• F0 k(an, ... , a0) F0(a) •
kc
•
•
c
•
•
.
F
is algebraic over k. Then a fortiori, is be a tower. Assume that Conversely, assume each step in algebraic over k and is algebraic over the tower to be algebraic. Let a e Then satisfies an equation
F0( a ) E
a a n + . . . + ao = 0 with ai e not all ai = 0. Let = Then is finite over k by Proposition 1 .6, and is algebraic over From the tower
kc = and the fact that each step in this tower is finite, we conclude that is finite over k, whence a is algebraic over k, thereby proving that is algebraic over k and proving condition (1) for algebraic extensions. Condition (2) has already been observed to hold, i.e. an element remains algebraic under lifting, and hence so does an extension. c
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Remark.
It is true that finitely generated extensions form a distinguished class, but one argument needed to prove part of (1) can be carried out only with more machinery than we have at present. cr. the chapter on transcen dental extensions.
§2. A LG E B RAI C C LO S U R E In this and the next section we shall deal with embeddings of a field into another. We therefore define some terminology. Let E be an extension of a field F and let
a: F + L be an embedding (i.e. an injective homomorphism) of F into L. Then a induces an isomorphism of F with its image aF, which is sometimes written Fa. An embedding t of E in L will be said to be over a if the restriction of t to F is equal to a. We also say that t extends a. If a is the identity then we say that t is an embedding of E over F. These definitions could be made in more general categories, since they depend only on diagrams to make sense : E
T
!
+
inc
F
(J
Remark.
L c
Let f(X) E F [X] be a polynomial, and let rx be a root of f in E. Say f(X) = a 0 + · · · + an x n . Then
a0 + a1a + + an a n . If t extends a as above, then we see that trx is a root of f a because 0 = t ( f(rx) ) = a g + a r (trx) + . . . + a: ( rrx)n. 0
=
f(a)
=
·
·
·
Here we have written a a instead of a (a) . This exponential notation is frequently convenient and will be used again in the sequel. Similarly, we write F a instead of a(F) or a F. In our study of embeddings it will also be useful to have a lemma concerning embeddings of algebraic extensions into themselves. For this we note that if a: E + L is an embedding over k (i.e. inducing the identity on k), then a can be viewed as a khomomorphism of vector spaces, because both E, L can be viewed as vector spaces over k. Furthermore a is injective.
230
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Lemma 2.1. Let E be an algebraic extension of k, and let a: E + E be an
embedding of E into itself over k. Then a is an automorphism.
Proof. Since a is injective, it will suffice to prove that a is surjective. Let rx be an element of E, let p(X) be its irreducible polynomial over k, and let E' be the subfield of E generated by all the roots of p(X) which lie in E. Then E ' is finitely generated, hence is a finite extension of k. Furthermore, a must map a root of p(X) on a root of p(X), and hence a maps E ' into itself. We can view a as a khomomorphism of vector spaces because a induces the identity on k. Since a is injective, its image a(E') is a subspace of E' having the same dimension [E' : k] . Hence u(E' ) = E' . Since a E E' , it follows that a is in the image of u, and our lemma is proved. Let E, F be extensions of a field k, contained in some bigger field L. We can form the rtng E[F] generated by the elements of F over E. Then E[F] = F[E] , and EF is the quotient field of this ring. It is clear that the elements of E[F] can be written in the form a 1 b 1 + . . . + a n bn with ai E E and bi E F. Hence EF is the field of quotients of these elements. Lemma 2.2. Let E 1 , E 2 be extensions of a field k, contained in some
bigger field E, and let a be an embedding of E in some field L. Then a(£ 1 £ 2 ) = a(E 1 ) a (£ 2 ).
Proof We apply a to a quotient of elements of the above type, say a 1 b 1 + . . . + an bn a r br + . . . + a: b: a = a ; b; + · · · + a�b� a�a b;a + · · · + a;: b;: ' and see that the image is an element of a(E 1 )a(E 2 ). It is clear that the image a(E 1 £ 2 ) is a(E 1 )a(£ 2 ). Let k be a field, f(X) a polynomial of degree > 1 in k [X]. We consider the problem of finding an extension E of k in which f has a root. If p(X) is an irreducible polynomial in k [X] which divides f(X), then any root of p(X) will also be a root of . f(X), so we may restrict ourselves to irreducible polynomials. Let p(X) be irreducible, and consider the canonical homomorphism
(
)
a: k [X] + k [X]j( p(X) ) . Then a induces a homomorphism on k, whose kernel is 0, because every nonzero element of k is invertible in k, generates the unit ideal, and 1 does not lie in the kernel. Let e be the image of X under a, i.e. e = a(X) is the residue class of X mod p( X). Then
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Hence e is a root of p a , and as such is algebraic over ak. We have now found an extension of ak, namely ak( e) in which p(l has a root. With a minor settheoretic argument, we shall have : Proposition 2.3. Let k be a field and f a polynomial in k[X] of degree > 1 . Then there exists an extension E of k in which f has a root.
Proof We may assume that f p is irreducible. We have shown that there exists a field F and an embedding a: k . F such that p a has a root e in F. Let S be a set whose cardinality is the same as that of F ak ( the complement of ak in F) and which is disjoint from k. Let E k u S. We can extend a: k + F to a bijection of E on F. We now define a field structure on E. If x, y E E we define xy a  1 ( a(x)a(y) ), x + y a  1 (a(x) + a(y) ) . =

=
=
=
=
Restricted to k, our addition and multiplication coincide with the given addition and multiplication of our original field k, and it is clear that k is a subfield of E. We let rx = a  1 (e). Then it is also clear that p(rx) = 0, as desired. Corollary 2.4. Let k be a field and let /1 , . . . , f, be polynomials in k [X] of degrees > 1 . Then there exists an extension E of k in which each fi has
a root, i
=
1 , . . . , n.
Proof Let E 1 be an extension in which /1 has a root. We may view f2 as a polynomial over E 1 • Let E 2 be an extension of E 1 in which /2 has a root. Proceeding inductively, our corollary follows at once. We define a field L to be algebraically closed if every polynomial in L[X] of degree > 1 has a root in L. Theorem 2.5. Let k be afield. Then there exists an algebraically closed field
containing k as a subfield. Proof We first construct an extension E1 of k in which every polyno mial in k [X] of degree > 1 has a root. One can proceed as follows (Artin). To each polynomial f in k [X] of degree > 1 we associate a letter X1 and we let S be the set of all such letters X1 (so that S is in bijection with the set of polynomials in k [X] of degree > 1 ). We form the polynomial ring k [S], and contend that the ideal generated by all the polynomials f(X1) in k[S] is not the unit ideal. If it is, then there is a finite combination of elements in our ideal which is equal to 1 : g t ft (XI. ) + . . . + gn fn (XI ) = 1 "
232
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§2
with g i E k [S]. For simplicity, write xi instead of xfi• The polynomials gi will involve actually only a finite number of variables, say X 1 , . . . , XN (with N > n). Our relation then reads n
g i (X l , . . . , XN )h (Xi ) = 1 . iL =1 Let F be a finite extension in which each polynomial f1 , • • • , f, has a root, say rt i is a root of fi in F, for i = 1 , . . . , n . Let rti = 0 for i > n . Substitute rt i for Xi in our relation. We get 0 = 1, contradiction. Let m be a maximal ideal containing the ideal generated by all polyno mials f(X1) in k [S]. Then k [S]/m is a field, and we have a canonical map a: k [S] + k [S]jm.
For any polynomial f E k [X] of degree > 1, the polynomial f a has a root in k [S]/m, which is an extension of ak. Using the same type of settheoretic argument as in Proposition 2.3, we conclude that there exists an extension E 1 of k in which every polynomial f E k [X] of degree > 1 has a root in E 1 . Inductively, we can form a sequence of fields
E1
c
E2
c
E3
c
···
c
En · · ·
such that every polynomial in En [X] of degree > 1 has a root in En + t · Let E be the union of all fields En , n = 1, 2, . . . . Then E is naturally a field, for if x, y E E then there exists some n such that x, y E En , and we can take the product or sum xy or x + y in En . This is obviously independent of the choice of n such that x, y E En , and defines a field structure on E. Every polynomial in E [X ] has its coefficients in some subfield En , hence a root in En + t , hence a root in E, as desired. Corollary 2.6. L et k be a field. There exists an extension ka which is algebraic over k and algebraically closed.
Proof Let E be an extension of k which is algebraically closed and let ka be the union of all subextensions of E, which are algebraic over k. Then ka is algebraic over k. If rt E E and rt is algebraic over ka then rt is algebraic over k by Proposition 1.7. If f is a polynomial of degree > 1 in k8 [X], then f has a root rt in E, and rt is algebraic over k8• Hence rt is in ka and ka is algebraically closed. We observe that if L is an algebraically closed field, and f E L[X] has degree > 1, then there exists c E L and rt 1 , . . . , rtn E L such that f(X) = c(X  rt 1 ) · · · (X  rtn ). Indeed, f has a root rt 1 in L, so there exists g(X) E L[ X ] such that f(X) = (X  rt 1 )g(X). If deg g
>
1, we can repeat this argument inductively, and express f as a
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product of terms (X  a i ) (i = 1 , . . . , n) and an element c E L. Note that c is the leading coefficient of f, i.e.
f(X)
=
eX"
+
terms of lower degree.
Hence if the coefficients of f lie in a subfield k of L, then c E k. Let k be a field and a : k + L an embedding of k into an algebraically closed field L. We are interested in analyzing the extensions of a to algebraic extensions E of k. We begin by considering the special case when E is generated by one element. Let E = k(a) where a is algebraic over k. Let p(X) = Irr(a, k, X). Let P be a root of p a in L. Given an element of k (a) = k [a], we can write it in the form f(a) with some polynomial f(X) E k [X]. We define an extension of a by mapping f(a) t+ f a (p).
This is in fact well defined, i.e. independent of the choice of polynomial f(X) used to express our element in k [a]. Indeed, if g(X) is in k[ X] and such that g(a) = f(a), then (g  f)(a) = 0, whence p (X) divides g(X)  f(X). Hence p a (X) divides g a (X)  f a (X), and thus g a (p) = f a ( p ). It is now clear that our map is a homomorphism inducing a on k, and that it is an extension of a to k(a). Hence we get : Proposition 2.7.
The number of possible extensions of a to k(a) is < the number of roots of p, and is equal to the number of distinct roots of p.
This is an important fact, which we shall analyze more closely later. For the moment, we are interested in extensions of a to arbitrary algebraic extensions of k. We get them by using Zorn' s lemma. Theorem 2.8. Let k be a field, E an algebraic extension of k, and
a: k + L an embedding of k into an algebraically closed field L. Then there exists an extension of a to an embedding of E in L. If E is algebraically closed and L is algebraic over ak, then any such extension of a is an isomorphism of E onto L.
Proof Let S be the set of all pairs (F, t) where F is a subfield of E containing k, and t is an extension of a to an embedding of F in L. If ( F, t) and (F ' , t ' ) are such pairs, we write (F, t) < (F', t ' ) if F c F' and t ' I F t . Note that S is not empty [it contains (k, a)], and is inductively ordered : If { (fi, t i )} is a totally ordered subset, we let F U .fi and define t on F to be equal to t i on each .fi. Then (F, t) is an upper bound for the totally ordered subset. Using Zorn 's lemma, let (K, A.) be a maximal element in S. Then A. is an extension of a, and we contend that K E. Otherwise, there exists a E E, =
=
=
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¢ K. By what we saw above, our embedding A. has an extension to K(rx), thereby contradicting the maximality of (K, A.). This proves that there exists an extension of a to E. We denote this extension again by a. If E is algebraically closed, and L is algebraic over ak, then a E is algebraically closed and L is algebraic over a E, hence L = a E. rx
As a corollary, we have a certain uniqueness for an " algebraic closure " of a field k. Corollary 2.9. Let
k be a field and let E, E' be algebraic extensions of k.
Assume that E, E' are algebraically closed. Then there exists an iso morphism t: E + E' of E onto E' inducing the identity on k. Proof Extend the identity mapping on k to an embedding of E into E' and apply the theorem. We see that an algebraically closed and algebraic extension of k is determined up to an isomorphism. Such an extension will be called an algebraic closure of k, and we frequently denote it by k8• In fact, unless otherwise specified, we use the symbol ka only to denote algebraic closure. It is now worth while to recall the general situation of isomorphisms and automorphisms in general categories. Let a be a category, and A, B objects in a. We denote by Iso(A, B) the set of isomorphisms of A on B. Suppose there exists at least one such isomorphism a: A + B, with inverse a  1 : B + A. If qJ is an automorphism of A, then a o qJ: A + B is again an isomorphism. If 1/J is an automorphism of B, then 1/J o a: A + B is again an isomorphism. Furthermore, the groups of automorphisms Aut(A) and Aut(B) are isomorphic, under the mappings qJ � a o lfJ o a  1 , a  1 0 1/1 0 a +1 1/J,
�l
luo � ou•
which are inverse to each other. The isomorphism a o qJ o a  1 is the one which makes the following diagram commutative : (J
A + B
A + B (J
We have a similar diagram for a  1 o 1/J o a. Let t: A + B be another isomorphism. Then t  1 o a is an automorphism of A, and t o a  1 is an automorphism of B. Thus two isomorphisms differ by an automorphism (of A or B). We see that the group Aut(B) operates on the
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set Iso (A, B) on the left, and Aut(A) operates on the set Iso(A, B) on the right. We also see that Aut(A) is determined up to a mapping analogous 'to a conjugation. This is quite different from the type of uniqueness given by universal objects in a category. Such objects have only the identity auto morphism, and hence are determined up to a unique isomorphism. This is not the case with the algebraic closure of a field, which usually has a large amount of automorphisms. Most of this chapter and the next is devoted to the study of such automorphisms. Examples.
It will be proved later in this book that the complex numbers are algebraically closed. Complex conjugation is an automorphism of C. There are many more automorphisms , but the other automorphisms =I= id . are not continuous . We shall discuss other possible automorphisms in the chapter on transcendental extensions . The subfield of C consisting of all numbers which are algebraic over Q is an algebraic closure Qa of Q . It is easy to see that Qa is denumerable . In fact, prove the following as an exercise: If k is a field which is not finite, then any algebraic extension of k has the same cardinality as k. If k is denumerable, one can first enumerate all polynomials in k, then enumerate finite extensions by their degree, and finall y enumerate the cardi nality of an arbitrary algebraic extension. We leave the counting details as . exerctses. In particular, Qa =F C. If R is the field of real numbers, then Ra = C. If k is a finite field, then algebraic closure k8 of k is denumerable. We shall in fact describe in great detail the nature of algebraic extensions of finite fields later in this chapter. Not all interesting fields are subfields of the complex numbers. For instance, one wants to investigate the algebraic extensions of a field C(X) where X is a variable over C. The study of these extensions amounts to the study of ramified coverings of the sphere (viewed as a Riemann surface), and in fact one has precise information concerning the nature of such extensions, because one knows the fundamental group of the sphere from which a finite number of points has been deleted. We shall mention this example again later when we discuss Galois groups.
§3. S P LITTI N G F I E L D S AN D N O R M A L EXT E N S I O N S
Let k be a field and let f be a polynomial in k [X] of degree > 1 . By a splitting field K of f we shall mean an extension K of k such that f splits into linear factors in K, i.e.
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c(X  � t ) · · · (X  �n ) 1, . . . , n, and such that K k(� 1 , . . . , �n ) is generated by all f(X)
with �i E K, i = the roots of f
§3
=
=
Theorem 3.1. Let K be a splitting field of the polynomial f(X) E k [X]. If E is another splitting field of f, then there exists an isomorphism a: E __.. K
inducing the identity on k. If k c K c k8, where ka is an algebraic closure of k, then any embedding of E in lc8 inducing the identity on k must be an isomorphism of E o nto K.
Proof Let K a be an algebraic closure of K. Then K a is algebraic over k, hence is an algebraic closure of k. By Theorem 2.8 there exists an embedding inducing the identity on k. We have a factorization
f(X) c(X  Pt ) · · · (X  Pn ) 1 , . . . , n. The leading coefficient c lies in k. We obtain f(X) f a (X) c(X  aP 1 ) · · · (X  af3n ). =
with pi E E, i =
=
=
We have unique factorization in K8 [X]. Since f has a factorization
c(X  � 1 ) · · (X  �n ) in K [X], it follows that (a/3 1 , , aPn ) differs from (� 1 , , �n ) by a permuta tion. From this we conclude that api E K for i 1 , . . . , n and hence that aE c K. But K k(� 1 , , �n ) k(aP 1 , , af3n ), and hence aE K, because f(X)
•
=
•
•
•
•
•
•
=
=
•
•
•
=
•
•
•
=
This proves our theorem. We note that a polynomial f(X) E k [X] always has a splitting field, namely the field generated by its roots in a given algebraic closure ka of k. Let I be a set of indices and let { .h }i e J be a family of polynomials in k[X], of degrees > 1 . By a splitting field for this family we shall mean an extension K of k such that every .h splits in linear factors in K [X], and K is generated by all the roots of all the polynomials _h, i E J. In most applica tions we deal with a finite indexing set I, but it is becoming increasingly important to consider infinite algebraic extensions, and so we shall deal with them fairly systematically. One should also observe that the proofs we shall give for various statements would not be simpler if we restricted ourselves to the finite case. Let ka be an algebraic closure of k, and let K i be a splitting field of .h in k8 • Then the compositum of the K i is a splitting field for our family,
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since the two conditions defining a splitting field are immediately satisfied. Furthermore Theorem 3. 1 extends at once to the infinite case :
Let K be a splitting field for the family { h } i e J and let E be another splitting field. Any embedding of E into K a inducing the identity on k gives an isomorphism of E onto K. Corollary 3.2.
Proof Let the notation be as above. Note that E contains a unique splitting field Ei of .h and K contains a unique splitting field Ki of };. Any embedding a of E into K a must map Ei onto K i by Theorem 3. 1 , and hence maps E into K. Since K is the compositum of the fields K i , our map a must send E onto K and hence induces an isomorphism of E onto K. Remark.
If I is finite, and our polynomials are /1 , • . . , fn , then a split ting field for them is a splitting field for the single polynomial f(X) = f1 (X) · · · f (X) obtained by taking the product. However, even when dealing with finite extensions only, it is convenient to deal simultaneously with sets of polynomials rather than a single one. n
Theorem 3.3. Let K be an algebraic extension of k, contained in an
algebraic closure ka of k. Then the following conditions are equivalent : NOR 1 . Every embedding of K in lc8 over k induces an automorphism of K. NOR 2. K is the splitting field of a family of polynomials in k [X]. NOR 3.
Every irreducible polynomial of k[X] which has a root in K splits into linear factors in K.
Proof. Assume NOR 1 . Let rx be an element of K and let Pa(X) be its irreducible polynomial over k. Let P be a root of Pa in k8• There exists an isomorphism of k(rx) on k(p) over k, mapping rx on p. Extend this iso morphism to an embedding of K in k8 • This extension is an automorphism a of K by hypothesis, hence arx P lies in K. Hence every root of Pa lies in K, and Pa splits in linear factors in K [X]. Hence K is the splitting field of the family { p« } « e as rx ranges over all elements of K, and NOR 2 is satisfied. Conversely, assume NOR 2, and let { h } i e J be the family of polynomials of which K is the splitting field. If rx is a root of some }; in K, then for any embedding a of K in ka over k we know that arx is a root of };. Since K is generated by the roots of all the polynomials /;,, it follows that a maps K into itself. We now apply Lemma 2. 1 to conclude that a is an automorphism. Our proof that NOR 1 implies NOR 2 also shows that NOR 3 is satisfied. Conversely, assume NOR 3. Let a be an embedding of K in ka over k. Let rx E K and let p(X) be its irreducible polynomial over k. If a is an embedding of K in ka over k then a maps rx on a root P of p (X), and by hypothesis p lies in K. Hence arx lies in K, and a maps K into itself. By Lemma 2. 1 , it follows that a is an automorphism. =
K
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An extension K of k satisfying the hypotheses NOR 1, NOR 2, NOR 3 will be said to be normal. It is not true that the class of normal extensions is distinguished. For instance, it is easily shown that an extension of degree 2 is normal, but the extension Q(.yl) of the rational numbers is not normal (the complex roots of X4 2 are not in it), and yet this extension is obtained by successive extensions of degree 2, namely 
E where
F
=
=
Q(�)
� y'2
Q(�),
=
=>
F
=>
Q,
and E = F( �).
Thus a tower of normal extensions is not necessarily normal. However, we still have some of the properties : Theorem 3.4.
Normal extensions remain normal under lifting. If K E k and K is normal over k, then K is normal over E. If K 1 , K 2 are normal over k and are contained in some field L, then K 1 K 2 is normal over k, and so is K 1 n K 2 =>
=>
•
Proof For our first assertion, let K be normal over k, let F be any extension of k, and assume K, F are contained in some bigger field. Let a be an embedding of KF over F (in F8). Then a induces the identity on F, hence on k, and by hypothesis its restriction to K maps K into itself. We get (KFY' K a F a KF whence KF is normal over F. Assume that K E k and that K is normal over k. Let a be an embedding of K over E. Then a is also an embedding of K over k, and our assertion follows by definition. Finally, if K 1 , K 2 are normal over k, then for any embedding a of K 1 K 2 over k we have =
=
=>
=>
and our assertion again follows from the hypothesis. The assertion concern ing the intersection is true because
a(K 1 n K 2 ) a(K 1 ) n a(K 2 ). We observe that if K is a finitely generated normal extension of k, say K k( � 1 ' . . . ' �n ), =
=
and p 1 , , Pn are the respective irreducible polynomials of � 1 , . . . , � n over k then K is already the splitting field of the finite family p 1 , , Pn · We shall investigate later when K is the splitting field of a single irreducible polynomial. •
•
•
•
•
•
S E PARAB LE EXTE N S I O N S
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§4. S E PA R A B LE EXT E N S I O N S Let E be an algebraic extension of a field F and let a:
F
.....
L
be an embedding of F in an algebraically closed field L. We investigate more closely extensions of a to E. Any such extension of a maps E on a subfield of L which is algebraic over aF. Hence for our purposes, we shall assume that L is algebraic over a F, hence is equal to an algebraic closure of a F. Let Sa be the set of extensions of a to an embedding of E in L. Let L' be another algebraically closed field, and let t: F L' be an embedding. We assume as before that L' is an algebraic closure of tF. By Theorem 2.8, there exists an isomorphism A.: L ..... L' extending the map t o a  1 applied to the field aF. This is illustrated in the following diagram : .....
L'
A.
E tF
I
F
u• (J
L aF
We let S1: be the set of embeddings of E in L' extending t. If a * E Sa is an extension of a to an embedding of E in L, then A. o a * is an extension of t to an embedding of E into L', because for the restriction to F we have A. o a * = t o a  1 o a = t. Thus A. induces a mapping from Sa into S1:. It is clear that the inverse mapping is induced by A.  1 , and hence that Sa , S1: are in bijection under the . mapptng a * 1+ A. 0 a * . In particular, the cardinality of Sa , S1: is the same. Thus this cardinality depends only on the extension E/F, and will be denoted by
[E : F]s. We shall call it the separable degree of E over F. It is mostly interesting when E/F is finite. Theorem 4.1. Let E ::J F ::J k be a tower. Then
[E : k]s = [E : F]s [F : k]s. Furthermore, if E is finite over k, then [E : k]s is finite and
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[E : k]s < [E : k]. The separable deg ree is at most equal to the degree. Proof. Let u : k L be an embedding of k in an algebraically closed field L. Let { ui } i e i be the family of distinct extensions of q to F, and for each i, let { t ii } be the family of distinct extensions of qi to E. By what we saw before, each qi has precisely [E : FJ s extensions to embeddings of E in L. The set of embeddings { t ii } contains precisely +
[E : F]s[F : k]s elements. Any embedding of E into L over q must be one of the t ii ' and thus we see that the first formula holds, i.e. we have multiplicativity in towers. As to the second, let us assume that E/k is finite. Then we can obtain E as a tower of extensions, each step being generated by one element :
k c k(cx 1 ) c k(cx 1 , � 2 ) c · · · c k (�1 , If we define inductively Fv + l
•
•
•
, cxr )
=
E.
Fv (cx v + l ) then by Proposition 2.7, [Fv (cx v +l ) : Fv J s < [Fv (� v +l ) : .fv ]. =
Thus our inequality is true in each step of the tower. By multiplicativity, it follows that the inequality is true for the extension E/k, as was to be shown. Corollary
4.2.
Let E be finite over k, and E ::J F ::J k. The equality [E : k]s = [E : k]
holds if and only if the corresponding equality holds in each step of the tower, i.e. for E/F and F/k. Proof. Clear. It will be shown later (and it is not difficult to show) that [E : k]s divides the degree [E : k] when E is finite over k. We define [E : k] i to be the quotient, so that [E : k]5 [E : k] i = [E : k]. It then follows from the multiplicativity of the separable degree and of the degree in towers that the symbol [E : k] i is also multiplicative in towers. We shall deal with it at greater length in §6. Let E be a finite extension of k. We shall say that E is separable over k if [E : k]s = [E : k]. An element ex algebraic over k is said to be separable over k if k(cx) is separable over k. We see that this condition is equivalent to saying that the irreducible polynomial Irr(cx, k, X) has no multiple roots. A polynomial f(X) e k [X] is called separable if it has no multiple roots.
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If rx is a root of a separable polynomial g(X) E k[X] then the irreducible polynomial of rx over k divides g and hence rx is separable over k. We note that if k C F C K and a E K is separable over k, then a is separable over F. Indeed , iff is a separable polynomial in k[X] such that f(a) = 0, then f also has coefficients in F, and thus a is separable over F. (We may say that a separable element remain's separable under lifting . ) Theorem 4.3. Let E be a finite extension of k. Then E is separable over k if and only if each element of E is separable over k.
Proof Assume E is separable over k and let rx E E. We consider the tower k c k( rx) c E. By Corollary 4 . 2 , we must have [k ( a) : k] = [k (a) : k]5 whence a is separable over k . Conversely , assume that each element of E is separable over k. We can write E = k ( a 1 , , an ) where each a; is separable over k. We consider the tower •
•
•
Since each rx i is separable over k, each rx i is separable over k( rx 1 , . . . , rx i _ 1 ) for i > 2. Hence by the tower theorem, it follows that E is separable over k. We observe that our last argument shows : If E is generated by a finite number of elements, each of which is separable over k, then E is separable over k. Let E be an arbitrary algebraic extension of k. We define E to be separable over k if every finitely generated subextension is separable over k, i.e., if every extension k( rx 1 , , rxn ) with rx 1 , , rxn E E is separable over k. •
•
•
•
•
•
\
/ Theorem 4.4. Let E be an algebraic extension of k, generated by a family of elements { rx i } i e i · If each rx i is separable over k then E is separable over k. Proof Every element of E lies in some finitely generated subfield k( rx i 1 , . . . , a,. ), n
and as we remarked above, each such subfield is separable over k. Hence every element of E is separable over k by Theorem 4.3, and this concludes the proof. Theorem 4.5. Separable extensions form a distinguished class of exten.
szons.
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Proof Assume that E is separable over k and let E ::J F ::J k. Every element of E is separable over F, and every element of F is an element of E, so separable over k. Hence each step in the tower is separable. Conversely, assume that E ::J F ::J k is some extension such that E/F is separable and Fjk is separable . If E is finite over k, then we can use Corollary 4 . 2 . Namely, we have an equality of the separable degree and the degree in eac h step of the tower, whence an equality for E over k by multiplicativity . If E is infinite, let CL E E. Then CL is a root of a separable polynomial f(X) with coefficients in F. Let these coefficients be an , . . . , a0• Let F0 = k (an , . . . , a 0 ). Then F0 is separable over k, and CL is separable over F0 • We now deal with the finite tower k c F0 c F0 (CL) and we therefore conclude that F0(CL) is separable over k, hence that CL is separable over k. This proves condition (1) in the definition of " distinguished. " Let E be separable over k. Let F be any extension of k, and assume that E, F are both subfields of some field. Every element of E is separable over k, whence separable over F. Since EF is generated over F by all the elements of E, it follows that EF is separable over F, by Theorem 4.4. This proves condition (2) in the definition of " distinguished, " and concludes the proof of our theorem. Let E be a finite extension of k. The intersection of all normal extensions K of k (in an algebraic closure E8 ) containing E is a normal extension of k which contains E, and is obviously the smallest normal extension of k containing E. If a 1 , , an are the distinct em beddings of E in E8, then the extension •
•
•
which is the compositum of all these embeddings, is a normal extension of k, because for any embedding of it, say t, we can apply t to each extension ai E. Then (ta1 , . . . , tan ) is a permutation of (a 1 , . . . , an ) and thus t maps K into itself. Any normal extension of k containing E must contain aiE for each i, and thus the smallest normal extension of k containing E is precisely equal to the compositum If E is separable over k, then from Theorem 4.5 and induction we conclude that the smallest normal extension of k containing E is also separ able over k. Similar results hold for an infinite algebraic extension E of k, taking an infinite compositum.
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In light of Theorem 4.5, the compositum of all separable extensions of a field k in a given algebraic closure ka is a separable extension, which will be denoted by k5 or ksep, and will be called the separable closure of k. As a matter of terminology, if E is an algebraic extension of k, and a any embedding of E in ka over k, then we call aE a conjugate of E in ka . We can say that the smallest normal extension of k containing E is the compositum of all the conjugates of E in E a . Let rx be algebraic over k. If a 1 , , ar are the distinct em beddings of k(rx) into ka over k, then we call a 1 rx, . . . , ar rx the conjugates of rx in /(8. These elements are simply the distinct roots of the irreducible polynomial of rx over k. The smallest normal extension of k containing one of these conjugates is simply k( a 1 rx, . . . , ar rx). •
•
•
E be a finite extension of a field k. There exists an element rx E E such that E = k(rx) if and only if there exists only a finite number of fields F such that k F c E. If E is separable over k, then there exists such an element rx. Proof If k is finite, then we know that the multiplicative group of E is generated by one element, which will therefore also generate E over k. We assume that k is infinite. Theorem 4.6. (Primitive Element Theorem). Let
c
Assume that there is only a finite number of fields, intermediate between k and E. Let rx, p E E. As c ranges over elements of k, we can only have a finite number of fields of type k( rx + cp). Hence there exist elements c 1 , c 2 E k with c 1 =F c 2 such that k(rx + c 1 P) = k (rx + c 2 P). Note that rx + c 1 P and rx + c 2 P are in the same field, whence so is (c 1  c 2 ) P, and hence so is p. Thus rx is also in that field, and we see that k(rx, p) can be generated by one element. Proceeding inductively, if E = k (rx 1 , , rxn ) then there will exist elements c 2 , , cn E k such that •
•
•
•
•
•
where � = rx 1 + c 2 rx 2 + · · · + cn rxn . This proves half of our theorem. Conversely, assume that E = k(rx) for some rx, and let f(X) = Irr(rx, k, X). Let k c F c E. Let gF(X) = lrr(rx, F, X). Then gF divides f. We have unique factorization in E[X], and any polynomial in E [X] which has leading coefficient 1 and divides f(X) is equal to a product of factors (X  rxi) where a 1 , . . . , an are the roots off in a fixed algebraic closure . Hence there is only a finite number of such polynomials . Thus we get a mapping
F � gF from the set of intermediate fields into a finite set of polynomials. Let F0 be
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the subfield of F generated over k by the coefficients of gF(X). Then gF has coefficients in F0 and is irreducible over F0 since it is irreducible over F. Hence the degree of rx over F0 is the same as the degree of rx over F. Hence F = F0 • Thus our field F is uniquely determined by its associated poly nomials gF, and our mapping is therefore injective. This proves the first assertion of the theorem. As to the statement concerning separable extensions, using induction, we may assume without loss of generality that E = k( rx, p ) where rx, p are separable over k. Let a 1 , • • • , an be the distinct em beddings of k(rx, p) in ka over k. Let P(X) = 0 (ai rx + X ai p  ai rx  X ai p). i �j Then P( X) is not the zero polynomial, and hence there exists c E k such that P(c) =F 0. Then the elements ai(rx + cp) (i = 1 , . . . , n) are distinct, whence k(rx + cp) has degree at least n over k. But n = [k(rx, p) : k], and hence
k(rx, p) = k(rx + cp), as desired. If E
§5.
=
k(rx), then we say that rx is a primitive element of E (over k).
FI N ITE FI E L D S
We have developed enough general theorems to describe the structure of finite fields. This is interesting for its own sake, and also gives us examples for the general theory. Let F be a finite field with have a homomorphism
q
elements. As we have noted previously, we Z + F
sending 1 on 1 , whose kernel cannot be 0, and hence is a principal ideal generated by a prime number p since ZjpZ is embedded in F and F has no divisors of zero. Thus F has characteristic p, and contains a field isomorphic to ZjpZ. We remark that ZjpZ has no automorphisms other than the identity. Indeed, any automorphism must map 1 on 1 , hence leaves every element fixed because 1 generates Z/pZ additively. We identify ZjpZ with its image in F. Then F is a vector space over ZjpZ, and this vector space must be
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finite since F is finite. Let its degree be n. Let ro 1 , , ron be a basis for F over Z/pZ. Every element of F has a unique expression of the form •
•
•
with a i E Z/pZ. Hence q = p " . The multiplicative group F* of F has order q  1 . Every a E F* satisfies the equation xq1 = 1 . Hence every element of F satisfies the equation f(X) = Xq  X =
0.
This implies that the polynomial f(X) has q distinct roots in F, namely all elements of F. Hence f splits into factors of degree 1 in F, namely Xq  X =
0 (X  a).
a. e F
In particular, F is a splitting field for f. But a splitting field is uniquely determined up to an isomorphism. Hence if a finite field of order p " exists, it is uniquely determined, up to an isomorphism, as the splitting field of X P"  X over ZjpZ. As a matter of notation, we denote Z/pZ by FP . Let n be an integer > 1 and consider the splitting field of XP"  X = f(X)
in an algebraic closure F;. We contend that this splitting field is the set of roots of f(X) in F; . Indeed, let a, p be roots. Then (a
+ p)P"  ( a + p ) = a P" + P P"  a  P = 0,
whence a + p is a root. Also, " ( af3)P  ap = aP"P P"  ap = a/3  ap =
and ap is a root. Note that 0, 1 ( p 1 )P"
_
are
0,
roots of f(X). If P #= 0 then
p  1 = (p P") 1 _ p 1 =
0
so that p 1 is a root. Finally, (
_
f3)P"
_
(
_
/3 ) = (
_
1 )P" /3 P"
+
/3.
If p is odd, then (  1 )P" =  1 and we see that  P is a root. If p is even then  1 = 1 (in Z/2Z) and hence  p = f3 is a root. This proves our contention. The derivative of f(X) is /'(X) = p"X P"  1 
1
=
 1.
Hence f(X) has no multiple roots, and therefore has p " distinct roots in F; . Hence its splitting field has exactly p " elements. We summarize our results :
246
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ALG E B RA I C EXT E N S I O N S
§5
Theorem 5.1. For each prime p and each integer n > 1 there exists a finite field of order p" denoted by FP"' uniquely determined as a subfield of an algebraic closure F; . It is the splitting field of the polynomial " XP  X '
and its elements are the roots of this polynomial. Every finite field is isomorphic to exactly one field Fp"· We usually write p" = q and Fq instead of Fp" ·
Let Fq be a finite field. Let n be an integer > 1 . In a given algebraic closure F;, there exists one and only one extension of Fq of degree n, and this extension is the field Fq" · Proof Let q = p "' . Then q" = p "'". The splitting field of X q"  X is precisely Fpm " and has degree mn over ZjpZ. Since Fq has degree m over ZjpZ, it follows that Fq" has degree n over Fq . Conversely, any extension of degree n over Fq has degree mn over FP and hence must be Fpm"· This proves our corollary. Corollary 5.2.
Theorem 5.3.
The multiplicative group of a finite field is cyclic. Proof. This has already been proved in Chapter IV, Theorem 1 .9.
We shall determine all automorphisms of a finite field. Let q = p n and let Fq be the finite field with q elements. We consider the
Frobenius mapping
cp : Fq + Fq such that cp(x) = x P. Then cp is a homomorphism, and its kernel is 0 since Fq is a field. Hence cp is injective. Since Fq is finite, it follows that cp is surjective, and hence that cp is an isomorphism. We note that it leaves FP fixed. Theorem 5.4.
generated by cp.
The group of automorphisms of Fq is cyclic of degree n,
Proof. Let G be the group generated by cp. We note that cp n = id because cp"(x) = x P" = x for all x E Fq. Hence n is an exponent for cp. Let d be the period of cp, so d > 1 . We have cp 4 (x) = x P d for all x E Fq. Hence each x E Fq is a root of the equation X Pd  X = 0.
This equation has at most p 4 roots. It follows that d > n, whence d = n. There remains to be proved that G is the group of all automorphisms of Fq. Any automorphism of Fq must leave FP fixed. Hence it is an auto
V,
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I N S E PA RAB LE EXTE N S I O N S
247
morphism of Fq over FP . By Theorem 4. 1 , the number of such auto morphisms is < n. Hence Fq cannot have any other automorphisms except fo r those of G. Theorem 5.5. Let m, n be integers > 1 . Then in any algebraic closure of FP ' the sub.field Fp" is contained in Fp'" if and only if n divides m. If that is the
case, let q = p n , and let m = nd. Then Fp'" is normal and separable o ver Fq ' and th e group of a u to mo rph isnzs ofFpm over Fq is cyclic of order d, generated by cpn. Proof. All the statements are trivi a l consequences of what has already been proved and will be left to the reader.
§6 .
I N S E PA R A B L E EXT E N S I O N S
This section is of a fairly technical nature, and can be omitted without impairing the understanding of most of the rest of the book. We begin with some remarks supplementing those of Proposition 2.7. Let f(X) = (X  �)mg(X) be a polynomial in k [X], and assume X  � does not divide g(X). We recall that m is called the multiplicity of � in f We � ay that � is a multiple root of f if m > 1 . Otherwise, we say that � is a simple root. Proposition 6. 1. Let � be algebraic over k, � E k8, and let
f(X) = lrr( �, k, X). If char k = 0, then all roots off have multiplicity 1 (f is separable). If char k = p > 0,
then there exists an integer p"'. We have
J.l
> 0 such that every root of f has multiplicity
[k(�) : k]
=
p"' [k(�) : kJs,
and � pll is separable over k. Proof Let � 1 , , �r be the distinct roots of f in ka and let � = � 1 . Let m be the multiplicity of � in f. Given 1 < i < r, there exists an isomorphism •
•
•
a:
over k such that a�
=
k(�) � k(�i)
�i · Extend
a
to an automorphism of k a and denote
248
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ALG E B RA I C EXTE N S I O N S
this extension also by note that
a.
§6
Since f has coefficients in k we have f a = f We r
/(X) = IT (X ua) mi j= I
is the multiplicity of ai in f. By unique factorization, we conclude that m i = m 1 and hence that all m i are equal to the same integer m. Consider the derivative f ' (X). If f and f ' have a root in common, then � is a root of a polynomial of lower degree than deg f. This is impossible unless deg f' = oo, in other words, f' is identically 0. If the characteristic is 0, this cannot happen. Hence if f has multiple roots, we are in characteris tic p, and f(X) = g(X P ) for some polynomial g(X) E k[X]. Therefore � P is a root of a polynomial g whose degree is < deg f Proceeding inductively, we take the smallest integer Jl > 0 such that � pll is the root of a separable polynomial in k[X], namely the polynomial h such that if
mi
f(X) = h(X pll ) . Comparing the degree of f and g , we conclude that Inductively, we find Since h has roots of multiplicity 1, we know that il il P P [k(� ) : k]s = [k(� ) : k], and comparing the degree of f and the degree of h, we see that the num ber of distinct roots of f is equal to the number of distinct roots of h. Hence From this our formula for the degree follows by multiplicativity, and our proposition is proved. We note that the roots of h are
� p1 ll '
Corollary
pll • � • · ·' r
For any finite extension E of k, the separable degree [E : k]s divides the degree [E : k]. The quotient is 1 if the characteristic is 0, and a power of p if the characteristic is p > 0. 6.2.
Proof We decompose E/k into a tower, each step being generated by one element, and apply Proposition 6. 1, together with the multiplicativity of our indices in towers. If E/K is finite, we call the quotient
V,
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I N S E PA RAB LE EXTE N S I O N S
249
[E : k] [E : kJs the inseparable degree (or degree of inseparability), and denote it by [E : k] i as in §4. We have [E : k]s [E : k] i = [E : k]. Corollary 6.3.
A finite extension is separable if and only if [E : k] i = 1 .
Proof By definition. Corollary 6.4 If E
=>
F
=>
k are two finite extensions, then
Proof. Immediate by Theorem 4 . 1 . We now assume throughout that k is a field of characteristic p > 0. An element rx algebraic over k is said to be purely inseparable over k if there exists an integer n > 0 such that rx P" lies in k. Let E be an algebraic extension of k. We contend that the following conditions are equivalent : P. Ins. 1 . We have [E : kJs = 1. P. Ins. 2.
Every element rx of E is purely inseparable over k.
P. Ins. 3.
For every rx E E, the irreducible equation of rx over k is of type X P"  a = 0 with some n > 0 and a E k.
P. Ins. 4. There exists a set of generators { rx i } i
each rx i is purely inseparable over k.
e
1
of E over k such that
To prove the equivalence, assume P. Ins. 1 . Let rx E E. By Theorem 4. 1 , we conclude that [k(rx) : kJs = 1. Let f(X) = l rr (rx, k, X). Then f has only one root since is equal to the number of distinct roots of f(X). Let m = [k(rx) : k]. Then deg f = m, and the factorization of f over k(rx) is f(X) = (X  rx)m. Write n m = p r where r is an integer prime to p. Then
f(X) = (X P"  rx P" t t = X P"r  rrx P"X P"< r  >
+
lo wer term s.
Since the coefficients of f(X) lie in k, it follows that
rrx P"
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ALG E B RAI C EXT E N S I O N S
V,
§6
lies in k, and since r =F 0 (in k), then rx, P" lies in k. Let a = rx P". Then rx is a root of the polynomial X P"  a, which divides f(X). It follows that f(X) = X P"  a. Essentially the same argument as the preceding one shows that P. Ins. 2 implies P. Ins. 3. It is trivial that the third condition implies the fourth. Finally, assume P. Ins. 4. Let E be an extension generated by purely inseparable elements rxi (i E J). Any embedding of E over k maps rxi on a root of fi(X) = Irr(rxi, k, X). But /i(X) divides some polynomial X P"  a, which has only one root. Hence any embedding of E over k is the identity on each rxi, whence the identity on E, and we conclude that [E : k]s = 1, as desired. An extension satisfying the above four properties will be called purely
inseparable.
Proposition
6.5.
of extensio ns.
Purely inseparable extensions form a distinguished class
Proof The tower theorem is clear from Theorem 4. 1 , and the lifting property is clear from condition P. Ins. 4. Proposition
Let E be an algebraic extension of k. Let E 0 be the compositum of all subfields F of E such that F ::J k and F is separable over k. Then E 0 is separable over k, and E is purely inseparable over Eo . 6.6.
Proof Since separable extensions form a distinguished class, we know that E0 is separable over k. In fact, E 0 consists of all elements of E which are separable over k. By Proposition 6. 1 , given rx E E there exists a power of p, say p n such that rx, P" is separable over k. Hence E is purely inseparable over E 0 , as was to be shown. Corollary
6.7.
Corollary
6.8.
If an algebraic extension E of k is both separable and purely inseparable, then E = k. Proof Obvious. Let K be normal over k and let K 0 be its maximal separa ble subextension. Then K 0 is also normal over k. Proof Let a be an embedding of K 0 in K a over k and extend a to an embedding of K. Then a is an automorphism of K. Furthermore, aK 0 is separable over k, hence is contained in K 0 , since K 0 is the maximal separa ble subfield. Hence aK 0 = K 0 , as contended.
V,
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Corollary
I N S E PA RAB LE EXTE N S I O N S
251
Let E, F be two finite extensions of k, and assume that E/k is separable, F/k is purely inseparable. Assume E, F are subfields of a common field. Then 6.9.
[EF : F]
=
[E : k]
=
[EF : E]
=
[F : k]
=
[EF : kJs, [EF : k]i.
Proof. The picture is as follows :
The proof is a trivial juggling of indices, using the corollaries of Proposition 6. 1. We leave it as an exercise.
Let E P denote the field of all elements x P, x E E. Let E be a finite extension of k. If E Pk = E, then E is separable over k. If E is separable over k, then E P" k = E for all n > 1 . Proof. Let E0 be the maximal separable subfield of E. Assume E Pk = E. Let E = k(rx 1 , , rxn ). Since E is purely inseparable over E 0 there exists m such that rxf '" E E0 for each i = 1 , . . . , n. Hence E P'" c E0 • But E P'" k = E whence E = E 0 is separable over k. Conversely, assume that E is separable over k. Then E is separable over EPk. Since E is also purely inseparable over £ P k we conclude that E = E Pk . Similarly we get E = E P n k for n > 1 , as was to be shown . Corollary
6. 1 0.
•
•
•
Proposition 6.6 shows that any algebraic extension can be decomposed into a tower consisting of a maximal separable subextension and a purely inseparable step above it. Usually, one cannot reverse the order of the tower. However, there is an important case when it can be done. Proposition 6. 1 1 . Let K be normal over k. Let G be its group ofautomorphisms
over k. Let J(G be the fixed field of G (see Chapter VI, § 1 ) . Then K G is purely inseparable over k, and K is sep arable over K G . If Ko is the maximal separa ble subextension of K, then K == K G Ko and Ko n K G == k. Proof Let rx E K G . Let t be an embedding of k(rx) over k in K a and extend t to an embedding of K, which we denote also by t. Then t is an automorphism of K because K is normal over k. By definition, t rx = rx and hence t is the identity on k(rx). Hence [k(rx) : kJs = 1 and rx is purely in separable. Thus K G is purely inseparable over k. The intersection of K 0
252
k.
V,
ALG E B RAI C EXT E N S I O N S
KG
k,
§6
and is both separable and purely inseparable over and hence is equal to To prove that is separable over assume first that is finite over Let a 1 , . . . , ar be a and hence that G is finite, by Theorem 4. 1 . Let rx E maximal subset of elements of G such that the elements
k,
K G,
K
K.
are distinct, and such that a 1 is the identity, and r
f(X) = 0 (X

i=l
rx
K
is a root of the polynomial
ai rx).
K G.
For any t E G we note that fr: = f because t permutes the roots. We note that f is separable, and that its coefficients are in the fixed field Hence rx The reduction of the infinite case to the finite case is is separable over done by observing that every rx E is contained in some finite normal subextension of We leave the details to the reader. We now have the following picture :
K G. K.
K
K
/ K 0 K G""
"K G
Ko
� � K0 n K G k
K K0K G. K K0K G. K K0K G,
=
K0, K G,
By Proposition 6.6, is purely inseparable over hence purely insepara ble over Furthermore, is separable over hence separable over Hence = thereby proving our proposition. We see that every normal extension decomposes into a compositum of a purely inseparable and a separable extension. We shall define a Galois ex tension in the next chapter to be a normal separable extension. Then is Galois over and the normal extension is decomposed into a Galois and a purely inseparable extension. The group G is called the Galois group of the extension A field is called perfect if = (Every field of characteristic zero is also called perfect.)
Kjk. k
Corollary
K0
k
kP k.
k
k
If is perfect, then every algebraic extension of is separable, and every algebraic extension of k is perfect. Proof Every finite algebraic extension is contained in a normal exten sion, and we apply Proposition 6. 1 1 to get what we want. 6. 1 2.
V, Ex
EXE R C I S ES
253
EX E R C I S ES 1 . Let E = Q(�), where � is a root of the equation �3 + � 2 + � + 2 = 0. Express (�2 + � + 1 ) (�2 + �) and (�  1)  1 in the form
a�2 + ba. + c with a, b, c e Q.
2. Let E
=
F(a.) where � is algebraic over F, of odd degree. Show that E = F(a. 2 ).
3. Let � and f3 be two elements which are algebraic over F. Let f(X) = Irr(a., F, X) and g(X) = Irr(/3, F, X). Suppose that deg f and deg g are relatively prime. Show that g is irreducible in the polynomial ring F(a.) [X]. 4. Let � be the real positive fourth root of 2. Find all intermediate fields in the extension Q(a.) of Q. 5. If a. is a complex root of X6 + X3 + 1 , find all homomorphisms [Hint : The polynomial is a factor of X9  1 .] 6. Show that
a:
Q(a.) + C.
.j2 + J3 is algebraic over Q, of degree 4.
7. Let E, F be two finite extensions of a field k, contained in a larger field K. Show that
[EF : k] < [E : k] [F : k]. If [E : k] and [F : k] are relatively prime, show that one has an equality sign in the above relation . 8 . Let f(X ) E k[X] be a polynomial of degree n . Let K be its splitting field . Show that [K : k] divides n !
9. Find the splitting field of X Ps  1 over the field Z/pZ. 10. Let a. be a real number such that a.4 = 5. (a) Show that Q(ia. 2 ) is normal over Q. (b) Show that Q(� + i�) is normal over Q(i� 2 ). (c) Show that Q(� + ia.) is not normal over Q.
1 1 . Describe the splitting fields of the following polynomials over Q, and find the degree (a) X 2 (c) X3 (e) X 2 (g) X5
of each such splitting field. 2 (b) X 2  1 (d) (X3  2) (X 2  2) 2 (f) X6 + X3 + 1 +x+ 1
7
1 2. Let K be a finite field with p" elements. Show that every element of K has a unique pth root in K.
254
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ALG E B RA I C EXT E N S I O N S
1 3 . If the roots of a monic polynomial f(X) E k[X] in some splitting field are distinct , and form a field , then char k = p and f(X) = XP n  X for some n > 1 .
14. Let char K
p. Let L be a finite extension of K, and suppose [L : K] prime to p. Show that L is separable over K. =
1 5. Suppose char K = p. Let a E K. If a has no pth root in K, show that X P"  a is irreducible in K [ X ] for all positive integers n. 1 6. Let char K = p. Let cx be algebraic over K. Show that cx is separable if and only if K (cx) = K (cx P" ) for all positive integers n. 1 7. Prove that the following two properties are equivalent : (a) Every algebraic extension of K is separable. (b) Either char K = 0, or char K = p and every element of K has a pth root in
K.
1 8. Show that every element of a finite field can be written as a sum of two squares in that field. 1 9. Let E be an algebraic extension of F. Show that every subring of E which contains F is actually a field . Is this necessarily true if E is not algebraic over F? Prove or give a counterexample.
20. (a) Let E
F(x) where x is transcendental over F. Let K :/; F be a subfield of E which contains F. Show that x is algebraic over K. (b) Let E = F(x). Let y = f(x)jg(x) be a rational function, with relatively prime polynomials f, g E F [x]. Let n = max(deg f, deg g). Suppose n > 1 . Prove =
that
[F(x) : F(y)]
=
n.
2 1 . Let z + be the set of positive integers, and A an additive abelian group. Let f: z + A and g : z + A be maps. Suppose that for all n, +
+
f(n) = L g(d). din
Let J.1. be the Mobius function (cf. Exercise 1 2 of Chapter II). Prove that
g(n)
=
L J.J.(njd)f(d).
din
22. Let k be a finite field with q elements. Let f(X) E k [X] be irreducible. Show that f(X) divides X q"  X if and only if deg f divides n. Show the multiplication formula
X q"

X = 0 0 fd( X ), din
f d irr
where the inner prod uct is over all irreducible polynomials of degree d with leading coefficient 1 . Counting degrees, show that qn
=
I dl/J (d),
din
where l/J(d) is the number of irreducible polynomials of degree d. Invert by
V, Ex
EX E R C I S ES
255
Exercise 2 1 and find that
nt/J (n) = L f.l(d)q nfd . din
23. (a) Let k be a finite field with q elements. Define the zeta function Z( t)
=
( 1  t)  1 0 ( 1 p

t deg p )  1 ,
where p ranges over all irreducible polynomials p = p(X) in k [X] with leading coefficient 1 . Prove that Z(t) is a rational function and determine this rational function. (b) Let nq (n) be the number of primes p as in (a) of degree < n. Prove that q
qm
q (m) "' q  1 m
n
for m +
oo .
Remark. This is the analogue of the prime number theorem i n number theory, but it is essentially trivial in the present case, because the Riemann hypothesis is trivially verified. Things get more interesting fast after this case. Consider an equation y2 = x 3 + ax + b over a finite field Fq of characteristic :/; 2, 3, and having q elements. Assume  4a 3  2 7b 2 :/; 0, in which case the curve defined by this equation is called an elliptic curve. Define Nn by Nn 
1
=
number of points (x, y) satisfying the above equation with x, y E Fq" (the extension of Fq of degree n).
Define the zeta function Z(t) to be the unique rational function such that Z(O) and Z'/Z(t)
=
=
1
L Nn t "  1 •
A famous theorem of Hasse asserts that Z(t) is a rational function of the form
Z(t) =
( 1  � t) ( l  �t) , ( 1  t) ( 1  qt)
where � is an imaginary quadratic number (not real, quadratic over Q), � is its complex conj ugate, and �� = q, so 1 � 1 = q 1 12• See Hasse, " Abstrakte Bergrund ung der komplexen Multiplikation und Riemannsche Vermutung in Funktionen korpern, " Abh. Math. Sem. Univ. Hamburg 10 ( 1934) pp. 325348. 24. Let k be a field of characteristic p and let t, u be algebraically independent over k. Prove the following : (a) k(t, u) has degree p2 over k(tP, uP). (b) There exist infinitely many extensions between k(t, u) and k(tP, uP). 25. Let E be a finite extension of k and let p r = [E : k] ;. We assume that the characteristic is p > 0. Assume that there is no exponent ps with s < r such that £Psk is separable over k (i.e., such that a, r is separable over k for each a. in E). Show that E can be generated by one element over k . [H in t : Assume first that E is purely inseparable.]
256
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ALG E B RA I C EXT E N S I O N S
26 . Let k be a field , f(X ) an irreducible polynomial in k[X] , and let K be a finite normal extension of k. If g , h are monic irreducible factors of f(X) in K[X] , show that there exists an automorphism u of K over k such that g = hu . Give an example when this conclusion is not valid if K is not normal over k.
, x,. be algebraically independent over a field k. Let y be algebraic over k(x) = k(x 1 , , x,.). Let P(X,. + 1 ) be the irreducible polynomial of y over k(x). Let qJ(x) be the least common multiple of the denominators of the coefficients of P. Then the coefficients of qJ(x)P are elements of k[x]. Show that the polynomial
27. Let x1 ,
•
•
•
•
•
.
f(X 1 , . . . , X,. + 1 ) = qJ(X 1 , . . • , X,.)P(X,. + 1 ) is irreducible over k, as a polynomial in n + 1 variables. Conversely, let f(X 1 , , X,. + 1 ) be an irreducible polynomial over k. x 1 , . . . , x,. be algebraically independent over k. Show that •
•
•
Let
is irreducible over k(x 1 , . . . , x,.). If f is a polynomial in n variables, and (b) = (b1 , . . . , b,.) is an ntuple of elements such that f(b) = 0, then we say that (b) is a zero of f. We say that (b) is nontrivial if not all coordinates b; are equal to 0. 28. Let f(X 1 , , X,.) be a homogeneous polynomial of degree 2 (resp. 3) over a field k. Show that if f has a nontrivial zero in an extension of odd degree (resp. degree 2) over k, then f has a nontrivial zero in k. •
•
•
29. Let f(X , Y) be an irreducible polynomial in two variables over a field k. Let t be transcendental over k, and assume that there exist integers m, n :/; 0 and elements a, b e k, ab :/; 0, such that f(at", btm) = 0. Show that after inverting possibly X or Y, and up to a constant factor, f is of type
with some c e k. The answer to the following exercise is not known.
30. (Artin conjecture). Let f be a homogeneous polynomial of degree d in n vari ables, with rational coefficients. If n > d, show that there exists a root of unity ,, and elements
x1 , . . . , x,. e Q ['] not all 0 such that f(x 1 ,
•
.
•
, x,.) = 0.
3 1 . D ifference equations. Let u 1 , . . . , ud be elements of a field K. We want to solve for infinite vectors (x0, x 1 , . . . , x,., . . . ) satisfying
for Define the characteristic polynomial of the system to be
n > d.
V, Ex
EXE R C IS ES
257
Suppose ex is a root of f. (a) Show that x,. = ex" (n > 0) is a solution of ( * ) . (b) Show that the set of solutions of ( • ) is a vector space of dimension d. (c) Assume that the characteristic polynomial has d distinct roots ex 1 , . . . , exd . Show that the solutions (ex� ), . . . , (ex;) form a basis for the space of solutions. (d) Let x,. = b 1 ex� + · · · + bd ex; for n > 0, show how to solve for b 1 , . . . , bd in terms of ex 1 , . . . , exd and x 0, , xd _ 1 . (Use the Vandermonde determinant.) (e) Under the conditions of (d), let F(T) = L x,. T". Show that F(T) represents a rational function, and give its partial fraction decomposition. •
•
•
32. Let d = 2 for simplicity. Given a0, a1 , u, of the system
v,
w,
t e K, we want to find the solutions for
n
>
2.
Let ex 1 , ex 2 be the root of the characteristic polynomial, that is
Assume that ex1 , ex 2 are distinct, and also distinct from t. Let 00
F (X ) = L a,.X". n =O
(a) Show that there exist elements A, B, C of K such that
F ( X) =
A B C . + + 1  ex 1 X 1  ex 2 X 1  tX
(b) Show that there is a unique solution to the difference equation given by for n > 0. (To see an application of this formalism to modular forms, as in the work of Manin, Mazur, and SwinnertonDyer, cf. my Introduction to Modular Forms, SpringerVerlag, New York, 1976, Chapter XII, §2.) 33. Let R be a ring which we assume entire for simplicity. Let
g ( T)
=
Td  ad 1 yd  1  . . .  a o
be a polynomial in R [T], and consider the equation
Td = ao + a1 T + . . . + ad  1 yd  1 . Let x be a root of g( T). (a) For any integer n > d there is a relation with coefficients ai , i in Z[a0, , ad  1 ] c R. (b) Let F(T ) e R [T] be a polynomial. Then •
•
•
F (x) = a0 ( F) + a 1 (F)x + · · · + ad _ 1 (F)x d  1
where the coefficients a;(F) lie in R and depend linearly on F.
258
ALG EB RA I C EXTEN S I O N S
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(c) Let the Vandermonde determinant be 1 1
V(x1 , . . , xd ) = .
1
x2
x dl 1 x d2  1
xd
1 xdd 
Xt
= n (xi  X; ) . i <j
Suppose that the equation g ( T) = 0 has d roots and that there is a factoriza tion
g ( T) =
d
n 1 . If the characteristic of k is p, then the equation has only one root, namely 1, and hence there is no pmth root of unity except 1 .
VI, §3
ROOTS OF U N ITY
277
Let n be an integer > 1 and not divisible by the characteristic . The polynomial xn  1
is separable because its derivative is n x n  1 =F 0, and the only root of the deriva tive is 0, so there is n o common root. Hence in ka the polynomial x n  1 has n distinct roots, which are roots of unity. They obviously form a group, and we know that every finite multiplicative group in a field is cyclic (Chapter IV, Theorem 1 . 9) . Thus the group of nth roots of unity is cyclic . A generator for this group is called a primitive nth root of unity . If Jln denotes the group of all nth roots of unity in ka and m, n are relatively prime integers, then Jlm n
�
Jlm
X
Jln ·
This follows because Jlm , Jln cannot have any element in common except 1 , and because Jlm Jln consequently has mn elements, each of which i s an mnth root of unity. Hence Jlm Jln = J.lmn ' and the decomposition is that of a direct product. As a matter of notation , to avoid double indices , especially in the prime power case , we write J1[n] for Jln · So if p is a prime , J1[p r] is the group of p r th roots of unity . Then J1[p00] denotes the union of all J1 (p r ] for all positive integers r. See the comments in § 14. Let k be any field. Let n be not divisible by the characteristic p . Let ( = ( be a primitive nth root of unity in ka. Let cr be an embeddin g of k(() in ka over k. Then n
(a()n
=
a((n)
=
1
so that a( is an nth root of unity also. Hence a( = ( i for some integer i = i( a), uniquely determined mod n. It follows that a maps k(() into itself, and hence that k(() is normal over k. If r is another automorphism of k ( () over k then ar(
=
( i(G) i (r)
.
Since a and r are automorphisms, it follows that i( a) and i( r) are prime to n (otherwise, a( would have a period smaller than n). In this way we get a homo morphism of the Galois group G of k( () over k into the multiplicative group ( Z/n Z ) * of integers prime to n, mod n. Our homomorphism is clearly injective since i(a) is uniquely determined by a mod n, and the effect of a on k(() is determined by its effect on ( . We conclude that k(() is abelian over k. We know that the order of (ZjnZ)* is qJ{ n). Hence the degree [k(() : k] divides qJ(n ) . For a specific field k, the question arises whether the image of G K( { )tK in (ZinZ) * is all of (ZinZ) * . Looking at K = R or C , one sees that this is not always the case . We n ow give an important e xample when it is the case .
278
VI, §3
GALOIS TH EORY
Theorem 3. 1 .
Let ( be a primitive nth root of unity. Then [Q(() : Q] = qJ(n) ,
where
cp
is the Euler function. The map u � i(u) gives an isomorphism GQIQ + (Z/nZ) * .
Proof. Let j'(X) be the irreducible polynomial of ( over Q. Then f (X) divides X "  1 , say X "  1 = f(X)h(X ), where bothf, h have leading coefficient 1. By the Gauss lemma, it follows that f, h have integral coefficients. We shall now prove that if p is a prime number not dividing n, then (P is also a root off
Since (P is also a primitive nth root of unity, and since any primitive nth root of unity can be obtained by raising ( to a succession of prime powers, with primes not dividing n , this will imply that all the primitive nth roots of unity are roots off, which must therefore have degree > qJ(n), and hence precisely qJ(n). Suppose (P is n ot a root off Then (P is a root of h, and ( itself is a root of h(X P ) . Hence f(X) divides h(X P ), and we can write
h(X P ) = f(X)g(X). Since f has integral coefficients and leading coefficient 1 , we see that g has integral coefficients. Since aP  a (mod p) for any integer a, we conclude that and hence
h(X P ) h(X )P
=
=
h(X ) P { mod p),
f(X)g(X) (mod p).
In particular, if we denote by f and h the polynomials in ZjpZ obtained by reducing f and h respectively mod p, we see that J and Ji are not relatively prime, i.e. have a factor in common. But x n  I = .f(X)Ii(X), and hence x n  T has multiple roots. This is impossible, as one sees by taking the de rivative, and our theorem is proved. Corollary 3.2.
If n, m are relative prime integers > 1 , then n
m
mn
n
Proof We note that ( and ( are both contained in Q( ( ) since (:Z is a primitive mth root of unity. Furthermore, ( ( is a primitive mnth root of m
n
unity. Hence
Our assertion follows from the multiplicativity qJ( mn ) = qJ( m ) qJ(n). Suppose that n is a prime number p (having nothing to do with the character istic). Then XP  1 = (X  1 )(X p  1 + . . . + 1 ).
VI, §3
ROOTS OF UN ITY
279
Any primitive pth root of unity is a root of the second factor on the right of this equation. Since there are exactly p  1 primitive pth roots of unity, we con clude that these roots are precisely the roots of xp  1 + . . . + 1 .
We saw in Chapter I V , §3 that this polynomial could be transformed into an Eisenstein polynomial over the rationals . This gives another proof that [Q((p ): Q] = p  1 . We investigate more closely the factorization of X"  1 , and suppose that we are in characteristic 0 for simplicity. We have
x n  1 = Il (X  () , '
where the product is taken over all nth roots of u nity. Collect together all terms belonging to roots of unity having the same period. Let
Then
x n  1 = Il d(X) . di n We see that 1 (X) = X  1 , and that
Il (X) din d d< n From this we can compute (X) recursively , and we see that n (X) is a polynomial in Q[X] because we divide recursively by polynomials having coefficients in Q . All our polynomials have leading coefficient 1 , so that in fact n (X) has integer coefficients by Theorem 1 . 1 of Chapter IV . Thus our construction is essentially universal and would hold over any field (whose characteristic does not divide
n) .
We call n (X) the nth cyclotomic polynomial . The roots of n are precisely the primitive nth roots of unity , and hence deg n = cp(n) .
From Theorem
3.1
we conclude that n is irreducible over Q, and hence
280
VI, §3
GALOIS TH EORY
We leave the proofs of the following recursion formulas as exercises : 1 . If p is a prime number , then
p (X)
=
xp  t
+
xp  2 +
and for an integer r > 1 , p r(X) 2.
Let n
p;•
=
· · ·
=
· · · + 1,
r 1
p (XP ) .
p�� be a positive integer with its prime factorization. Then
n (X)
=
t · · ps(X P'i �  t P�s  • ). P
3. If n is odd > 1 , then 2n (X) = n < X) . 4. If p is a prime number, not dividing n, then
n (XP) pn (X)  n (X) . On the other hand , if p I n , then pn (X) = n (XP) .
5. We have
n (X)
=
IT (X nld  1 ) J.L(d ) . di n
As usual , J1 is the Mobius function : if n is divisible by p 2 for some prime p, ( 1 )r if n = p 1 • • • P r is a product of distinct primes , 1 if n =
0 Jl(n)
=

1.
As an exercise, show that
L din
JL(d)
=
{1
if n 0 if n
= >
1.
1,
Example. In light of Exercise 2 1 of Chapter V , we note that the association n � n (X) can be viewed as a function from the positive integers into the multiplicative group of nonzero rational functions . The multiplication formula x n  1 = n d (X) can therefore be inverted by the general formalism of convolutions . Computations of a number of cyclotomic polynomials show that for low values of n , they have coefficients equal to 0 or + 1 . However, I am indebted to Keith Conrad for bringing to my attention an extensive literature on the subject , starting with Bang in 1 895 . I include only the first and last items: A.
S . B ANG ,
pp .
6 1 2
Om Ligningen m (X )
=
0, Nyt Tidsskrift for Matematik (B) 6 ( 1 895) ,
H . L . MONTGOMERY and R . C. VA UGHN , The order of magnitude of the mth coef ficients of cyclotomic polynomials, Glasgow Math . J. 27 ( 1 985 ) , pp . 1 43 1 59
VI, §3
ROOTS OF U N ITY
281
In particular, if n (X) = �anjXi , define L(j) = log maxn I a ni j . Then Montg omery and Vaughn prove that · I /2
J
(log j) 1 14
•
•
.
Corollary 4.2. Let a b . . , an be distinct nonzero elements of a field K. If , an are elements of K such that for all integers v > 0 we have a1, .
•
•
a then a;
=
1
a�
+
· · ·
+ an a�
0
=
0 for all i.
Proof We apply the theorem to the distinct homomorphisms of Z> o into K* . Another interesting application will be given as an exercise (relative in variants).
§5.
T H E N O R M AN D TRACE
Let E be a finite extension of k. Let [E : k]s p
ll
=
=
r, and let
[E : k] ;
, ar be the distinct if the characteristic is p > 0, and 1 otherwise. Let a 1 , embeddings of E i n an algebraic closure ka of k. If a is an element of E, we define its norm from E to k to be •
NE/ k ( a)
=
N f(rx)
=
•
•
.D/r.a.P" CD! rE:k);. =
u. rx
Similarly, we define the trace
TrE1k( a)
=
Tr f( a )
=
[E : k] i
r
L a v a.
v= l
The trace is equal to 0 if [E : k] ; > 1 , in other words, if Ejk is not separable.
VI,
THE NORM AND TRAC E
§5
285
Thus if E is separable over k, we have (I
where the product is taken over the distinct embeddings of E in ka over k. Similarly, if Ejk is separable, then Trf(rx) = L arx. (I
Theorem 5. 1 . Let E/k be a finite extension. Then the norm Nf is a multi
plicative homomorphism of E * into k * and the trace is an additive homo morphism of E into k. If E F k is a tower offields, then the two maps are transitive, in other words, =>
=>
Nf = Nf o N: and Trf = Trf o Tr: .
If E
=
k(a), and f(X) = Irr{a, k, X) = x n + an _ 1xn  t + . . . + a 0 , then N� < cx > (a) = {  l )n a 0 and Tr� (a) =  a n 1 · 
Proof For the first assertion, we note that rxP"' is separable over k if p11 = [E : k] ; . On the other hand, the product r
n1 (] v a P"'
v =
is left fixed under any isomorphism into ka because applying such an iso morphism simply permutes the factors. Hence this product must lie in k since rxP"' is separable over k. A similar reasoning applies to the trace. For the seco!ld assertion, let { r i} be the family of distJ? ct em beddings of F into ka over k. Extend each ri to an automorphism of lc8 , and denote this extension by ri also. Let { a; } be the family of em beddings of E in ka over F. (Without loss of generality, we may assume that E c k8.) If a is an embedding of E over k in k8, then for some j, r i 1 a leaves F fixed, and hence rj 1 a = a; for some i. Hence a = r i O"; and consequently the family {r i a;} gives all distinct em beddings of E into ka over k. Since the inseparability degree is multiplicative in towers, our assertion concerning the transitivity of the norm and trace is obvious, because we have already shown that N� maps E into F, and similarly for the trace. Suppose now that E = k(rx). We have
f {X)
=
{ {X  a 1 ) . • • { X
_
lir )) [E : k 1 ,
if a b . . . , a, are the distinct roots off Looking at the constant term off gives us the expression for the norm, and looking at the next to highest term gives us the expression for the trace . We observe that the trace is a klinear map of E into k, namely Trf(crx) = c Trf(a )
286
VI, §5
GALOIS TH EORY
for all rx E E and c E k. This is clear since c is fixed under every embedding of E over k. Thus the trace is a klinear functional of E into k. For simplicity , we write Tr = Tr f . Theorem 5.2.
Let E be a .finite separable extension of k. Then Tr : E __.. k is a nonzero functional. The map (x, y) � Tr( xy) of E x E __.. k is bilinear, and identifies E with its dual space. Proof. That Tr is nonzero follows from the theorem on linear indepen dence of characters. For each x E E, the map Trx : E __.. k such that Trx(Y) = Tr(xy) is obviously a klinear map, and the map is a khomomorphism of E into its dual space Ev. (We don't write E* for the dual space because we use the star to denote the multiplicative group of E . ) If Trx is the zero map, then Tr(xE) = 0. If x =F 0 then x E = E. Hence the kernel of x 1+ Trx is 0. Hence we get an injective homomorphism of E into the dual space E. Since these spaces have the same finite dimension, it follows that we get an isomorphism. This proves our theorem. Corollary 5.3. Let w 1 , , wn be a basis of E over k. Then there exists a basis w'1 , , w� of E over k such that Tr(W; wj) = J ii . •
•
•
•
•
•
Proof The basis
w '1 , • • . , w� is none other than the dual basis which we
defined when we considered the dual space of an arbitrary vector space.
Let E be a finite separable extension of k, and let a , an be the distinct set of embeddings of E into k8 over k. Let w 1 , , wn be ele ments of E. Then the vectors
Corollary 5.4.
1,
•
are linearly independent over E if w 1 ,
•
•
•
ments of E such that
Then we see that
•
•
•
, wn form a basis of E over k.
Proof Assume that w . . . , wn form a basis of Ejk. Let b
•
•
rx 1 , . . . , rxn be ele
VI, §5
THE NORM AND TRAC E
287
applied to each one of w 1 , , an are linearly , wn gives the value 0. But a 1 , ind ependent as characters of the multiplicative group E* into k8*. It follows that rx; = 0 for i = 1 , . . , n, and our vectors are linearly independent. •
•
•
•
•
•
.
Remark . In characteristic 0, one sees much more trivially that the trace is not identically 0. Indeed, if c E k and c =F 0, then Tr( c ) = nc where n = [E : k], and n =F 0. This argument also holds in characteristic p when n is prime to p. Propo�i tion
5.5.
Let E
=
k(rx) be a separable extension. Let f (X)
Irr( rx, k, X),
=
and let f ' (X) be its derivative. Let ( ) = Po + f3 t X + ( rx.)
{�
·
·
·
Pn  l x n 
+
l
1
with Pi E E. Then the dual basis of 1 , rx, . . . , rxn  is Po
f ' (rx) Proof Let rx
1,
.
•
•
'
·
·
Pn  1
f ' (rx) ·
·'
, rxn be the distinct roots off Then for 0 < r < n

1.
To see this , let g(X) be the difference of the left and righthand side of this equality. Then g has degree < n  1 , and has n roots rx b . . . , rxn . Hence g is identically zero. The polynomials
f(X) ( X  rxi ) f '(rx i )
are all conjugate to each other. If we define the trace of a polynomial with coefficients in E to be the polynomial obtained by applying the trace to the coefficients, then
[ f(X) Tr (
X
_
rxr rx.) f ' ( rx.)
J
=
r
X.
Looking at the coefficients of each power of X in this equation, we see that
thereby proving our proposition. Finally we establish a connection with determinants , whose basic propertie s we now assume . Let E be a finite extension of k, which we view as a finite dimensional vector space over k. For each a E E we have the klinear map
288
VI, §6
GALOIS THEORY
multiplication by a, ma :
E � E such that ma(x)
= ax .
Then we have the determinant det(ma) , which can be computed as the determinant of the matrix Ma representing ma with respect to a basis . Similarly we have the trace Tr(ma) , which is the sum of the diagonal elements of the matrix Ma. Proposition 5. 6. Let E be a finite extension of k and let a E E. Then
det(ma)
=
NE1k(a) and Tr(ma)
=
TrE1k(a) .
Let F = k(a) . If [F : k] = d , then 1 , a, . . . , atL 1 is a basis for F over k. Let { w b . . . , wr } be a basis for E over F. Then { a iwi } (i = 0, . . . , d  1 ; j = 1 , . . . , r) is a basis for E over k. Let Proof.
+ ad _ 1 Xd  1 + . . . + ao be the irreducible polynomial of a over k. Then NFlk( a) = (  1 )da0 , and by the j (X)
=
Xd
transitivity of the norm , we have
NElk( a)
=
NFlk ( a ) r .
The reader can verify directly on the above basis that NF;k (rx) ' is the determinant of ma on F, and then that NFlk ( a) d is the determinant of ma on E, thus concluding the proof for the determinant . The trace is handled exactly in the same way , except that TrE1k(a) = r · Trp1k(a) . The trace of the matrix for ma on F i s equal to  ad _ 1 • From this the statement identifying the two traces is immediate , as it was for the norm .
§ 6.
CYC LI C EXTE N S IO N S
We recall that a finite extension is said to be cyclic if it is Galois and its Galois group is cyclic . The determination of cyclic extensions when enough roots of unity are in the ground field is based on the following fact . Theorem 6.1 . (Hilbert's Theorem 90). Let K/k be cyclic of degree n with Galois group G. Let q be a generator of G. Let f3 E K. The norm N:( f3) = N({J) is equal to 1 if and only if there exists an element rx =F 0 in K such that f3
=
rxfqrx.
Proof Assume such an element rx exists. Taking the norm of f3 we get N(rx)/N(qrx). But the norm is the product over all automorphisms in G. Inserting
q
just permutes these automorphisms. Hence the norm is equal to 1 . It will be convenient to use an exponential notation as follows. If r, t' and � E K we write
E
G
VI, §6
CYC LIC EXTENSIONS
289
B y Artin's theorem on characters, the map given by
on K is not identically zero. Hence there exists (} E K such that the element
is n ot equal to 0. It is then clear that prxa = rx using the fact that N(p) = 1 , and hence that when we apply a to the last term in the sum, we obtain 0. We divide by rxa to conclude the proof. Theorem 6.2.
Let k be a field, n an integer > 0 prime to the characteristic of k, and assume that there is a primitive nth root of unity in k. (i) Let K be a cyclic extension of degree n. Then there exists rx E K such that n K = k(rx), and rx satisfies an equation x  a = 0 for some a E k. (ii) Conversely, let a E k. Let rx be a root of x n  a. Then k(rx) is cyclic over k, of degree d, d I n, and rxd is an element of k. Proof Let ( be a primitive nth root of unity in k, and let K/k be cyclic with
group G. Let a be a generator of G. We have N((  1 ) = ((  1)n = 1 . By Hilbert's theorem 90, there exists rx E K such that arx = (rx. Since ( is in k, we have ai rx = ( irx for i = 1 , . . . , n. Hence the elements ( i rx are n distinct conj ugates of rx over k, whence [k(rx) : k] is at least equal to n. Since [K : k] = n, i t follows that K = k(rx). Furthermore,
Hence rxn is fixed under a, hence is fixed under each power of a, hence is fixed under G. Therefore rxn is an element of k, and we let a = rxn . This proves the first part of the theorem. Conversely , let a E k. Let a be a root of x n  a. Then a( ; is also a root for each i = 1 , . . . , n, and hence all roots lie in k( a) which is therefore normal over k. All the roots are distinct so k ( a) is Galois over k. Let G be the Galois group. If a is an automorphism of k(rx)/k then arx is also a root of xn  a. Hence arx = wa rx where wa is an nth root of unity, not necessarily primitive. The map a �+ wa is obviously a homomorphism of G into the group of nth roots of unity, and is injective. Since a subgroup of a cyclic group is cyclic , we conclude that G is cyclic , of order d, and d i n . The image of G is a cyclic group of order d. If u is a generator of G , then wu is a primitive dth root of unity . Now we get
Hence rxd is fixed under a, and therefore fixed under G. It is an element of k, and our theorem is proved.
290
VI, §6
GALOIS TH EORY
We now pass to the analogue of Hilbert ' s theorem 90 in characteristic p for cyclic extensions of degree p. Theorem 6.3. (Hilbert's Theorem 90, Additive Form). Let k be a field and K/k a cyclic extension of degree n with group G. Let o be a generator of G. Let p E K. The trace Trf (p) is equal to 0 if and only if there exists an element rx E K such that p = rx  arx.
Proof If such an element rx exists, then we see that the trace is 0 because the trace is equal to the sum taken over all elements of G, and applying a per
mutes these elements. Conversely, assume Tr(/3) Tr( B) =F 0. Let
=
0. There exists an element (J E K such that
From this it follows at once that p Theorem 6.4. (ArtinSchreier)
rx
=

arx.
Let k he a field of characteristic p.
(i) Let K be a cyclic extension of k of degree p. Then there exists rx E K such that K = k(rx) and rx satisfies an equation X P  X  a = 0 with some
a E k.
(ii) Conversely, given a E k, the polynomial f ( X )
=
X P  X  a either has
one root in k, in which case all its roots are in k, or it is irreducible. In this latter case, if rx is a root then k( rx) is cyclic of degree p over k.
Proof Let K/k be cyclic of degree p. Then Tr f(  1 ) = 0 (it is just the sum of  1 with itself p times). Let a be a generator of the Galois group. By the additive form of Hilbert ' s theorem 90, there exists rx E K such that arx  rx = 1 , o r in other words , arx = rx + 1 . Hence a irx = rx + i for all integers i = 1 , . . . , p and rx has p distinct conjugates. Hence [k(rx) : k] > p. It follows that K = k(rx). We note that
a(rx P  rx)
=
a(rx)P  a(rx)
=
(rx + 1 ) P  (rx + 1 )
=
rxP
 rx.
Hence rxP  rx is fixed under a, hence it is fixed under the powers of a, and therefore under G. It lies in the fixed field k. If we let a = rxP  rx we see that our first assertion is proved. Conversely, let a E k. If rx is a root of X P  X a then rx + i is also a root for i = 1 , . . . , p. Thus f(X) has p distinct roots. If one root lies in k then all roots lie in k. Assume that no root lies in k. We contend that the 
VI, §7
SOLVABLE AND RAD ICAL EXTENSIONS
291
polynomial is irreducible. Suppose that
f(X) with g, h E k[X] and 1 < deg g
1, and we let M range over positive integers divid ing N. We let P be the set of primes dividing N. We let G be a group, and let : A
=
Gmodule such that the isotropy group of any element of A is of finite index in G . We also assume that A is divisible by the primes p I N, that is pA = A
for all p E P .
r = finitely generated subgroup of A such that r is pointwise fixed by G.
We assume that A N is finite. Then
� r is also finitely generated.
Note that
Example.
For our purposes here, the above situation summarizes the properties which hold in the following situation. Let K be a finitely generated field over the rational numbers, or even a finite extension of the rational numbers. We let A be the mu ltiplicative group of the algebraic closure Ka. We let G = GK be the Galois group Gal(Ka/K). We let r be a finitely generated subgroup of the multiplicative group K * . Then all the above properties are satisfied. We see that A N
J!N is the group of Nth roots of unity. The group written in additive notation is written r t ; N in multiplicative notation. =
�
r
Next we define the appropriate groups analogous to the Galois groups of Kummer theory, as follows. For any Gsubmodule B of A, we let : G(B)
=
image of G in Aut(B),
G(N)
=
G(AN)
H (N)
=
subgroup of G leaving A N pointwise fixed,
=
image of G in Aut(A N ),
Hr(M, N) (for M I N) = image of H(N) in Aut
(� } r
VI, § 1 1
NONABELIAN KUM M E R EXTENSIONS
Then we have an exact sequence : 0 + H r(M, N) + G
(� r + A N) +
} }
Example.
G( N )
305
+ 0.
In the concrete case mentioned above, the reader will easily recognize these various groups as Galois groups. For instance, let A be the multiplicative group. Then we have the following lattice of field extensions with corresponding Galois groups : K( N , r t ; M ) JI
I
K(pN ) I
K
Hr(M, N) G(N)
In applications, we want to know how much degeneracy there is when we trans late K (p M , r11 M ) over K(p N ) with M I N . This is the reason we play with the pair M, N rather than a single N. Let us return to a general Kummer representation as above. We are in terested especially in that part of (Z/NZ)* contained in G(N), namely the group of integers n (mod N) such that there is an element [n] in G(N) such that [n]a
=
na
for all a E A N
.
Such elements are always contained in the center of G(N), and are called homotheties.
Write N
=
fJ pn(p)
Let S be a subset of P. We want to make some nondegeneracy assumptions about G( N ). We call S the special set. There is a product decomposition (Z/NZ)*
=
n (Z/pn
Z)*.
PIN
If 2 1 N we suppose that 2 E S. For each p E S we suppose that there is an integer c( p) = pf < P > with f(p) > 1 such that G(A N)
::::::>
n Uc (p)
peS
X
n (z/pn(p)Z)*,
pfS
where Uc < P > is the subgroup of Z(pn
) consisting of those elements = 1 mod c(p).
306
VI, §1 1
GALOIS TH EORY
The product decomposition on the right is relative to the direct sum decom position AN
EBN A pn(p) .
=
pi
The above assumption will be called the nondegeneracy assumption. The integers c(p) measure the extent to which G(AN) is degenerate. Under this assumption, we observe that [2] E G(A M ) if M I N and M is not divisible by primes of S ; [ 1 + c] E G(A M ) if M I N and M is divisible only by primes of S, where
c
=
c(S)
=
n c(p).
peS
We can then use [2]  [ 1 ] = [ 1 ] and [ 1 + c]  [ 1 ] Lemma 10.2, since [1] and [c] are in the center of G. For any M we define
c(M)
=
=
[c] in the context of
n c(p). PIM peS
Define
r'
=
_!_ r n AG N
and the exponent
e(r'jr)
=
smallest positive integer e such that er ' c r.
It is clear that degeneracy in the Galois group H r(M, N) defined above can arise from lots of roots of unity in the ground field, or at least degeneracy in the Galois group of roots of unity ; and also if we look at an equation
XM 
a =
0,
from the fact that a is already highly divisible in K. This second degeneracy would arise from the exponent e(r 'jr), as can be seen by looking at the Galois group of the divisions of r. The next theorem shows that these are the only sources of degeneracy. We have the abelian Kummer pairing for M I N, H r(M, N)
X
r;Mr __.. A M given by {t, x) I+ ty
where y is any element such that M y
=
 y,
x. The value of the pairing is indepen
NONABELIAN KU M M E R EXTENSIONS
VI, §1 1
307
dent of the choice of y. Thus for X E r, we have a homomorphism
CfJx : H r(M, N) __.. AM such that
CfJ x{ t) = ty  y, Theorem
11.1 .
where My = x.
Let M I N. Let cp be the homomorphism cp :
r __.. Hom ( H r(M, N), AM)
and let r be its kernel. Let eM(r) = g.c.d. (e(r 'jr), M). Under the non degeneracy assumption, we have (/)
c(M)eM(r)r c Mr. q>
Proof Let X E r and suppose CfJx = 0. Let My =
X.
For a E G let
Ya = ay  Y · Then { Ya } is a 1cocycle of G in AM , and by the hypothesis that CfJx = 0, this cocycle depends only on the class of a modulo the subgroup of G leaving the elements of AN fixed. In other words, we may view { Ya } as a cocycle of G(N) in AM . Let c = c(N). By Lemma 1 0.2, it follows that {cya } splits as a cocycle of G(N) in AM . In other words, there exists t 0 E A M such that and this equation in fact holds for a E G. Let t be such that ct = t 0 • Then
cay  cy = act  cy, whence c(y  t) is fixed by all (j E G, and therefore lies in _!_ r. Therefore N
e(r'jf)c(y  t) E r. We multiply both sides by M and observe that cM(y  t) = cMy = ex . This shows that
c(N)e(r'jr)r c Mr. qJ
Since r;M r has exponent M, we may replace e(r 'jr) by the greatest common divisor as stated in the theorem, and we can replace c(N) by c(M) to conclude the proof. Corollary 1 1 .2.
Assume that M is prime to 2(f ' : r) and is not divisible by any primes of the special set S. Then we have an injection cp :
r;Mr __.. Hom(Hr(M, N), AM ).
308
VI, §1 2
GALO IS TH EORY
/fin addition r isfree with basis {a b . . . ' ar }, and we let C{J; = CfJ a; ' then the map H r(M, N) � A 1 is additive, and given by the additive polynomial aX Pm. We shall see later that this is a typical example. Theorem
(Artin). Let A. b . . . , A.n : A __.. K be additive homomorph isms of an additive group into a field. If these homomorphisms are alge braically dependent over K, then there exists an additive polynomial 1 2. 1 .
in K[ X ] such that
for all x E A. Proof Let f( X ) = f(X b . . . , X n) E K[X] be a reduced polynomial of lowest possible degree such that f =F 0 but for all x E A, f(i\(x)) = 0, where i\(x) is the vector (A. 1 (x), . . . , A.n (x)) . We shall prove that f is additive. Let g( X , Y) = f(X + Y )  f( X ) f( Y ). Then 
g( i\(x), i\(y))
= f(i\(x
+ y))  f(i\(x))  f(i\(y))
=
0
for all x, y E A. We shall prove that g induces the zero function on K < n > x K . Assume otherwise. We have two cases. Case 1 . We have g(�, i\(y)) = 0 for all � E K and all y E A. By hypothesis, there exists �' E K such that g(� ' , Y) is not identically 0. Let P( Y) = g( � ' , Y). Since the degree of g in ( Y) is strictly smaller than the degree of f, we have a contradiction. Case 2. There exist �' E K and y' E A such that g( �', i\ ( y ' )) =F 0. Let P( X ) = g( X , 1\(y ' )). Then P is not the zero polynomial, but P(i\(x)) = 0 for all x E A, again a contradiction.
31 0
VI, §1 2
GALOIS TH EORY
We conclude that g induces the zero function on K what we wanted, namely that f is additive.
x
K 0, and assume that a 1 , , a" are algebraically dependent. There exists an additive polynomial f(X 1 , , X" ) in K[X] which is reduced, j' =F 0, and such that •
•
•
•
f(a 1 (x ), . . . , an( x ))
=
•
•
0
for all x E K. By what we saw above, we can write this relation in the form m
n L L a; r U; (x )Pr
i= 1 r= l
=
0
for all x E K , and with not all coefficients a ir equal to 0. Therefore by the linear independence of characters, the automorphisms { (JP1•r }
•
WIt h
•
z =
1 , . . . , n an d r
=
1, . . . , m
cannot be all distinct. Hence we have with either i =F j or r =F s. Say r < s. For all x E K we have
Extracting pth roots in characteristic p is unique. Hence for all x E K. Let a
=
aj 1a; . Then U( X )
for all x E K. Taking a"
=
s
r = Xp 
id shows that X = XP
n( s  r )
for all x E K. Since K is infinite, this can hold only if s = r. But in that case, a; = ai , contradicting the fact that we started with distinct automorphisms.
31 2
§1 3.
VI, §1 3
GALOIS TH EORY
T H E N O R M A L BAS IS T H E O R E M
Let K/k be afinite Galois extension of degree n. Let a an be the elements of the Galois group G. Then there exists an elemen t w E K such that a 1 w, . . . , an w form a basis of K over k. Proof. We prove this here only when k is infinite. The case when k is finite can be proved later by methods of linear algebra, as an exercise. For each a E G, let X a be a variable, and let tu, r = Xa  l r · Let Xi = Xaj · Let Theorem
1 3. 1 .
1,
•
•
.
,
Then f is not identically 0, as one sees by substituting 1 for Xid and 0 for Xa if a =F id. Since k is infinite,f is reduced. Hence the determinant will not be 0 for all x E K if we substitute a i(x) for Xi in f. Hence there exists w E K such that Suppose ab . . . , an E k are such that
Apply a; 1 to this relation for each i = 1 , . . . , n. Since a ; E k we get a system of linear equations, regarding the a ; as unknowns. Since the determinant of the coefficients is =F 0, it follows that aJ· = 0
for j
=
1, . . . , n
and hence that w is the desired element. Remark.
In terms of representations as in Chapters III and XVIII , the normal basis theorem says that the representation of the Galois group on the additive group of the field is the regular representation . One may also say that K is free of dimension 1 over the group ring k[ G] . Such a result may be viewed as the first step in much more subtle investigations having to do with algebraic number theory . Let K be a number field (finite extension of Q) and let o K be its ring of algebraic integers , which will be defined in Chapter VII , § 1 . Then one may ask for a description of o K as a Z[ G] module , which is a much more difficult problem. For fundamental work about this problem , see A. Frohlich , Galois Module Structures of Algebraic Integers, Ergebnisse der Math . 3 Folge Vol . 1 , Springer Verlag ( 1 983) . See also the reference [CCFT 9 1 ] given at the end of Chapter III , § 1 .
VI, § 1 4
§ 1 4.
I N F I N ITE GALO IS EXTENSIONS
31 3
I N F I N I T E G A LO I S EXT E N S I O N S
Although we have already given some of the basic theorems of Galois theory already for possibly infinite extensions , the nonfiniteness did not really appear in a substantial way . We now want to discuss its role more extensively . Let KI k be a Galois extension with group G . For each finite Galois subex tension F, we have the Galois groups GKIF and GF!k · Put H = G KIF · Then H has finite index , equal to #(GF1k ) = [F : k] . This just comes as a special case of the general Galois theory . We have a canonical homomorphism G � G IH = G F/k ·
Therefore by the universal property of the I nverse limit, we obtain a homomorphism G � lim G IH , HE �
where the limit is taken for H in the family Theorem 14. 1 .
�
of Galois groups G KIF as above .
The homomorphism G � lim G IH is an isomorphism.
Proof. First the kernel is trivial , because if u is in the kernel , then u restricted to every finite subextension of K is trivial , and so is trivial on K. Recall that an element of the inverse limit is a family { uH } with uH E G IH, satisfying a certain compatibility condition . This compatibility condition means that we may define an element u of G as follows . Let a E K. Then a is contained in some finite Galois extension F C K. Let H = Gal(KIF) . Let ua = uHa . The compatibility condition means that uH a is independent of the choice ofF. Then it is immediately verified that u is an automorphism of K over k, which maps to each uH in the canonical map of G into G I H. Hence the map G � li!!l G IH is surjective , thereby proving the theorem. Remark. For the topological interpretation , see Chapter I, Theorem 1 0 . 1 , and Exercise 43 . Example. Let JJ,[poc] be the union of all groups of roots of unity Jl [p n ] , where p is a prime and n = 1 , 2, . . . ranges over the positive integers . Let K = Q(J1 [p00] ) . Then K is an abelian infinite extension of Q. Let ZP be the ring of padic integers , and z; the group of units . From §3 , we know that (Zip n Z) * is isomorphic to Gal(Q(JJ, [p n ]IQ)) . These isomorphisms are compatible in the tower of pth roots of unity , so we obtain an isomorphism
z; � Gal(Q(J1[p00] 1Q)).
31 4
VI, §1 4
GALOIS TH EORY
Towers of cyclotomic fields have been extensively studied by Iwasawa. Cf. a systematic exposition and bibliography in [La 90] . For other types of representations in a group GL 2 (Zp ) , see Serre [ Se 68 ] , [Se 72] , Shimura [Shi 7 1 ] , and LangTrotter [LaT 75] . One general framework in which the representation of Galois groups on roots of unity can be seen has to do with commutative algebraic groups , starting with elliptic curves . Specif ically , consider an equation
y2
=
4x 3  g2x  g3
with g2 , g3 E Q and nonzero discriminant: d = g�  27 g� =I= 0 . The set of solutions together with a point at infinity is denoted by E . From complex analysis (or by purely algebraic means) , one sees that if K is an extension of Q , then the set of solutions E(K) with x, y E K and oo form a group , called the group of rational points of E in K. One is interested in the torsion group, say E (Qa) tor of points in the algebraic closure , or for a given prime p , in the group E ( Qa ) [p r] and E ( Qa ) [ p oc ] . As an abelian group, there is an isomorphism E ( Qa ) [ p r] = (Z/p rZ)
X
(Z/p r z) ,
so the Galois group operates on the points of order p r via a representation in GL 2 (Z/p rz) , rather than GL 1 (Z/p rz) = (Z/p rZ) * in the case of roots of unity . Passing to the inverse limit, one obtains a representation of Gal ( Q a /Q ) = G Q in GL 2 (Zp ) . One of Serre ' s theorems is that the image of G Q in GL 2 (Zp ) is a subgroup of finite index , equal to GL 2 (Zp ) for all but a finite number of primes p, if End C (E) = Z. More generally , using freely the language of algebraic geometry , when A is a commutative algebraic group, say with coefficients in Q , then one may consider its group of points A (Qa) to P and the representation of G Q in a similar way . Developing the notions to deal with these situations leads into algebraic geometry . Instead of considering cyclotomic extensions of a ground field , one may also consider extensions of cyclotomic fields . The following conjecture is due to Shafarevich . See the references at the end of §7 . Conjecture 14. 2. Let k0 = Q(Ji) be the compositum of all cyclotomic exten
sions of Q in a given algebraic closure Qa. Let k be a finite extension of k0. Let Gk = Gal(Qa/k) . Then Gk is isomorphic to the completion of a free group on countably many generators.
If G is the free group, then we recall that the completion is the inverse limit lim G /H , taken over all normal subgroups H of finite index . Readers should view this conjecture as being in analogy to the situation with Riemann surfaces , as mentioned in Example 9 of §2. It would be interesting to investigate the extent to which the conjecture remains valid if Q( Ji) is replaced by Q(A (Qa) tor ) , where A is an elliptic curve . For some results about free groups occurring as Galois groups, see also Wing berg [Wi 9 1 ] .
VI, § 1 5
TH E MODULAR CONNECTION
31 5
Bibliography S. LANG , Cyclotomic Fields I and II, Second Edition, Springer Verlag , 1 990 (Combined edition from the first editions , 1 978 , 1 980) [LaT 75 ] S . LANG and H . TROTTER , Distribution of Frobenius Elements in GL2 Extensions of the Rational Numbers , Springer Lecture Notes 504 ( 1 97 5) [Se 68] J . P. S ERRE , Abelian 1adic Representations and Elliptic Curves , Benjamin , 1 968 [Se 72] J . P. S ERRE , Proprietes galoisiennes des points d' ordre fini des courbes ellip tiques , Invent. Math. 15 ( 1 972) , pp . 25933 1 [Shi 7 1 ] G. S H IMURA , Introduction to the arithmetic theory of Automorphic Functions, Iwanami Shoten and Princeton University Press, 1 97 1 [Wi 9 1 ] K . WINGB ERG , On Galois groups of pclosed algebraic number fields with restricted ramification , I , J. reine angew. Math. 400 ( 1 989) , pp. 1 85202 ; and I I , ibid. , 416 ( 1 99 1 ) , pp . 1 87 1 94 [La 90]
§1 5 . T H E M O D U L A R C O N N E CT I O N This final section gives a major connection between Galois theory and the theory of modular forms, which has arisen since the 1 960s . One fundamental question is whether given a finite group G , there exists a Galois extension K of Q whose Galois group is G . In Exercise 23 you will prove this when G is abelian . Already in the nineteenth century , number theorists realized the big difference between abelian and nonabelian extensions, and started understanding abelian extensions . Kronecker stated and gave what are today considered incomplete arguments that every finite abelian extension of Q is contained in some extension Q( () , where ( is a root of unity . The difficulty lay in the peculiarities of the prime 2 . The trouble was fixed by Weber at the end of the nineteenth century . Note that the trouble with 2 has been systematic since then . It arose in Artin' s conjecture about densities of primitive roots as mentioned in the remarks after Theorem 9. 4. It arose in the Grunwald theorem of class field theory (corrected by Wang , cf. ArtinTate [ArT 68] , Chapter 1 0) . It arose in Shafarevich ' s proof that given a solvable group, there exists a Galois extension of Q having that group as Galois group, mentioned at the end of §7 . Abelian extensions of a number field F are harder to describe than over the rationals , and the fundamental theory giving a description of such extensions is called class field theory (see the above reference) . I shall give one significant example exhibiting the flavor. Let RF be the ring of algebraic integers in F. I t can be shown that RF is a Dedekind ring . (Cf. [La 70] , Chapter I , §6 , Theorem 2 . ) Let P be a prime ideal of R F . Then P n Z = (p) for some prime number p.
31 6
VI, §1 5
GALO IS TH EORY
Furthermore , RF/ P is a finite field with q elements . Let K be a finite Galois extension of F. It will be shown in Chapter VII that there exists a prime Q of RK such that Q n RF = P. Furthermore , there exists an element FrQ E G = Gal(K/F ) such that FrQ (Q) = Q and for all a E RK we have FrQ a a'l mod Q . =
We call FrQ a Frobenius element i n the Galois group G associated with Q . (See Chapter VII , Theorem 2 . 9 . ) Furthermore , for all but a finite number of Q, two such elements are conjugate to each other in G . We denote any of them by Frp . I f G is abelian, then there is only one element Frp in the Galois group . Theorem 15 . 1 .
There exists a unique finite abelian extension K of F having the following property. If P1 , P2 are prime ideals of RF, then Frp 1 = Frp 2 if and only if there is an element a of K such that aP1 = P2 • In a similar but more complicated manner, one can characterize all abelian extensions of F. This theory is known as class field theory , developed by Kro necker, Weber, Hilbert, Takagi , and Artin . The main statement concerning the Frobenius automorphism as above is Artin ' s Reciprocity Law . ArtinTate ' s notes give a cohomological account of class field theory . My Algebraic Number Theory gives an account following Artin ' s first proof dating back to 1 927 , with later simplifications by Artin himself. Both techniques are valuable to know . Cyclotomic extensions should be viewed in the light of Theorem 1 5 . 1 . Indeed , let K = Q( �) , where � is a primitive nth root of unity . For a prime ptn , we have the Frobenius automorphism FrP , whose effect on � is Frp (�) = �P . Then FrP 1 = Frp 2 if and only if p 1 p 2 mod n . =
To encompass both Theorem 1 5 . 1 and the cyclotomic case in one framework , one has to formulate the result of class field theory for generalized ideal classes , not just the ordinary ones when two ideals are equivalent if and only if they differ multiplicatively by a nonzero field element. See my Algebraic Number Theory for a description of these generalized ideal classes . The nonabelian case is much more difficult . I shall indicate briefly a special case which gives some of the flavor of what goes on . The problem is to do for nonabelian extensions what Art in did for abelian extensions . Artin went as far as saying that the problem was not to give proofs but to formulate what was to be proved . The insight of Langlands and others in the sixties shows that actually Artin was mistaken . The problem lies in both . Shimura made several computations in this direction involving "modular forms" [Sh 66] . Langlands gave a number of conjectures relating Galois groups with "automorphic forms" , which showed that the answer lay in deeper theories, whose formulations , let alone their proofs , were difficult. Great progress was made in the seventies by Serre and Deligne , who proved a first case of Langland ' s conjecture [DeS 74] .
TH E MODULAR CON N ECTION
VI, § 1 5
31 7
The study of nonabelian Galois groups occurs via their linear "representa tions" . For instance , let I be a prime number. We can ask whether GL n (F1) , or GL 2 (F1) , or PGL 2 (F1) occurs as a Galois group over Q , and "h o w . The problem is to find natural objects on which the Galois group operates as a linear map , such that we get in a natural way an isomo rphism of this Galois group with one of the above linear groups . The theories which indicate in which direction to find such objects are much beyond the level of this course, and lie in the theory of modular functions , involving both analysis and algebra , which form a back ground for the number theoretic applications . Again I pick a special case to give the flavor. Let K be a finite Galois extension of Q, with Galois group "
G
=
Gal(K/Q) .
Let p : G � GL 2 (F1) be a homomorphism of G into the group of 2 x 2 matrices over the finite field F1 for some prime /. Such a homomorphism is called a representation of G . From elementary linear algebra, if M= is a 2
x
(: �)
2 matrix , we have its trace and determinant defined by tr(M) = a + d and det M = ad  be .
Thus we can take the trace and determinant tr p( u) and det p( u) for u E G . Consider the infinite product with a variable q: oc
ll(q)
=
q nn ( 1 =l

q n) 2 4
oc
=
L a nq n . n= l
The coefficients a n are integers , and a 1 = 1 . Theorem 15.2. For each prime I there exists a unique Galois extension K of Q, with Galois group G, and an injective homomorphism
p : G � GL 2 (F1)
having the following property. For all but a finite number of primes p, if aP is the coefficient of q P in ll(q), then we have tr p(Frp ) aP mod I and det p(Frp ) p 1 1 mod l. Furthermore, for all primes I =I= 2 , 3 , 5, 7, 23, 691 , the image p(G) in GL2 (F1) consists of those matrices M E GL 2(F1) such that det M is an eleventh power in Ff. =
=
31 8
GALOIS TH EORY
VI, § 1 5
The above theorem was conjectured by Serre in 1 968 [Se 68 ] . A proof of the existence as in the first statement was given by Deligne [De 68] . The second statement , describing how big the Galois group actually is in the group of matrices GL 2 (F1) is due to Serre and SwinnertonDyer [Se 72] , [SwD 73 ] . The point of ll ( q ) is that if we put q = e 2 7Tiz , where z is a variable in the upper halfplane, then ll is a modular form of weight 1 2 . For definitions and an introduction, see the last chapter of [Se 73] , [La 73] , [La 76] , and the followin g comments . The general result behind Theorem 1 5 . 2 for modular forms of weight > 2 was given by Deligne [De 73] . For weight 1 , it is due to DeligneSerre [DeS 7 4] . We summarize the situation as follows . Let N be a positive integer. To N we associate the subgroups f(N) C f 1 (N) C f0(N) of SL 2 (Z) defined by the conditions for a matrix a =
(: !)
E
SL 2 (Z):
d
a E f (N) if and only if a = = 1 mod N and b = c = 0 mod N; a E f 1 (N) if and only if a = d = 1 mod N and c = 0 mod N; a E f0(N) if and only if c = 0 mod N. Let / be a function on the upper halfplane Sj = {z E C , l m ( z ) > 0 } . Let k be an integer. For
'Y
=
(: !)
E
SL 2(R) ,
define f o [ 'Y] k (an operation on the right) by f o [ 'Y] k (z) = ( cz
+
d)  '1< yz) where yz
=
az cz
+
b
+ d.
Let r be a subgroup of SL 2 (Z) containing f (N) . We define f to be modular of weight k on r if: Mk 1 . f is holomorphic on Sj; Mk 2. f is holomorphic at the cusps , meaning that for all a E SL 2 (Z) , the function f o [a ] k has a power series expansion 00
Mk 3. We have f o [ 'Y] k = J for all 'Y E r . One says that f i s cuspidal i f in Mk 2 the power series has a zero; that is , the power starts with n > 1 .
T H E MODULAR CON N ECTION
VI, § 1 5
31 9
Suppose that f is modular of weight k on f (N) . Then f is modular on f 1 (N ) if and only if f(z + 1 ) = f(z) , or equivalently f has an expansion of the form
This power series is called the qexpansion of f. suppose f has weight k on r I (N) . If 'Y E ro(N) and 'Y is the above written matrix , then f [ 'Y] k depends only on the image of d in (Z/NZ ) * , and we then denote f [ 'Y] k by f [ d ] k . Let o
o
o
e:
(Z/NZ) * � C *
be a homomorphism (also called a Dirichlet character) . One says that e is odd if e (  1 ) =  1 , and even if e (  1 ) = 1 . One says that f is modular of type (k , E) on fo (N) if f has weight k on r l (N) , and o
f [d ]k
= e (d)f for all
d E (Z/NZ)* .
It is possible to define an algebra of operators on the space of modular forms of given type . This requires more extensive background , and I refer the reader to [La 76] for a systematic exposition . Among all such forms , it is then possible to distinguish some of them which are eigenvectors for this Heeke algebra , or, as one says , eigenfunctions for this algebra. One may then state the Deligne Serre theorem as follows.
Let f =I= 0 be a modular form of type ( 1 ' E) on ro(N) ' so f has weight 1 . Assume that e is odd. Assume that f is an eigenfunction of the Heeke algebra, with q expansion fx = L anq n , normalized so that a 1 = 1 . Then there exists a unique finite Galois extension K of Q with Galois group G, and a representation p : G � GL 2 (C ) (actually an injective homomorphism) , such that for all primes p % N the characteristic polynomial of p ( Frp ) is X2  ap x + e (p) . The representation
p
is irreducible if and only iff is cuspidal.
Note that the representation p has values in GL 2 (C ) . For extensive work of Serre and his conjectures concerning representations of Galois groups in GL 2 (F) when F is a finite field , see [Se 87] . Roughly speaking, the general philosophy started by a conjecture of TaniyamaShimura and the Langlands conjectures is that everything in sight is "modular" . Theorem 1 5 . 2 and the DeligneSerre theorem are prototypes of results in this direction . For "modular" representations in GL 2 (F) , when F is a finite field , Serre ' s conjectures have been proved, mostly by Ribet [Ri 90] . As a result, following an idea of Frey , Ribet also showed how the TaniyamaShimura conjecture implies Fermat ' s last theorem [Ri 90b] . Note that Serre ' s conjectures that certain representations in GL 2 (F) are modular imply the TaniyamaShimura conjecture .
320
GALOIS TH EORY
VI, Ex
Bibliography [ArT 68] [De 68] [De 73] [DeS 74] [La 70] [La 73] [La 76] [Ri 90a] [Ri 90b] [Se 68] [Se 72] [Se 73] [Se 87] [Shi 66] [Shi 7 1 ] [SwD 73]
E. ARTIN and J. TATE , Class Field Theory, BenjaminAddisonWesley , 1 968 (reprinted by AddisonWesley , 1 99 1 ) P. DELIGNE , Formes modulaires et representations /adiques , Seminaire Bourbaki 1 968 1 969 , exp . No . 355 P. DELIGNE , Formes modulaires et representations de GL(2) , Springer Lecture Notes 349 ( 1 973) , pp . 55 1 05 P. DELIGNE and J. P. SERRE , Formes modulaires de poids 1 , Ann . Sci. ENS 7 ( 1 974) , pp . 507530 S . LANG , Algebraic Number Theory, Springer Verlag , reprinted from AddisonWesley ( 1 970) S. LANG , Elliptic functions , Springer Verlag , 1 973 S. LANG , Introduction to modular forms , Springer Verlag , 1 976 K . RIBET , On modular representations of Gal(Q / Q) arising from modular forms , Invent. Math. 100 ( 1 990) , pp . 43 1476 K. RIBET , From the TaniyamaShimura conjecture to Fermat' s last theorem , Annales de Ia Fac. des Sci. Toulouse ( 1 990) , pp . 1 1 6 1 39 J . P. SERRE , Une interpretation des congruences relatives a Ia fonction de Ramanujan , Seminaire DelangePisotPoitou , 1 967 1 968 J . P. SERRE , Congruences et formes modulaires (d' apres SwinnertonDyer) , Seminaire Bourbaki , 1 97 1 1 972 J . P. SERRE , A course in arithmetic , Springer Verlag , 1 973 J . P. SERRE , Sur les representations modulaires de degre 2 de Gal(Q/Q) , Duke Math . J. 54 ( 1 987) , pp . 1 79230 G. S HIMURA , A reciprocity law in nonsolvable extensions , J. reine angew. Math . 22 1 ( 1 966) , pp . 209220 G . SHIMURA , Introduction to the arithmetic theory of automorphic functions , Iwanami Shoten and Princeton University Press , 1 97 1 H. P. SwiNNERTONDYER , On /adic representations and congruences for coefficients of modular forms , (Antwerp conference) Springer Lecture Notes 350 ( 1 973)
EX E R C I S ES 1 . What is the Galois group of the following polynomials? (a) X3  X  1 over Q. (b) X3  1 0 over Q. (c) X3  1 0 over Q(j2). (d ) X 3  10 over Q(j=3). (e) X 3  X  1 over Q(J=2) ). (f) X4  5 over Q, Q(J5), Q(j=5), Q(i). (g) X4  a where a 1s any integer # 0, # + 1 and 1s square free. Over
Q.
V I , §Ex
EXERC ISES
(h) (i) (j) (k)
321
X 3  a where a is any squarefree integer > 2. Over Q. X4 + 2 over Q, Q(i). ( X 2  2 )( X 2  3)( X 2  5)( X 2  7) over Q.
Let p 1 , , Pn be distinct prime numbers. What is the Galois group of (X 2  p 1 ) ( X2  Pn ) over Q ? (I) ( X 3  2 )( X3  3)( X 2  2 ) over Q( j  3). (m) X "  t , where t is transcendental over the complex numbers C and n is a positive integer. Over C( t). (n) X4  t, where t is as before. Over R(t). •
•
•
•
•
•
2 . Find the Galois groups over Q of the following polynomials. (a) X 3 + x + 1 (b) X3  x + 1 (g) x3 + x2  2X  1 (c) X3 + 2X + 1 (d) X3  2x + 1 (e) X 3  x  1 (f) X 3  1 2X + 8 3 . Let k = C( t) be the field of rational functions in one variable. Find the Galois group over k of the following polynomials : (a) X 3 + x + t (b) X3  x + t (c) X 3 + t X + 1 (d) X 3  2t X + t (e) X3  x  t (f) X3 + r 2 x  t 3
4 . Let k be a field of characteristic =I= 2 . Let c E k , c ft. k2 • Let F = k(Vc ) . Let a = a + b Vc with a , b E k and not both a , b = 0 . Let E = F (Ya) . Prove that the following conditions are equivalent . ( 1 ) E is Galois over k. (2) E = F( W) , where a' = a  b Vc . (3) Either aa' = a 2  cb2 E k2 or caa' E k2 . Show that when these conditions are satisfied , then E is cyclic over k of degree 4 if and only if caa' E k2 •
5 . Let k be a field of characteristic =I= 2 , 3 . Let f(X ) , g(X) = X 2  c be irreducible polynomials over k, of degree 3 and 2 respectively . Let D be the discriminant of f. Assume that Let a be a root of f and {3 a root of g in an algebraic closure . Prove : (a) The splitting field of fg over k has degree 1 2 . (b) Let y = a + {3 . Then [k( y) : k] = 6 . 6 . (a) Let K Let E exists zE E
be cycl ic over k of degree 4 , and of characteristic =I= 2 . Let GKtk = (u) . be the unique subfield of K of degree 2 over k . Since [ K : £] = 2 , there a E K such that a2 = y E E and K = E(a) . Prove that there exists such that zuz =  1 , ua = z a , z2 = uy/ y.
(b) Conversely , let E be a quadratic extension of k and let GEtk = ( T) . Let z E E be an element such that zn =  1 . Prove that there exists y E E such that z2 = Ty/ y. Then E = k( y) . Let a2 = y, and let K = k ( a) . Show that K is Galois , cyclic of degree 4 over k. Let u be an extension of T to K. Show that u is an automorphism of K which generates GK1k , satisfying u2 a =  a and ua = + za. Replacing z by  z originally if necessary , one can then have ua = za.
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7 . (a) Let K = Q( Va ) where a E Z , a < 0. Show that K cannot be embedded in a cyclic extension whose degree over Q is divisible by 4 . (b) Let f(X) = X4 + 30X2 + 45 . Let a be a root of F . Prove that Q(a) is cyclic of degree 4 over Q . (c) Let f(X) = X4 + 4x 2 + 2 . Prove that f i s irreducible over Q and that the Galois group is cyclic . 4 2 8 . Let f(X) = X + aX + b be an irreducible polynomial over Q, with roots + ex, + {J, and splitting field K. (a) Show that Gal(K/Q) is isomorphic to a subgroup of D8 (the nonabelian group of order 8 ot her than the quaternion group), and thus is isomorphic to one of the following : {i) Z/4Z (ii) Z/2Z x Z/2Z (iii) D8 . (b) Show that the first case happens if and only if a {3  E Q. a {3 Case (ii) happens if and only if a{3 E Q or a2  {32 E Q . Case (iii) happens otherwise . (Actually, in (ii) , the case a2  {32 E Q cannot occur. lt corresponds to a subgroup D8 C S4 which is isomorphic to Z/2Z x Z/2Z , but is not transitive on { 1 , 2 , 3 , 4 }) . (c) Find the splitting field K in C of the polynomial X 4  4X 2  1 . Determine the Galois group of this splitting field over Q , and describe fully the lattices of subfields and of subgroups of the Galois group. 9. Let K be a finite separable extension of a field k, of prime degree p. Let 0 E K be such that K = k(8), and let (} b . . . , ()P be the conjugates of (} over k in some algebraic closure. Let (} = 0 1 • If 8 E k( 0), show that K is Galois and in fact cyclic over k. 2 1 0. Let f(X) e Q[X] be a polynomial of degree n, and let K be a splitting field off over Q. Suppose that Gal(K/Q) is the symmetric group Sn with n > 2 . (a) Show that f is irreducible over Q. (b) If ex is a root ofJ, show that the only automorphism of Q(ex) is t he identity. (c) If n > 4, show that ex" ¢ Q.
A polynomial f(X) is said to be reciprocal if whenever r:x is a root� then I I� IS also a root. We suppose that f has coefficients in a real subfield k of the complex numbers. If j is irreducible over k, and has a nonreal root of absolute value 1 , show that j t s reciprocal of even degree. 1 2. What is the Galois group over the rationals of X 5  4X + 2 ?
I I.
1 3. What is the Galois group over the rationals of the following polynomials : (a) X 4 + 2X 2 + X + 3 (b) X 4 + 3X 3  3X  2 (c) X 6 + 22X 5  9X 4 + 1 2X 3  37X 2  29X  1 5 [Hint : Reduce mod 2, 3, 5.] 14. Prove that given a symmet ric group Sn , there extsts a polynomial f(X) E Z [ X ] with leading coefficient 1 whose Galois group over Q is Sn . [Hint : Red ucing mod 2, 3, 5, show that there exists a polynomial whose reductions are such that the Galois group
V I , §Ex
EXERC ISES
323
contains enough cycles to generate S, . U se the Chtnese remainder theorem, also to be able to apply Eisenstein's criterion.] 1 5 . Let K/k be a Galois extension , and let F be an intermediate field between k and K . Let H be the subgroup of Gal(K/k) mapping F into itself. S how that H is the normal izer of Gal(K/F) in Gal(K/k) . 1 6 . Let K/k be a finite Galois extension with group G . Let a E K be such that {aa} uE G is a normal basis . For each subset S of G let S( a) = 2: uE s aa. Let H be a subgroup of G and let F be the fixed field of H. Show that there exists a basis of F over k consisting of elements of the form S( a) .
Cyclotomic fields 1 7 . (a) Let k be a field of characteristi c f2n , for some odd integer n > 1 , and let ' be a primitive nth root of unity , in k. S how that k also contain s a primitive 2nth root of unity . (b) Let k be a finite extension of the rationals . Show that there is only a finite number of roots of unity in k. 1 8 . (a) Determine which roots of unity lie in the following fields : Q (i ) , Q( V2) ,
Q( V2) , Q( V3) , Q(V3) , Q( v=5) .
(b) For which integers
m
does a primitive mth root of unity have degree 2 over Q?
1 9 . Let ( be a primitive nth root of unity . Let K = Q( () . (a) If n = p r (r > 1 ) is a prime power , show that NK;Q( l  ( ) = p . (b) If n is composite (divisible by at least two primes) then NK1Q( l  ' ) = 1 .
20. Let f(X) E Z[X ] be a nonconstant polynomial with integer coefficients . S how that the values f(a) with a E z+ are divisible by infinitely many primes .
[Note : This is trivial . A much deeper question is whether there are infinitely many a such that f(a) is prime . There are three necessary conditions: The leading coefficient of f is positive . The polynomial is irreducible. The set of values f(Z+) has no common divisor > 1 . A conjecture of Bouniakowski [Bo 1 854] states that these conditions are sufficient. The conjecture was rediscovered later and generalized to several polynomials by Schinzel [ Sch 58] . A special case is the conjecture that X 2 + 1 represents infinitely many primes . For a discussion of the general conj ecture and a quantitative version giving a conjectured asymptotic estimate , see B ateman and Horn [BaH 62] . Also see the comments in [HaR 74] . More precisely , let /. , . . . , fr be polynomials with integer coefficients satisfying the first two conditions (positive leading coefficient, irre ducible) . Let f = /1 · · · fr be their product , and assume that f satisfies the third condition . Define :
1r(f) (x) = number of positive integers n < x such that f1 (n) , . . . fr( n) are all primes. ,
(We ignore the finite number of values of n for which some /;(n) is negative . ) The
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BatemanHom conjecture is that X
1T 1 define •
•
•
x = px <  1 >
xCn> = (Fx)
and
+ p"xn .
Also from the definition, we have
x< n ) = x�" + px� n  l + . . . + p n.x
n·
4 8 . Let k be a field of characteristic p, and consider W(k ). Then V is an additive endomorph ism of W(k), and F is a ring homomorphism of W(k) into itself. Furthermore, if x e W(k ) then px = VFx.
If x , y e W(k ), then ( V ix)( Viy) = y i + i(F Pix FPiy ) . For a e k denote by {a} the Witt vector (a, 0, 0, . . . ). Then we can write symbolically ·
00
i
x = L V {x;}. i=O
Show that if x e W(k) and x0 # 0 then x is a unit in W(k). Hint : One has
and then 00
x { x o 1 } L ( Vy ) i = ( 1 0

Vy )
00
L ( Vy ) i = 1 . 0
49 . Let n be an integer > 1 and p a prime number again. Let k be a field of characteristic p. Let W,(k) be the ring of truncated Witt vectors (x0 , , xn  1 ) with components in k. We view W,(k) as an additive group. If x e W,(k ) define p(x) = Fx  x. Then fcJ is a homomorphism . If K is a Galois extension of k, and u E G(K/k) , and x E W (K ) we can define ux to have component (ux0 , ux n _ 1 ). Prove the analogue of Hilbert's Theorem 90 for Witt vectors, and prove that the first cohomology group is trivial. (One takes a vector whose trace is not 0, and finds a co boundary the same way as in the proof of Theorem 1 0. 1). .
•
•
,
,
•
•
n
•
50. If x e W,(k), show that there exists � e W,(k) such that p( � ) = x. Do this inductively , , cxn  1 ) is solving first for the first component, and then showing that a vector (0, cx 1 , in the image of fcJ if and only if (cx . , . . . , cxn  1 ) is in the image of fcJ. Prove inductively that if � ' �' E W,(k') for some extension k' of k and if fcJ� = p � then �  �' is a vector with components in the prime field. Hence the solutions of fcJ� = x for given x e W, (k ) all differ by the vectors with components in the prime field, and there are p" such vectors. We define •
'
•
•
332
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GALOIS TH EORY
or symbolically, Prove that it is a Galois extension of k, and show that the cyclic extensions of k, of degree p" , are precisely those of type k(p  1 x) with a vector x such that x0 ¢ pk. 5 1 . Develop the Kummer theory for abelian extensions of k of exponent p" by using W,(k). In other words, show that there is a bijection between subgroups B of W,(k) containing fJ W,(k) and abelian extensions as above, given by
B � K8 where K8 = k(p 1 B) . All of this is due to Witt , cf. the references at the end of § 8 , especially [Wi 37] . The proofs are the same , mutatis mutandis , as those given for the Kummer theory in the text .
Further Progress and directions Major progress was made in the 90s concerning some problems mentioned in the chapter. Foremost was Wiles ' s proof of enough of the ShimuraTaniyama conjecture to imply Fermat's Last Theorem [Wil 95], [TaW 95]. [TaW 95] R. TAYLOR and A. WILES, Ringtheoretic properties or certain Heeke alge bras, Annals of Math. 141 ( 1 995) pp. 553 572 [Wil 95]
A. WILES, Modular elliptic curves and Fermat's last theorem, A nnals. of Math. 141 ( 1 995) pp. 44355 1
Then a proof of the complete ShimuraTaniyama conjecture was given in [BrCDT 0 1 ]. [BrCDT 0 1 ] C. BREUIL, B . CONRAD, F. DIAMOND, R. TAYLOR, On the modularity of el liptic curves over Q: Wild 3adic exercises, J. Amer. Math. Soc. 1 4 (200 1 ) pp. 843839 In a quite different direction, Neukirch started the characterization of n um ber fields by their absolute Galois groups [Ne 68], [Ne 69a], [Ne 69b], and proved it for Galois extensions of Q. His results were extended and his subsequent conjectures were proved by Ikeda and Uchida [Ik 77], [Uch 77], [Uch 79], [Uch 8 1 ]. These results were extended to finitely generated extensions of Q (function fields) by Pop [Pop 94], who has a more extensive bibliography on these and related questions of algebraic geometry. For these references, see the bibliography at the end of the book.
C H A PT E R
VI I
E xte nsio ns of Rings
It is not always desirable to deal only with field extensions . Sometimes one wants to obtain a field extension by reducing a ring extension modulo a prime ideal . This procedure occurs in several contexts , and so we are led to give the basic theory of Galois automorphisms over rings , looking especially at how the Galois automorphisms operate on prime ideals or the residue class fields . The two examples given after Theorem 2 . 9 show the importance of working over rings, to get families of extensions in two very different contexts . Throughout this chapter, A , B, C will denote commutative rings.
§1 .
I NT E G R A L R I N G EXT E N S I O N S
In Chapters V and VI we have studied algebraic extensions of fields . For a number of reasons , it is desirable to study algebraic extensions of rings . For instance , given a polynomial with integer coefficients say X 5  X  1, one can reduce this polynomial mod p for any prime p, and thus get a poly nomial with coefficients in a finite field. As another example, consider the polynomial X n + Sn  1 xn  1 + · · · + S o ,
where s n 1 , , s0 are algebraically independent over a field k. This poly nomial has coefficients in k[s0 , . . . , sn  1 ] and by substituting elements of k for s 0 , . . . , sn  t one obtains a polynomial with coefficients in k. One can then get _
•
.
•
333
334
EXTENSION O F R I N G S
VII, § 1
information about polynomials by taking a homomorphism of the ring in which they have their coefficients. This chapter is devoted to a brief descriptio n of the basic facts concerning polynomials over rings. Let M be an Amodule . We say that M is faithful if, whenever a E A is such that aM = 0, then a = 0 . We note that A is a faithful module over itself since A contains a unit element . Furthermore , if A =I= 0 , then a faithful module over A cannot be the 0module . Let A be a subring of B . Let a E B . The following conditions are equivalent: INT 1 . The element a is a root of a polynomial xn + an  l x n  1 + . . . + a o with coefficients a i E A, and degree n > 1 . (The essential thing here is that the leading coefficient is equal to 1 . ) INT 2. The subring A [a] is a finitely generated Amodule. INT 3. There exists a faithful module over A[a] which is a finitely gener ated Amodule. We prove the equivalence. Assume INT 1 . Let g(X) be a polynomial In A [ X] of degree > 1 with leading coefficient 1 such that g(a) = 0. If f(X) E A [ X] then
f(X) = q(X)g(X) + r(X) with q , r E A [X ] and deg r < deg g. Hence f(a) = r(a), and we see that if deg g = n, then 1 , a , . . . , an  1 are generators of A [a] as a modu le over A. An equation g(X) = 0 with g as above, such that g(a) = 0 is called an integral equation for a over A. Assume INT 2. We let the module be A[ a] itself. Assume INT 3, and let M be the faithful module over A[a] which is finitely generated over A, say by elements w , wn . Since aM c M there exist ele ments a ii E A such that 1,
•
•
•
Transposing aw b . . . , aw n to the righthand side of these equations, we con clude that the determinant
a
d=
a
.
.
l)
.
.
l)
VI I , §1
I NTEG RAL RING EXTENSIONS
335
is such that dM = 0. (This will be proved in the chapter when we deal with determinants.) Since M is faithful, we must have d = 0. Hence rx is a root of the polynomial det(X b;i  au), which gives an integral equation for rx over A. An element rx satisfying the three conditions INT 1, 2, 3 is called integral over A. Proposition 1 . 1 .
Let A be an entire ring and K its quotient field. l£t rx be algebraic over K. Then there exists an element c =F 0 in A such that crx is integral over A. Proof. There exists an equation an rxn + an  1 rxn  1 + . . . + a o = 0 with a; E A and an "# 0. Multiply it by a�  1 • Then (a n rx )n + · · · + a0 a:  1 = 0 '
is an integral equation for an rx over A. This proves the proposition . Let A C B be subrings of a commutative ring C, and let a E C. If a is integral over A then a is a fortiori integral over B . Thus integrality is preserved under lifting . In particular, a is integral over any ring which is intermediate between A and B . Let B contain A as a subring . We shall say that B is integral over A if every element of B is integral over A . Pro position 1 .2.
If B is integral over A and finitely generated as an Aalgebra, then B is finitely generated as an Amodule.
Proof. We may prove this by induction on the number of ring generators, and thus we may assume that B = A[rx] for some element rx integral over A, by considering a tower But we have already seen that our assertion is true in that case, this being part of the definition of integrality. Just as we did for extension fields, one may define a class e of extension rings A c B to be distinguished if it satisfies the analogous properties, namely : (1)
Let A c B c C be a tower of rings. The extension A c C is in e if and only if A c B is in e and B c C is in e . (2) If A c B is in e , if C is any extension ring of A , and if B, C are both subrings of some ring, then C c B[C] is in e . (We note that B[C] = C[B] is the smallest ring containing both B and C.)
336
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VII, §1
As with fields, we find formally as a consequence of ( 1 ) and (2) that (3) holds, namely : (3) If A c B and A c C are in e , and B, C are subrings of some ring, then A c B[C] is in e . Proposition 1 . 3.
Integral ring extensions form a distinguished class. Proof. Let A C B C C be a tower of rings . lf C is integral over A , then it is clear that B is integral over A and C is inte g ral over B. Conversely, assume that each step in the tower is integral. Let a E C. Then a satisfies an integral eq uation an + bn 1 an  1 + . . . + b 0 = 0 
,
with b i E B. Let B 1 = A [b 0 , bn  1 ] . Then B 1 is a finitely generated A module by Proposition 1 .2, and is obviously faithful. Then B 1 [a] is finite over B b hence over A, and hence a is integral over A . Hence C is integral over A. F inally let B, C be extension rings of A and assume B integral over A. Assume that B, C are subrings of some ring. Then C[B] is generated by elements of B over C, and each element of B is integral over C. That C[B] is integral over C will follow immediately from our next proposition. .
•
.
Proposition 1 .4. Let A be a subring of C. Then the elements of C which are integral over A form a sub ring of C.
Proof. Let a, P E C be integral over A. Let M = A [a] and N = A [fJ]. Then M N contains 1, and is therefore faithful as an A module. Furthermore, aM c M and PN c N. Hence M N is mapped into itself by multiplication with a + P and a{J. F u rthermore M N is finitely generated over A (if { w i } are generators of M and {vi } are generators of N then { w; vi } are generators of M N). This proves our proposition. In Proposition 1 .4, the set of elements of C which are integral over A is called the integral closure of A in C Example. Consider the integers Z . Let K be a finite extension of Q. We call K a number field . The integral closure of Z in K is called the ring of algebraic integers of K . This is the most classical example .
In algebraic geometry , one considers a finitely generated entire ring R over Z or over a field k . Let F be the quotient field of R . One then considers the integral closure of R in F, which is proved to be finite over R . If K is a finite extension of F, one also considers the integral closure of R in K. Proposition
Let A c B b e an extension ring, and let B be integral over A. Let a b e a homomorphism of B. Then a(B) is integral over a ( A). Proof. Let a E B, and let an + a n  an  1 + . . . + a o = 0 1 .5.
1
V I I , §1
I NTEG RAL R I NG EXTENSIONS
be an integral equation for rx over A. Applying a yields a(rx)" + a(an _ 1 )a(rx)"  1 + · · · + a(a 0 ) there by proving our assertion.
=
337
0,
Corollary 1 .6. Let A be an entire ring, k its quotient field, and E a finite extension of k. Let rx E E be integral over A. Then the norm and trace of rx
(from E to k) are integral over A, and so are the coefficients of the irreducible polynomial satisfied by rx over k. Proof. For each embedding a of E over k, arx is integral over A. Since the norm is the product of arx over all such a (raised to a power of the characteristic), it follows that the norm is integral over A . Similarly for the trace, and similarly for the coefficients of Irr(rx, k, X), which are elementary symmetric functions of the roots. Let A be an entire ring and k its quotient field. We say that A is integrally closed if it is equal to its integral closure in k. Proposition 1 .7.
Let A be entire and factorial. Then A is integrally closed. Proof. Suppose that there exists a quotient ajb with a, b E A which is integral over A, and a prime element p in A which divides b but not a. We have, for some integer n > 1, and ai E A, (a/b)" + an _ 1 (a/b)"  1 + · · · + a 0 0 whence a " + an _ 1 ba"  1 + · · · + a0 b" 0. Since p divides b, it must divide a" , and hence must divide a, contradiction. =
=
Let f : A __.. B be a ringhomomorphism (A , B being commutative rings). We recall that such a homomorphism is also called an Aalgebra. We may view B as an Amodule. We say that B is integral over A (for this ringhomo morphism f) if B is integral over f(A). This extension of our definition of integrality is useful because there are applications when certain collapsings take place, and we still wish to speak of integrality. Strictly speaking we should not say that B is integral over A , but that f is an integral ringhomomorphism , or simply that f is integral. We shall use this terminology frequently. Some of our preceding propositions have immediate consequences for integral ringhomomorphisms ; for instance, if f : A __.. B and g : B __.. C are integral, then g o f : A __.. C is integral. However, it is not necessarily true that if g o f is integral, so is f. Let f : A __.. B be integral, and let S be a multiplicative subset of A. Then we get a homomorphism s  1 / : s 1 A __.. s  1 B, where strictly speaking, s  1 B = (f(S))  1 B, and s  1 / is defined by (S  1/)(x/s) = f(x)/f(s). 
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It is trivially verified that this is a homomorphism. We have a commutative diagram
the horizontal maps being the canonical ones : x
__..
xjl.
Proposition 1 .8.
Let f : A __.. B be integral, and let S be a multiplicative subset of· A. Then s  1j' : s  1 A __.. s  1 B is integral. Proof. If rx E B is integral over f ( A), then writing rxp instead of f(a)(J, for a E A and P E B we have rxn + an  1 rxn  1 + . . . + a o = 0 with a i E A. Taking the canonical image in s  1 A and s  1 B respectively, we see that this relation proves the integrality of rx/1 over S  1 A, the coefficients being now aJl . Proposition 1 .9.
Let A be entire and integrally closed. Let S be a multipli cative subset of A, 0 ¢ S. Then s  1 A is integrally closed. Proof. Let rx be an element of the quotient field, integral over S  1 A. We have an equation o rxn + an  1 rxn  1 + · · · + a = 0 Sn 

So
1
'
a; E A and s i E S. Let s be the product sn  1 • • s 0 • Then it is clear that srx is integral over A, whence in A. Hence rx lies in s  1 A, and s  1 A is integrally closed. •
Let p be a prime ideal of a ring A and let S be the complement of p in A. We write S = A p . Iff : A __.. B is an Aalgebra (i.e. a ringhomomorphism), we shall write B" instead of s  1 B. We can view B" as an A " = s  1 Amodule. Let A be a subring of B. Let p be a prime ideal of A and let � be a prime ideal of B. We say that � lies above p if � n A = p. If that is the case, then the injection A __.. B induces an injection of the factor rings 
I
A/p
__..
I
B/�,
and in fact we have a commutative diagram : B
B/�
A
A/p
INTEGRAL RING EXTENSIONS
VII, §1
339
the horizontal arrows being the canonical homomorphisms, and the vertical arrows being injections. If B is integral over A, then B/� is integral over Ajp by Proposition 1 .5.
Let A be a subring of B, let p be a prime ideal of A, and assume B integral over A. Then pB =F B and there exists a prime ideal � of B lying above p. Proposition 1 . 10.
Proof. We know that B., is integral over A., and that A., is a local ring with maximal ideal m ., = s  1 p, where S = A p. Since we obviously have 
pB.,
pA., B.,
=
=
m.,
B., ,
it will suffice to prove our first assertion when A is a local ring. (Note that the existence of a prime ideal p implies that 1 =F 0, and pB = B if and only if 1 E pB.) In that case, if pB = B, then 1 has an expression as a finite linear combination of elements of B with coefficients in p, 1
=
a b + . . . + an b n 1
1
with a; E p and b; E B. We shall now use notation as if A., c B., . We leave it to the reader as an exercise to verify that our arguments are valid when we deal only with a canonical homomorphism A., __.. B., . Let B 0 = A [ b h . . . , bn] Then pB 0 = B0 and B 0 is a finite Amodule by Proposition 1 .2. Hence B 0 = 0 by Nakayama's lemma, contradiction. (See Lemma 4. 1 of Chapter X . ) To prove our second assertion, note the following commutative diagram : .
We have j ust proved m ., B., =F B., . Hence m ., B., is contained in a maximal ideal 9Jl of B., . Taking inverse images, we see that the inverse image of 9Jl in A., is an ideal containing m ., (in the case of an inclusion A., c B., the inverse image is 9Jl n A.,). Since m., is maximal, we have 9J1 n A., = m ., . Let � be the inverse image of 9J1 in B (in the case of inclusion, � = 9Jl n B). Then � is a prime ideal of B. The inverse image of m ., in A is simply p. Taking the inverse image of 9Jl going around both ways in the diagram, we find that
�
n
A = p,
as was to be shown. Proposition 1 . 1 1 .
Let A be a subring of B, and assume that B is integral over A. Let � be a prime ideal of B lying over a prime ideal p of A. Then � is maximal if and only if p is maximal.
340
VII, §2
EXTENSION O F R I NGS
Proof. Assume p maximal in A. Then A/p is a field, and B/� is an entire ring, integral over Ajp. If rx E Bj�, then rx is algebraic over A/p, and we know that A/p[rx] is a field. Hence every nonzero element of B/ � is invertible in B/� , which is therefore a field. Conversely, assume that � is maximal in B. Then B/� is a field, which is integral over the entire ring Ajp. If A/p is not a field, it has a nonzero maximal ideal m. By Proposition 1 . 10, there exists a prime ideal 9Jl of B/� lying above m, 9Jl =F 0, contradiction.
§2.
I NTEG R A L G A LO I S EXT E N S I O N S
We shall now investigate the relationship between the Galois theory of a polynomial, and the Galois theory of this same polynomial reduced modulo a prime ideal. Proposition 2. 1.
Let A be an entire ring, integrally closed in its quotient field K. Let L be a finite Galois extension of K with group G. Let p be a maximal ideal of A , and let �' .Q be prime ideals of the integral closure B of A in L lying above p. Then there exists a E G such that a� .0. Proof. Suppose that .Q =F a� for any a E G. Then t.O =F a� for any pair of elements a, t E G. There exists an element x E B such that x = 0 (mod a�), all a E G all a E G x = 1 (mod a.O), =
(use the Chinese remainder theorem). The norm Ni(x) = n ax ti E
G
lies in B n K = A (because A is integrally closed), and lies in � n A = p. But x ¢ a.O for all a E G, so that ax ¢ .Q for all a E G. This contradicts the fact that the norm of x lies in p = .Q n A. If one localizes, one can eliminate the hypothesis that p is maximal ; just assume that p is prime. Corollary 2.2 Let A be integrally closed in its quotient field K. Let E be a finite separable extension of K, and B the integral closure of A in E. Let p be
a maximal ideal of A . Then there exists only a finite number of prime ideals of B lying above p.
Proof. Let L be the smallest Galois extension of K containing E. If .Q 1 , .0 2 are two distinct prime ideals of B lying above p, and �h �2 are two prime ideals of the integral closure of A in L lying above .0 1 and .0 2 respectively, then � 1 =F � 2 • This argument reduces our assertion to the case that E is Galois over K, and it then becomes an immediate consequence of the proposition.
VI I , §2
INTEG RAL GALOIS EXTENSIONS
341
Let A be integrally closed in its quotient field K, and let B be its integral closure in a finite Galois extension L, with group G . Then aB = B for every a E G. Let p be a maximal ideal of A, and � a maximal ideal of B lying above p. We denote by G'll the subgroup of G consisting of those automorphisms such that a� = � Then G'll operates in a natural way on the residue class field Bj�, and leaves Ajp fixed. To each a E G'll we can associate an automorphism ii of B/� over Ajp , and the map given by (J I+ (J
induces a homomorphism of G'll into the group of automorphisms of B/� over Ajp. The group G'll will be called the decomposition group of � Its fixed field will be denoted by L d e c, and will be called the decomposition field of � Let Bd ec be the integral closure of A in Ld ec , and .Q = � n.Q Bd ec. By Proposition 2. 1, we know that � is the only prime of B lying above . Let G = U ai G'll be a coset decomposition of G� in G. Then the prime ideals ai � are precisely the distinct primes of B lying above p. Indeed, for two elements a, r E G we have a � = r� if and only if r  1 a� = �' i.e. r  1 a lies in G'll . Thus r, a lie in the same coset mod G� . It is then immediately clear that the decomposition group of a prime a� is aG'll a  1 •
The field Ld e c is the smallest subfield E of L containing K such that � is the only prime of B lying above � n E (which is prime in B n E). Proof. Let E be as above, and let H be the Galois group of L over E. Let q = � n E. By Proposition 2 . 1 , all primes of B lying above q are conjugate by elements of H . Since there is only one prime, namely � ' it means that H leaves � invariant. Hence G c G'll and E L d ec. We have already observed that Ld e c has the required property. Proposition 2.3.
::J
Proposition 2.4.
Notation being as above, we have A jp the canonical injection Ajp Bd e cj.Q ). Let
+
=
Bd ec;.a (under
Proof. If a is an element of G, not in G'll , then a� =F � and a  1 � =F � .Q
Then .Q u =F .0. Let such that
x
a 1�
u =
n
Bd ec.
be an element of Bd e c. There exists an element y of Bd e c
y
= x
(mod .Q )
y
=
(mod .Oa)
1
342
VI I , §2
EXTENS ION OF R I N G S
for each a in G, but not in G'll . Hence in particular,
y = x (mod �) y = 1 (mod a  1 �) for each a not in G'll . This second congruence yields
ay =
(mod �)
1
for all a ¢ G'll . The norm of y from L d ec to K is a product of y and other factors ay with a ¢ G'll . Thus we obtain Ldec{ N K y) = X _
(mod �).
But the norm lies in K, and even in A, since it is a product of elements integral over A. This last congruence holds mod .Q, since both x and the norm lie in Bd ec. This is precisely the meaning of the assertion in our proposition. If x is an element of B, we shall denote by x its image under the homo morphism B + B/�. Then u is the automorphism of B/� satisfying the relation
ux
= ( ax ).
If f ( X ) is a polynomial with coefficients in B, we denote by f(X) its natural image under the above homomorphism. Thus, if
then
Proposition 2.5.
Let A be integrally closed in its quotient field K, and let B be its integral closure in a finite Galois extension L of K, with group G. Let p be a maximal ideal of A, and � a maximal ideal of B lying above p. Then B/� is a normal extension of Ajp, and the map a 1+ a induces a homo morphism of G'll onto the Galois group of B/� over Ajp. Proof. Let B = B/� and A = Ajp. Any element of B can be written as x for some x E B. Let x generate a separable subextension of B over A, and let f be the irreducible polynomial for x over K. The coefficients of f lie in A because x is integral over A, and all the roots off are integral over A. Thus f(X)
m
=
0 (X
i= 1

x i)
I NTEG RAL GALOIS EXTENSIONS
VI I , §2
343
splits into linear factors in B. Since f(X)
m
=
L (X =1
i

X ;)
and all the X ; lie in B, it follows thatfsplits into linear factors in B. We observe that f(x) = 0 implies f(x) = 0. Hence B is normal over A, and
[A(x) : A] < [K(x) : K ] < [L : K ] . Thts implies that the maximal separable subextension of A in B is of finite degree over A (using the primitive element theorem of elementary field theory). This degree is in fact bounded by [L : K]. There remains to prove that the map a �+ ii gives a surjective homo morphism of G'll onto the Galois group of B over A. To do this, we shall give an argument which reduces our problem to the case when � is the only prime ideal of B lying above p . Indeed, by Proposition 2.4, the residue class fields of the ground ring and the ring Bd ec in the decomposition field are the same. This means that to prove our surjectivity, we may take Ld e c as ground field. This is the desired reduction, and we can assume K = Ld e c , G = G'll . This being the case, take a generator of the maximal separable subextension of B over A, and let it be .X, for some element x in B. Let f be the irreducible polynomial of x over K. Any automorphism of B is determined by its effect on x, and maps .X on some root of f. Suppose that x = x 1 • Given any root X; off, there exists an element a of G = G'll such that ax = X; . Hence iix = xi . Hence the automorphisms of B over A induced by elements of G operate transitively on the roots of f. Hence they give us all automorphisms of the residue class field, as was to be shown. Corollary 2.6. Let A be integrally closed in its quotient field K. Let L be a
finite Galois extension of K, and B the integral closure of A in L . Let p be a maximal ideal of A . Let cp : A � A/p be the canonical homomorphism, and let l/11 , l/12 be two homomorphisms of B extending cp in a given algebraic closure of A/p . Then there exists an automorphism u of L over K such that
t/1 1
= t/1 2
o
a.
Proof. The kernels of 1/J 1 , 1/J 2 are prime ideals of B which are conjugate by Proposition 2. 1 . Hence there exists an element r of the Galois group G such that t/1 1 , 1/J 2 o r have the same kernel. Without loss of generality, we may therefore assume that 1/J 1 , 1/1 2 have the same kernel �. Hence there exists an automorphism w of 1/J 1 (B) onto 1/J 2 ( B ) such that w o 1/J 1 = 1/1 2 • There exists an element a of G'll such that w o 1/J 1 = 1/J 1 o a, by the preceding proposition. This proves what we wanted.
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Remark.
In all the above propositions, we could assume p prime instead of maximal. In that case, one has to localize at p to be able to apply our proofs . In the above discussions, the kernel of the map
is called the inertia group of �. It consists of those automorphisms of G'll which induce the trivial automorphism on thei residue class field. Its fixed field is called the inertia field , and is denoted by L ".
Let the assumptions be as in Corollary 2.6 and assume that � is the only prime of B lying above p. Let f(X) be a polynomial in A [X] with leading coefficient 1. Assume that f is irreducible in K [X], and has a root rx in B. Then the reduced polynomial J is a power of an irreducible poly nomial in A[X]. Proof. By Corollary 2.6, we know that any two roots of 1 are conjugate under some isomorphism of B over A, and hence that1cannot split into relative prime polynomials. Therefore, 1 is a power of an irreducible polynomial. Corollary 2. 7.
Proposition 2.8. Let A be an entire ring, integrally closed in its quotient field K. Let L be a finite Galois extension of K. Let L = K ( rx), where rx is
integral over A, and let
f(X)
n = x
+
an  t x n  1
+ . . . + ao
be the irreducible polynomial of rx over k, with a i E A. Let p be a maximal ideal in A, let � be a prime ideal of the integral closure B of A in L, � lying above p. Let 1(X) be the reduced polynomial with coefficients in A/p. Let G'll be the decomposition group. If J has no multiple roots, then the map a 1+ a has trivial kernel, and is an isomorphism of G'll on the Galois group of f over A/p. Proof. Let f(X) = n ( x  X; ) be the factorization of f in L. We know that all x i E B. If a E Gt� , then we denote by u the homomorphic image of a in the group G'll , as before. We have J(x ) =
n (x
 X ;).
Suppose that ux; = X ; for all i. Since (ax;) = ux; , and since 1 has no multiple roots, it follows that a is also the identity. Hence our map is injective, the in ertia group is trivial. The field A[x 1 , , xn] is a subfield of B and any auto•
•
.
V I I , §2
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345
morphism of B over A which restricts to the identity on this subfield m ust be the identity, because the map G'll + G'll is onto the Galois group of B over A. Hence B is purely inseparable over A[x 1 , , xn ] and therefore G'll is iso morphic to the Galois group of f over A. •
.
•
Proposition 2.8 is only a special case of the moregeneral situation when the root of a polynomial does not ne cessarily generate a Galois extension. We state a version usefu l to compute Galois groups. Theorem 2.9. Let A be an entire ring, integrally closed in its quotient field K. Let f(X) E A [ X ] have leading coefficient 1 and be irreducible over K (or A, it ' s the same thing). Let p be a maximal ideal of A and let f = f mod p. Suppose that f has no multiple roots in an algebraic closure of Ajp. Let L be a splitting field for f over K, and let B be the integral closure of A in
L. Let � be any prime of B above p and let a bar denote reduction mod p. Then the map G'll G'll +
is an isomorphism of G'll with the Galois group of f over A. Proof. Let ( rx , rxn ) be the roots off in B and let reductions mod � · Since 1,
.
•
•
(a 1 ,
•
•
•
, an) be their
n
f(X)
=
0 (X
i
= 1

rx;),

&;).
it follows that f( X )
n
=
0 (X 1
i=
Any element of G is determined by its effect as a permutation of the roots, and for a E G'll , we have Hence if u = id then a = id, so the map G'll + G'll is injective. It is surjective by Proposition 2.5, so the theorem is proved. This theorem justifies the statement used to compute Galois groups in Chapter
VI , § 2 .
Theorem 2.9 gives a very efficient tool for analyzing polynomials over a ring. Example.
Consider the '' generic " polynomial
346
EXTENSION OF R I N G S
VI I , §3
, wn _ 1 are algebraically independent over a field k. We know that where w0, , wn _ 1 ) is the the Galois group of this polynomial over the field K = k(w0, symmetric group . Let t 1 , , tn be the roots . Let a be a generator of the splitting field L; that is , L = K( a) . Without loss of generality , we can select a to be integral over the ring k[ w0, . . . , wn  I ](multiply any given generator by a suitably chosen polynomial and use Proposition 1 . 1 ) . Let gw(X) be the irreducible poly nomial of a over k(w0, , wn _ 1 ) . The coefficients of g are polynomials in (w) . If we can substitute values (a ) for (w) with a0, , a n  1 E k such that 9a remains irreducible , then by Proposition 2 . 8 we conclude at once that the Galois group of 9a is the symmetric group also . Similarly , if a finite Galois extension of k(w0 , , wn _ 1 ) has Galois group G, then we can do a similar substitution to get a Galois extension of k having Galois group G , provided the special polynomial 9a remains irreducible . •
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Let K be a number field; that is , a finite extension of Q . Let o be the ring of algebraic integers . Let L be a finite Galois extension of K and () the algebraic integers in L . Let p be a prime of o and 'l' a prime of () lying above p . Then o/p is a finite field , say with q elements . Then ()j� is a finite extension of o/p , and by the theory of finite fields , there is a unique element in G'll , called the Frobenius element Fr'll , such that Fr'll(i) = _xq for i E ()/�. The conditions of Theorem 2 . 9 are satisfied for all but a finite number of primes p, and for such primes, there is a unique element Fr'll E G'll such that Fr'll(x) xq mod 'l' for all x E (). We call Fr'll the Frobenius element in G'll . Cf. Chapter VI , § 1 5 , where some of the significance of the Frobenius element is explained . Example.
=
§3 .
EXT E N S I O N O F H O M O M O R P H I S M S
When we first discussed the process of localization, we considered very briefly the extension of a homomorphism to a local ring. In our discussion of field theory, we also described an extension theorem for embeddings of one field into another. We shall now treat the extension question in full generality. First we recall the case of a local ring . Let A be a commutative rin_g and p a prime ideal. We know that the local ring A v is the set of all fractions x/y, with x, y E A and y ¢. p . Its maximal ideal consists of those fractions with x E p . Let L be a field and let cp : A � L be a homomorphism whose kernel is p . Then we can extend cp to a homomorphism of A" into L by letting
qJ(xjy)
=
qJ{x)/({J(y)
if xjy is an element of A" as above. Second, we have integral ring extensions. Let o be a local ring with maximal ideal m, let B be integral over o, and let lfJ : o __.. L be a homomorphism of o
V I I , §3
EXTENS ION OF HOMOMORPH ISMS
347
into an algebraically closed field L. We assume that the kernel of qJ is m. By Proposition 1 . 1 0, we know that there exists a maximal ideal 9J1 of B lying above m, i.e. such that 9Jl n o = m. Then B/9Jl is a field, which is an algebraic exten sion of o/m, and o/m is isomorphic to the subfield qJ(o) of L because the kernel of qJ is m. We can find an isomorphism of o/m onto qJ{o) such that the composite homomorphism
1
o + o /m + L
1
is equal to qJ. We now embed B/9Jl into L so as to make the following diagram commutative :
B o
B/9Jl ojm
�L
and in this way get a homomorphism of B into L which extends qJ. Proposition 3. 1 .
Let A be a subring of B and assume that B is integral over A. Let qJ : A + L be a homomorphism into a field L which is algebraically closed. Then qJ has an extension to a homomorphism of B into L.
Proof. Let p be the kernel of qJ and let S be the complement of p in A. Then we have a commutative diagram
and qJ can be factored through the canonical homomorphism of A into s  1 A . Furthermore, s  1 B is integral over s  1 A. This reduces the question to the case when we deal with a local ring, which has j ust been discussed above. Theorem 3.2.
Let A be a subring of a field K and let x E K, x =F 0. Let qJ : A + L be a homomorphism of A into an algebraically closed field L. Then qJ has an extension to a homomorphism of A[x] or A[x  1 ] into L.
Proof. We may first extend qJ to a homomorphism of the local ring A" , where p is the kernel of qJ. Thus without loss of generality, we may assume that A is a local ring with maximal ideal m. Suppose that
348
Vl l , §3
EXTENS ION OF R I NGS
Then we can write with a; E m. Multiplying by xn we obtain ( 1  a o )X n + bn  1 X n  1 + · · · + b o = 0 with suitable elements b i E A. Since a 0 E m, it follows that 1  a0 ¢ m and hence 1  a0 is a unit in A because A is assumed to be a local ring. Dividing by 1  a0 we see that x is integral over A, and hence that our homomorphism has an extension to A [x] by Proposition 3. 1. If on the other hand we have mA [x  1 ] =F A[x  1 ] then mA [x  1 ] is contained in some maximal ideal � of A[x  1 ] and � n A contains m. Since m is maximal, we must have � n A = m. Since qJ and the canonical map A + A/m have the same kernel, namely m, we can find an embedding ljJ of Ajm into L such that the composite map
A + Ajm � L is eq ual to qJ. We note that A/m is canonically embedded in B/� where B = A[x  1 ], and extend ljJ to a homomorphism of B/� into L, which we can do whether the image of x  1 in B/� is transcendental or algebraic over Ajm. The composite B B/� + L gives us what we want. +
Corollary 3.3.
Let A be a subring of a field K and let L be an algebraically closed field. Let qJ : A + L be a homomorphism. Let B be a maximal subring of K to which qJ ha s an extension homomorphism into L. Then B is a local ring and if x E K, x =F 0, then x E B or x  1 E B. Proof. Let S be the set of pairs (C, l/J) where C is a subring of K and l/1 : C + L is a homomorphism extending qJ. Then S is not empty (containing (A, qJ)], and is partially ordered by ascending inclusion and restriction. In other words, (C, l/J) < (C', l/J') if C c C' and the restriction of l/J' to C is equal to l/J. It is clear that S is inductively ordered, and by Zorn ' s lem m a there exists a maximal element, say (B, ljJ 0 ). Then first B is a local ring, otherwise ljJ 0 extends to the local ring arising from the kernel, and second, B has the desired property according to Theorem 3.2. Let B be a subring of a field K having the property that given x E K, x =I= 0 , then x E B or x  1 E B . Then we call B a valuation ring in K. We shall study such rings in greater detail in Chapter XII. However, we shall also give some applications in the next chapter, so we make some more comments here .
VI I , §3
EXTENSION OF HOMOMORPH ISMS
349
Let F be a field. We let the symbol oo satisfy the usual algebraic rules. If a E F, we define
a + oo
=
oo,
a · OO = OO
1 0
00 · 00 = 00 '
= 00
if a # 0, 1 and = 0. 
00
The expressions oo + oo, 0 · oo, 0/0, and oojoo are not defined. A place lfJ of a field K into a field F is a mapping
cp : K __.. {F, oo } of K into the set consisting of F and oo satisfying the usual rules for a homo morphism, namely
qJ(a + b)
=
lfJ(a) + qJ(b ),
cp( ab)
=
lfJ( a)lfJ( b)
whenever the expressions on the righthand side of these formulas are defined, and such that lfJ( 1 ) = 1 . We shall also say that the place is Fvalued. The elements of K which are not mapped into oo will be called finite under the place, and the others will be called infinite. The reader will verify at once that the set o of elements of K which are finite under a place is a valuation ring of K. The maximal ideal consists of those elements x such that qJ{x) = 0. Conversely, if o is a valuation ring of K with maximal ideal m, we let cp : o __.. o jm be the canonical homomorphism, and define qJ{x) = oo for x E K, x ¢ o. Then it is trivially verified that lfJ is a place. If qJ 1 : K __.. {F b oo } and qJ 2 : K __.. {F 2 , oo } are places of K, we take their restrictions to their images. We may therefore assume that they are surjective. We shall say that they are equivalent if there exists an isomorphism A. : F 1 __.. F 2 such that qJ 2 = qJ 1 o A.. (We put A.( oo) = oo.) One sees that two places are equivalent if and only if they have the same valuation ring. It is clear that there is a bijection between equivalence classes of places of K, and valuation rings of K. A place is called trivial if it is injective. The valuation ring of the trivial place is simply K itself. As with homomorphisms, we observe that the composite of two places is also a place (trivial verification). It is often convenient to deal with places instead of valuation rings, just as it is convenient to deal with homomorphisms and not always with canonical homo morphisms or a ring modulo an ideal. The general theory of valuations and valuation rings is due to Krull , All gemeine Bewertungstheorie , J. reine angew. Math . 167 ( 1 932) , pp . 1 69 1 96 . However, the extension theory of homomorphisms as above was realized only around 1 945 by Chevalley and Zariski .
350
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We shall now give some examples of places and valuation rings . Example 1 . Let p be a prime number. Let Z (p ) be the ring of all rational numbers whose denominator is not divisible by p . Then Z (p ) is a valuation ring . The maximal ideal consists of those rational numbers whose numerator is divisible by p . Example 2. Let k be a field and R = k[X] the polynomial ring in one variable . Let p = p (X ) be an irreducible polynomial . Let o be the ring of rational functions whose denominator is not divisible by p . Then o is a valuation ring , similar to that of Example 1 . Example 3. Let R be the ring of power series k[ [X ] ] in one variable . Then R is a valuation ring , whose maximal ideal consists of those power series divisible by X. The residue class field is k itself. Example 4. Let R = k[ [X I , . . . , Xn 1 1 be the ring of power series in several variables . Then R is not a valuation ring , but R is imbedded in the field of repeated power series k((XI ))((X2 )) • • • ((Xn )) = Kn . By Example 3 , there is a place of Kn which is Kn  I valued . By induction and composition , we can define a kvalued place of Kn . Since the field of rational functions k(X I , . . , Xn ) is contained in Kn , the restriction of this place to k(X I , . . . , Xn ) gives a kvalued place of the field of rational functions in n variables . .
Example 5. In Chapter XI we shall consider the notion of ordered field . Let k be an ordered sub field of an ordered field K. Let o be the subset of elements of K which are not infinitely large with respect to k. Let m be the subset of elements of o which are infinitely small with respect to k. Then o is a valuation ring in K and m is its maximal ideal .
The following property of places will be used in connection with projective space in the next chapter.
Let cp : K � {L, oo } be an Lvalued place of K. Given a finite number of nonzero elements x I , . . . , Xn E K there exists an index j such that cp is finite on x,/xj for i = 1 , . . . , n. Proof. Let B be the valuation ring of the place . Define x, < xj to mean that x, /xj E B . Then the relation < is transitive , that is if x, < xj and xj < xr then x, < xr  Furthermore , by the property of a valuation ring , we always have x, < xj or xj < x, for all pairs of indices i , j. Hence we may order our ele ments , and we select the index j such that x, < xj for all i . This index j Proposition 3.4.
satisfies the requirement of the proposition .
We can obtain a characterization of integral elements by means of val uation rings. We shall use the following terminology . If o , 0 are local rings with maximal ideals m, 9Jl respectively, we shall say that 0 lies above o if o c 0 and 9Jl n o = m. We then have a can o nical injection ojm � Oj9Jl.
EXTENSION OF HOMOMORPHISMS
VI I, §3 Proposition
35 1
3.5. Let o be a local ring contained in a field L. An element x o.f
L is integral over o if. and only if x lies in every valuation ring above o.
0
of· L lying
Proof Assume that x is not integral over o. Let m be the maximal ideal of o. Then the ideal (m, 1/x) of o[1/x] cannot be the entire ring, otherwise we can write
with y E m and a; E o. From this we get
{ 1 + y)x n +
· · ·
+ an
=
0.
But 1 + y is not in m, hence is a unit of o. We divide the equation by 1 + y to conclude that x is integral over o , contrary to our hypothesis. Thus {m, 1/x) is not the entire ring, and is contained in a maximal ideal � , whose intersection with o contains m and hence must be equal to m. Extending the canonical homo morphism o[1/x] __.. o[1/x]/� to a homomorphism of a valuation ring 0 of L, we see that the image of 1/x is 0 and hence that x cannot be in this valuation ring. Conversely, assume that x is integral over o , and let
be an integral equation for x with coefficients in o . Let 0 be any valuation ring of L lying above o. Suppose x ¢. 0 . Let cp be the place given by the canonical homomorphism of 0 modulo its maximal ideal . Then cp(x) = oo so cp( 1 I x) = 0 .· D ivide the above equation by x n , and apply cp. Then each term except the first maps to 0 under cp, so we get cp( 1 ) = 0, a contradiction which proves the proposition . .. Proposition 3.6. Let A be a ring contained in a field L . An element x of L
is integral over A if and o n ly if x lies in every valuation ring 0 of L containing A . In terms of places , x is integral over A if and o n ly if every place of L finite on A is finite on x. Proof. Assume that every place finite on A is finite on x. We may assume x =I= 0 . If 1 1 x is a unit in A[ 1 1 x] then we can write
with c; E A and some n . Multiplying by x n  I we conclude that x is integral over A . If 1 I x is not a unit in A [ 1 I x] , then 1 I x generates a proper principal ideal . By Zorn ' s lemma this ideal is contained in a maximal ideal IDl . The homomorphism A [ 1 I x] � A [ 1 I x] I IDl can be extended to a place which is a finite on A but maps
352
EXTENSION O F R I N G S
V I I , Ex
1 I x on 0 , so x on oo , which contradicts the possibility that 1 I x is not a unit in A [ 1 I x] and proves that x is integral over A . The converse implication is proved just as in the second part of Proposition 3 . 5 . Remark. Let K be a subfield of L and let x E L . Then x is integral over K if and only if x is algebraic over K. So if a place cp of L is finite on K, and x is algebraic over K, then cp is finite on K(x) . Of course this is a trivial case of the integrality criterion which can be seen directly . Let
be the irreducible equation for x over K. Suppose x =I= 0 . Then a0 =t= 0 . Hence cp(x) =I= 0 immediately from the equation , so cp is an isomorphism of K(x) on its I mage . The next result is a generalization whose technique of proof can also be used in Exercise 1 of Chapter IX (the HilbertZariski theorem) . Theorem 3. 7.
General Integrality Criterion . Let A be an entire ring. Let z 1 , , zm be elements of some extension field of its quotientfield K. Assume that each z5 (s = 1 , . . . , m) satisfies a polynomial relation .
•
•
where g5(Z 1 , , Zm ) E A [Z 1 , . . . , Zm ] is a polynomial of total degree < d5, and that any pure power of Z5 occuring with nonzero coefficient in g5 occurs with a power strictly less than d Then z 1 , , zm are integral over A . •
•
•
5•
•
•
•
Proof. We apply Proposition 3 . 6 . Suppose some z5 is not integral over A . There exists a place cp of K, finite on A , such that cp(z5) = oo for some s . By Proposition 3 . 4 we can pick an index s such that cp(zi l z5) =t= oo for all j. We
divide the polynomial relation of the hypothesis in the lemma by z'js and apply the place . By the hypothesis on g5, it follows that cp(g5(z) l z'js) = 0 , whence we get 1 = 0, a contradiction which proves the theorem.
EX E R C I S ES 1.
Let K be a Galois extension of the rationals Q, with group G. Let B be the I ntegral closure of Z in K , and let a E B be such that K = Q(a). Let f(X) = Irr(a, Q, X). Let p be a pnme number, and assume that f remains irreducible mod p over Z/pZ. What can you say about the Galois group G ? (Arttn asked this question to Tate on his qualify ing exam.)
2. Let A be an entue ring and K its quotient field. Let t be transcendental over K. If A is integrally closed, show that A [t] is integrally closed.
EXERCISES
V I I , Ex
353
For the following exercises, you can use §1 of Chapter X. 3. Let A be an entire nng, Integrally closed in its quotient field K. Let L be a finite separable extension of K , and let B be the integral closure of A in L. If A is Noetherian, show that B ts a fin ite A module. [Hint : Let {w 1 , , w,.} be a basis of L over K . Multiplying all elements of this basis by a suitable element of A, we may assume wtthout loss of generality that all w; are integral over A. Let { w'1 , w�} be the dual basis relative to the trace, so that Tr(w; wj) = bii · Write an element cx of L i ntegral over A in the form •
•
•
•
cx
= b 1 w'1 + · · ·
•
•
,
+ b,. w�
with b1 E K. Taking the trace Tr( aw; ) , for i = 1 , . . . , n , conclude that B is contained in the finite module Awi + · · · + A w� . ] Hence B is Noetherian .
4 . The preceding exercise applies to the case when A = Z and k = Q . Let L be a finite , an be extension of Q and let o L be the ring of algebraic integers in L. Let a1 , the distinct embeddings of L into the complex numbers . Embedded oL into a Euclidean space by the map •
•
•
Show that in any bounded region of space , there is only a finite number of elements of oL . [Hint : The coefficients in an integral equation for a are elementary symmetric functions of the conjugates of a and thus are bounded integers . ] Use Exercise 5 of Chapter III to conclude that o L is a free Zmodule of dimension < n . In fact , show that the dimension is n , a basis of oL over Z also being a basis of L over Q . 5 . Let E be a finite extension of Q , and let oE be the ring of algebraic integers of E. Let U be the group of units of o £ · Let u1 , , an be the distinct em beddings of E into C . Map U into a Euclidean space , by the map •
•
•
Show that /( U ) ts a free abelian group, finitely generated, by showing that in any fintte region of space, there is only a finite number of elements of I( U). Show that the kernel of I ts a fin ite group, and is therefore the group of roots of unity in E. Thus U itself is a finitely generated a belian group. 6. Generalize the results of §2 to infinite Galois extenstons, especially Propositions 2. 1 and 2.5, using Zorn's lemma.
7 . Dedekind rings. Let o be an entire ring which is Noetherian , integrally closed , and , such that every nonzero prime ideal is maximal . Define a fractional ideal a to be an o submodule =I= 0 of the quotient field K such that there exists c E o , c =I= 0 for which c a C o . Prove that the fractional ideals form a group under multiplication . Hint following van der Waerden : Prove the following statements in order: (a) Given an ideal a =I= 0 in o , there exists a product of prime ideals Pt · · · P r C a . (b) Every maximal ideal p is invertible , i . e . if we let p  1 be the set of elements x E K such that x p C o , then p  1 p = o . (c ) Every nonzero ideal is invertible , by a fractional ideal . (Use the Noetherian property that if this is not true , there exists a maximal noninvertible ideal a , and get a contradiction. )
354
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VII, Ex
8 . Using prime ideals instead of prime numbers for a Dedekind ring A , define the notion of content as in the Gauss lemma , and prove that if/(X ) , g(X) E A [X ] are polynomials of degree > 0 with coefficients in A , then cont( fg) = cont( j)cont(g) . Also if K is the quotient field of A , prove the same statement for f, g E K[X] . 9 . Let A be an entire ring , integrally closed . Let B be entire , integral over A . Let Q . , Q2 be prime ideals of B with Q I :J Q 2 but Q I =I= Q 2 . Let P; = Q ; n A . Show that PI =I= P2 . 1 0 . Let n be a positive integer and let ,, , be primitive nth roots of unity . (a) Show that ( 1  0/( 1  '' ) is an algebraic integer. (b) If n > 6 is divisible by at least two primes , show that 1  ' is a unit in the ring Z [ � . ,
1 1 . Let p be a prime and ' a primitive pth root of unity . Show that there is a principal ideal J in Z[(J such that JP  1 = (p) (the principal ideal generated by p) .
Symmetric Polynomials 1 2.
Let F be a field of characteristic 0. Let f1 , . . . , fn be algebraically independent over F. , sn be the elementary symmetric functions. Then R = F[fi , . . . , fn] is an Let s1 , integral extension of S = F[s1 , . . . , sn] , and actually is its integral closure in the rational field F(fJ , . . . , tn ) · Let W be the group of permutation of the variables l) , . . , tn . (a) Show that S = R w is the fixed subring of R under W. f�n with 0 < r; < n  i form a basis of R over (b) Show that the elements fr1 S, so in particular, R is free over S. •
•
•
.
•
•
•
I am told that the above basis is due to Kronecker. There is a much more interesting basis, w hich can be defined as follows. Let OJ ' . . . ' On be the partial derivatives with respect to f) ' . . . ' fn , so 0; = a I Of;. Let P E F [t] = F[fi , . . . , tn] . Substituting o; for f; (i = 1 , . . . , n) gives a partial differential operator P( o) = P( OJ , . . . , on ) on R. An element of S can also be viewed as an element of R. Let Q e R. We say that Q is Wharmonic if P( o) Q = 0 for all symmetric polynomials P e S with 0 constant term. It can be shown that the Wharmonic polynomials form a finite dimensional space. Furthermore, if { H1 , . . . , HN } is a basis for this space over F, then it is also a basis for R over S. This is a special case of a general theorem of Che valley. See [La 99b], where the special case is worked out in detail.
C H A PT E R
VI I I
Tra n sce n d e nta l Exte n s i o n s
Both for their own sake and for applications to the case of finite exten sions of the rational numbers, one is led to deal with ground fields which are function fields, i.e. finitely generated over some field k, possibly by elements which are not algebraic. This chapter gives some basic properties of such fields.
§1 . TR A N S C E N D E N C E BAS ES Let K be an extension field of a field k. Let S be a subset of K. We recall that S (or the elements of S) is said to be algebraically independent over k, if whenever we have a relation
with coefficients a E k, almost all a = 0, then we must necessarily have all a(v) = 0. We can introduce an ordering among algebraically independent subsets of K, by ascending inclusion. These subsets are obviously inductively ordered, and thus there exist maximal elements. If S is a subset of K which is algebraically independent over k, and if the cardinality of S is greatest among all such subsets, then we call this cardinality the transcendence degree or dimension of K over k. Actually, we shall need to distinguish only between finite transcendence degree or infinite transcendence degree. We observe that 355
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§1
the notion of transcendence degree bears to the notion of algebraic indepen dence the same relation as the notion of dimension bears to the notion of linear independence. We frequently deal with families of elements of K, say a family { x i}i e J' and say that such a family is algebraically independent over k if its elements are distinct (in other words, x i =F xi if i =F j) and if the set consisting of the elements in this family is algebraically independent over k. A subset S of K which is algebraically independent over k and is maximal with respect to the inclusion ordering will be called a transcendence base of K over k. From the maximality, it is clear that if S is a transcendence base of K over k, then K is algebraic over k(S). Theorem 1 . 1 . Let K be an extension of a field k. Any two transcendence
bases of K over k have the same cardinality. If r is a subset of K such that K is algebraic over k(f), and S is a subset of r which is algebraically indepen dent over k, then there exists a transcendence base of K over k such that s c CB c r. Proof. We shall prove that if there exists one finite transcendence base, say {x1 , , xm }, m > 1 , m minimal , then any other transcendence base must also have m elements . For this it will suffice to prove: If w b . . . , wn are elements of K which are algebraically independent over k then n < m (for we can then use symmetry) . By assumption , there exists a nonzero irreducible polynomial f1 in m + 1 variables with coefficients in k such that •
•
•
f1 ( W I ' X I ' ' Xm ) = 0 · After renumbering x I , . . . , xm we may write f1 = � gj( w I , x2 , , Xm ) x1 with some gN =I= 0 with some N > 1 . No irreducible factor of gN vanishes on ( w i , x2 , , xn ) , otherwise w 1 would be a root of two distinct irreducible polyno , xm ) and mials over k(xi , . . . , xm ) . Hence x 1 is algebraic over k(w i , x2 , w b x2 , , xm are algebraically independent over k, otherwise the minimal ity of m would be contradicted . Suppose inductively that after a suitable re numbering of x2 , , xm we have found wi , . . . , wr (r < n) such that K is algebraic over k( w I , . . . , w Xr+ I , . . . , xm ) . Then there exists a nonzero polynomial f in m + 1 variables with coefficients in k such that •
•
•
•
•
•
•
•
•
•
.
•
•
•
•
•
•
•
n
Since the w' s are algebraically independent over k, it follows by the same argument as in the first step that some xi , say xr+ I , is algebraic over k(w i , . . . , wr + I , xr+ 2 , , xm ) . Since a tower of algebraic extensions is algebraic , it follows that K is algebraic over k( w I , . . . , wr+ I , Xr+ 2 , . . . , xm ) . We can repeat the procedure , and if n > m we can replace all the x ' s by w' s , to see that K is algebraic over k(wi , . . . , wm ) . This shows that n > m implies n = m , as desired . •
.
.
VI I I,
§2
N O ETH E R N O R MA LIZATI O N TH EO R E M
357
We have now proved : Either the transcendence degree is finite, and is equal to the cardinality of any transcendence base, or it is infinite, and every transcendence base is infinite. The cardinality statement in the infinite case will be left as an exercise. We shall also leave as an exercise the statement that a set of algebraically independent elements can be completed to a transcendence base , selected from a given set I such that K is algebraic over k(f) . (The reader will note the complete analogy of our statements with those concerning linear bases . ) Note.
The preceding section is the only one used in the next chapter. The remaining sections are more technical, especially §3 and §4 which will not be used in the rest of the book. Even §2 and §5 will only be mentioned a couple of times, and so the reader may omit them until they are referred to again.
§2.
N O ET H E R N O R M A LIZATI O N T H EO R E M
Theorem
=
Let k [x 1 , , xn ] k[x] be a finitely generated entire ring over a field k, and assume that k(x) has transcendence degree r. Then there exist elements y 1 , , Yr in k[x] such that k [x] is integral over k [y] k [y 1 , · · · ' Yr ]. 2. 1 .
•
•
•
•
•
•
=
Proof If (x 1 , , xn ) are already algebraically independent over k, we are done. If not, there is a nontrivial relation L ax { • · · · x�" 0 •
•
•
=
with each coefficient a E k and a =F 0. The sum is taken over a finite number of distinct ntuples of integers (i 1 , , in ), iv > 0. Let m 2 , , mn be positive integers, and put •
=
=
.
•
•
•
.
•
•
Substitute xi Y i + x�i (i 2, . , n) in the above equation. Using vector notation, we put (m) ( 1, m 2 , , mn ) and use the dot product (i) · (m) to denote i 1 + m 2 i 2 + · · · + mn in · If we expand the relation after making the above substitution, we get m " yn )  0 � C j X (1j) · ( ) + f(X 1 ' y 2 () where f is a polynomial in which no pure power of x 1 appears. We now select d to be a large integer [say greater than any component of a vector (i) such that c =F 0] and take (m) ( 1, d, d 2 , , d n ). =
•
•
•
'
=
•
•
•
•
•
•
'
358
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§2
Then all (j) · (m) are distinct for those (j) such that cu) =F 0. In this way we obtain an integral equation for x 1 over k [ y 2 , • • • , Yn ]. Since each x i (i > 1 ) is integral over k [ x 1 , y2 , • • • , yn ], it follows that k [x] is integral over k [y 2 , • • • , Yn ]. We can now proceed inductively, using the transitivity of integral extensions to shrink the number of y's until we reach an alge braically independent set of y's. The advantage of the proof of Theorem 2. 1 is that it is applicable when k is a finite field. The disadvantage is that it is not linear in x 1 , . . . , x n . We now deal with another technique which leads into certain aspects of algebraic geometry on which we shall comment after the next theorem. We start again with k [x 1 , • • • , xn ] finitely generated over k and entire. Let (uii ) (i, j = 1, . . . , n) be algebraically independent elements over k(x), and let ku = k(u) = k( uii )a u i , j · Put n "' U u· ·XJ· " Y.' = � j= l
This amounts to a generic linear change of coordinates in nspace, to use geometric terminology. Again we let r be the transcendence degree of k(x) over k. Theorem
ku [ Y 1 '
2.2.
With the above notation, ku [ x] is integral over
' Yr J · Proof Suppose some xi is not integral over ku [y 1 , • • • , Yr ]. Then there exists a place qJ of ku( Y) finite on ku [y 1 , . . . , Yr ] but taking the value oo on some xi. Using Proposition 3.4 of Chapter VII, and renumbering the indices if necessary, say qJ(xi /x n ) is finite for all i. Let zj = qJ(xi /xn ) for j = 1 , . . . , n. Then dividing the equations Yi = L uiixi by xn (for i = 1 , . . . , r) and applying the place, we get • • •
The transcendence degree of k (z ') over k cannot be r, for otherwise, the place qJ would be an isomorphism of k(x) on its image. [Indeed, if, say, z� , . . . , z; are algebraically independent and zi = xi/xn , then z 1 , • • • , zr are also alge braically independent, and so form a transcendence base for k(x) over k. Then the place is an isomorphism from k(z 1 , • • • , zr) to k (z � , . . . , z;), and hence is an isomorphism from k(x) to its image.] We then conclude that with i = 1, . . . , r ; j = 1, . . . , n  1 . Hence the transcendence degree of k(u) over k would be < rn  1, which is a contradiction, proving the theorem.
VIII,
§2
N O ET H E R N O R MALIZATI O N TH EO R E M
359
Corollary 2.3. Let k be a field, and let k(x) be a finitely generated
extension of transcendence degree r. There exists a polynomial P(u) = P(uii ) E k [u] such that if (c) = (cii ) is a family of elements cii E k satisfying P(c) =F 0, and we let y; = L cii xi , then k [x] is integral over k [y� , . . . y; ]. Proof By Theorem 2.2, each xi is integral over ku [y 1 , . . . , Yr ]. The coefficients of an integral equation are rational functions in ku . We let P(u) be a common denominator for these rational functio ns. If P(c) =F 0, then there is a homomorphism qJ : k(x) [u, P(u)  1 ] k(x) ,
+
such that lfJ(u) = (c), and such that cp is the identity on k(x). We can apply cp to an integral equation for xi over ku [y] to get an integral equation for xi over k [y'], thus concluding the proof. Remark.
After Corollary 2.3, there remains the problem of finding ex plicitly integral equations for x 1 , , Xn (or Yr + 1 , . . . , Yn ) over ku [y 1 , . . . , Yr ]. This is an elimination problem, and I have decided to refrain from further involvement in algebraic geometry at this point. But it may be useful to describe the geometric language used to interpret Theorem 2.2 and further results in that line. After the generic change of coordinates, the map •
•
•
( y 1 , . . . , Yn ) �+ ( y 1 , . . . , Yr ) is the generic projection of the variety whose coordinate ring is k [x] on affine rspace. This projection is finite, and in particular, the inverse image of a point on affine rspace is finite. Furthermore, if k(x) is separable over k (a notion which will be defined in §4), then the extension ku( Y) is finite separable over ku( Y 1 , , Yr ) (in the sense of Chapter V). To determine the degree of this finite extension is essentially Bezout ' s theorem. Cf. [La 58], Chapter VIII, §6. The above techniques were created by van der Waerden and Zariski, cf., for instance, also Exercises 5 and 6. These techniques have unfortunately not been completely absorbed in some more recent expositions of algebraic geometry. To give a concrete example : When Hartshorne considers the intersection of a variety and a sufficiently general hyperplane, he does not discuss the " generic " hyperplane (that is, with algebraically independent coefficients over a given ground field), and he assumes that the variety is nonsingular from the start (see his Theorem 8. 1 8 of Chapter 8, [Ha 77] ). But the description of the intersection can be done without simplicity as sumptions, as in Theorem 7 of [La 58], Chapter VII, §6, and the corre sponding lemma. Something was lost in discarding the technique of the algebraically independent (u ii ). After two decades when the methods illustrated in Chapter X have been prevalent, there is a return to the more explicit methods of generic construc tions using the algebraically independent (u ii ) and similar ones for some •
•
•
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applications because part of algebraic geometry and number theory are returning to some problems asking for explicit or effective constructions, with bounds on the degrees of solutions of algebraic equations. See, for instance, [Ph 9 1 95] , [So 90], and the bibliography at the end of Chapter X, § 6. Return ing to some techniques, however, does not mean abandoning others ; it means only expanding available tools. Bibliography [Ha 77] [La 58]
Algebraic Geometry, SpringerVerlag, New York, 1 977 S. LA NG , Introduction to Algebraic Geometry, W ile y In t ers cien ce , New R. HARTSHORNE,

[Ph 9 1 95 ]
York, 1 958 P. PHILIPPON, Sur des h au te u rs alternatives, I Math. Ann. 289 ( 1 99 1 ) pp. 255283 ; II Ann. Jnst. Fourier 44 ( 1 994) pp. 1 043 1 065 ; III J. Math. Pures Appl. 74 ( 1 995) pp. 345365
[So 90]
C. SouLE, Geometric d 'Arakelov et theorie des nombres transccndants, Asterisque 1 98200 ( 1 99 1 ) pp. 35537 1
§3.
LI N EA R LY D I SJ O I NT EXT EN S I O N S
In this section we discuss the way in which two extensions K and L of a field k behave with respect to each other. We assume that all the fields involved are contained in one field n, assumed algebraically closed. K is said to be linearly disjoint from L over k if every finite set of elements of K that is linearly independent over k is still such over L. The definition is unsymmetric, but we prove right away that the property of being linearly disjoint is actually symmetric for K and L. Assume K linearly disjoint from L over k. Let 1 , Yn be elements of L linearly independent over k. Suppose there is a nontrivial relation of linear depen dence over K,
y
•
•
•
,
(1) Say x 1 , . . . , x, are linearly independent over k, and x, + 1 , , xn are linear combinations x i = L a illxll, i = r + 1 , . . . , n. We can write the relation ( 1 ) as •
r
follows :
IJ. = 1
and collecting terms, after inverting the second sum, we get
+ ( aip Yi )) x" y J. ( " i=� t
=
0.
.
.
VI II,
§3
L I N EA R LY D I SJ O I NT EXTE N S I O N S
361
The y' s are linearly independent over k, so the coefficients of xll are =F 0. This contradicts the linear disjointness of K and L over k. We now give two criteria for linear disjointness. Criterion 1. Suppose that K is the quotient field of a ring R and L the
quotient field of a ring S. To test whether L and K are linearly disjoint, it suffices to show that if elements y 1 , • • • , Yn of S are linearly independent over k, then there is no linear relation among the y' s with coefficients in R. Indeed, if elements y 1 , , Yn of L are linearly independent over k, and if there is a relation x 1 y1 + · · · + Xn Yn = 0 with x i E K, then we can select y in S and x in R such that xy =F 0, YYi E S for all i, and xx i E R for all i. Multiplying the relation by xy gives a linear dependence between elements of R and S. However, the YYi are obviously linearly independent over k, and this proves our criterion. •
•
•
Criterion 2. Again let R be a subring of K such that K is its quotient field and R is a vector space over k. Let {uti } be a basis of R considered as a
vector space over k. To prove K and L linearly disjoint over k, it suffices to show that the elements {uti } of this basis remain linearly independent over L. Indeed, suppose this is the case. Let x 1 , • • • , xm be elements of R linearly independent over "k. They lie in a finite dimension vector space generated by some of the uti, say u 1 , . . . , un. They can be completed to a basis for this space over k. Lifting this vector space of dimension n over L, it must conserve its dimension because the u ' s remain linearly independent by hy pothesis, and hence the x ' s must also remain linearly independent. Proposition
Let K be a field containing another field k, and let L ::J E be two other extensions of k. Then K and L are linearly disjoint over k if and only if K and E are linearly disjoint over k and KE, L are linear/y disjoint over E. 3. 1 .
1\ /\1 \ KL
KE
K
L
E
k
362
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§3
Proof Assume first that K, E are linearly disjoint over k, and KE, L are linearly disjoint over E. Let { K} be a basis of K as vector space over k (we use the elements of this basis as their own indexing set), and let { �} be a basis of E over k. Let {A.} be a basis of L over E. Then {�A.} is a basis of L over k. If K and L are not linearly disjoint over k, then there exists a relation
Changing the order of summation gives
contradicting the linear disjointness of L and KE over E. Conversely, assume that K and L are linearly disjoint over k. Then a fortiori, K and E are also linearly disjoint over k, and the field K E is the quotient field of the ring E[K] generated over E by all elements of K. This ring is a vector space over E, and a basis for K over k is also a basis for this ring E[K] over E. With this remark, and the criteria for linear disjointness, we see that it suffices to prove that the elements of such a basis remain linearly independent over L. At this point we see that the arguments given in the first part of the proof are reversible. We leave the formalism to the reader. We introduce another notion concerning two extensions K and L of a field k. We shall say that K is free from L over k if every finite set of elements of K algebraically independent over k remains such over L. If (x) and (y) are two sets of elements in n, we say that they are free over k (or independent over k) if k(x) and k(y) are free over k. Just as with linear disjointness, our definition is unsymmetric, and we prove that the relationship expressed therein is actually symmetric. Assume therefore that K is free from L over k. Let y 1 , , Yn be elements of L, algebraically independent over k. Suppose they become dependent over K. They become so in a subfield F of K finitely generated over k, say of transcendence degree r over k. Computing the transcendence degree of F(y) over k in two ways gives a contradiction (cf. Exercise 5). •
F(y)
F
7 � k(y)
"""
r """
k
/
•
•
VI I I ,
§4
S E PA RAB LE AN D R E G U LA R EXTE N S I O N S
363
Proposition 3.2. If K and L are linearly disjoint over k, then they are free
over k. Proof Let x 1 , . . . , xn be elements of K algebraically independe nt over k. Suppose they become algebraically dependent over L. We get a relation L Ya M« (x) = 0 between monomials M"(x) with coefficients y" in L. This gives a linear relation among the M"(x). But these are linearly independent over k because the x's are assumed algebraically independent over k. This is a contradiction. Proposition 3.3. Let L be an extension of k, and let (u)
(u1 , , ur ) be a set of quantities algebraically independent over L. Then the field k(u) is linear/y disjoint from L over k. Proof. According to the criteria for linear disjointness, it suffices to prove that the elements of a basis for the ring k[u] that are linearly indepen dent over k remain so over L. In fact the monomials M(u) give a basis of k [u] over k. They must remain linearly independent over L, because as we have seen, a linear relation gives an algebraic relation. This proves our proposition. Note finally that the property that two extensions K and L of a field k are linearly disjoint or free is of finite type. To prove that they have either property, it suffices to do it for all subfields K0 and L0 of K and L respectively which are finitely generated over k. This comes from the fact that the definitions involve only a finite number of quantities at a time.
§4.
=
•
•
•
S E PA R A B LE AN D R EG U LA R EXTE N S I O N S
Let K be a finitely generated extension of k, K = k(x). We shall say that it is separably generated if we can find a transcendence basis (t 1 , , tr ) of K/k such that K is separably algebraic over k(t). Such a transcendence base is said to be a separating transcendence base for K over k. We always denote by p the characteristic if it is not 0. The field obtained from k by adjoining all p mth roots of all elements of k will be denoted by k 1 1P m. The compositum of all such fields for m = 1 , 2, . . . , is denoted by k 1 1P00• •
Proposition 4.1.
•
•
The following conditions concerning an extension field K of k are equivalent : (i) K is linearly disjoint from k 1 'pOO · (ii) K is linearly disjoint from k 1 1Pm for some m.
364
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(iii) Every subfield of K containing k and finitely generated over k is separably generated.
Proof It is obvious that (i) implies (ii). In order to prove that (ii) implies (iii), we may clearly assume that K is finitely generated over k, say K
k(x) = k(x 1 ,
=
.
•
.
, Xn ).
Let the transcendence degree of this extension be r. If r = n, the proof is complete. Otherwise, say x 1 , , xr is a transcendence base. Then xr+ 1 is algebraic over k(x 1 , . . . , xr ). Let f(X 1 , . . . , Xr+ 1 ) be a polynomial of lowest degree such that .
.
•
Xr+ 1 ) = 0. Then f is irreducible. We contend that not all xi (i = 1 , . . . , r + 1 ) appear to the pth power throughout. If they did, we could write f(X) = L c(J M(J(X)P where M(J(X) are monomials in X 1 , , Xr+ 1 and c(J E k. This would imply that the M(J(x) are linearly dependent over k 1 1P (taking the pth root of the equation L c(JM(J(x)P = 0). However, the M(J(x) are linearly independent over k (otherwise we would get an equation for x 1 , , xr+ 1 of lower degree) and we thus get a contradiction to the linear disjointness of k(x) and k 11P. Say X 1 does not appear to the pth power throughout, but actually appears in f(X). We know that f(X) is irreducible in k[X 1 , , Xr+ 1 ] and hence f(x) = O is an irreducible equation for x 1 over k(x 2 , , xr + 1 ). Since X 1 does not appear to the pth power throughout, this equation is a separable equation for x 1 over k(x 2 , , xr+ 1 ), in other words, x 1 is separable algebraic over k(x 2 , , xr + t ). From this it follows that it is separable algebraic over k(x 2 , , xn ). If (x 2 , , xn ) is a transcendence base, the proof is complete. If not, say that x 2 is separable over k(x 3 , , xn ). Then k(x) is separable over k(x 3 , , xn ). Proceeding inductively, we see that the procedure can be continued until we get down to a transcendence base. This proves that (ii) implies (iii). It also proves that a separating transcendence base for k(x) over k can be selected from the given set of generators (x). To prove that (iii) implies (i) we may assume that K is finitely generated over k. Let (u) be a transcendence base for K over k. Then K is separably algebraic over k(u) . B y Proposition 3.3, k(u) and k 1 1Poo are linearly disjoint. Let L = k 11pOO . Then k(u)L is purely inseparable over k(u), and hence is linearly disjoint from K over k(u) by the elementary theory of finite algebraic extensions. Using Proposition 3. 1, we conclude that K is linearly disjoint from L over k, thereby proving our theorem. An extension K of k satisfying the conditions of Proposition 4. 1 is called separable. This definition is compatible with the use of the word for alge braic extensions. The first condition of our theorem is known as MacLane's criterion. It has the following immediate corollaries. f(x 1 ,
· · · '
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Corollary 4.2. If K is separable over k, and E is a subfield of K contain ing k, then E is separable over k. Corollary 4.3.
Let E be a separable extension of k, and K a separable extension of E. Then K is a separable extension of k. Proof Apply Proposition 3 . 1 and the definition of separability. Corollary 4.4.
If k is perfect, every extension of k is separable.
Corollary 4.5.
Let K be a separable extension of k, and free from an extension L of k. Then KL is a separable extension of L. Proof An element of KL has an expression in terms of a finite number of elements of K and L. Hence any finitely generated subfield of KL containing L is contained in a composite field FL, where F is a subfield of K finitely generated over k. By Corollary 4.2, we may assume that K is finitely generated over k. Let (t) be a transcendence base of K over k, so K is separable algebraic over k(t). By hypothesis, (t) is a transcendence base of KL over L, and since every element of K is separable algebraic over k(t), it is also separable over L(t). Hence KL is separably generated over L. This proves the corollary. Corollary 4.6.
Let K and L be two separable extensions of k, free from each other over k. Then KL is separable over k. Proof Use Corollaries 4.5 and 4.3. Corollary 4.7.
Let K, L be two extensions of k, linearly disjoint over k. Then K is separable over k if and only if KL is separable over L. Proof If K is not separable over k, it is not linearly disjoint from k 1 1P over k, and hence a fortiori it is not linearly disjoint from Lk 1 1P over k. By Proposition 4. 1, this implies that KL is not linearly disjoint from Lk 1 1P over L, and hence that KL is not separable over L. The converse is a special case of Corollary 4.5, taking into account that linearly disjoint fields are free. We conclude our discussion of separability with two results. The first one has already been proved in the first part of Proposition 4. 1, but we state it here explicitly. Proposition 4.8.
If K is a separable extension of k, and is finitely gener ated, then a separating transcendence base can be selected from a given set of generators. To state the second result we denote by K P m the field obtained from K by raising all elements of K to the p m th power. 
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Proposition 4.9. Let K be a finitely generated extension of a field k. If
K P mk = K for some m, then K is separably algebraic over k. Conversely, if K is separably algebraic over k, then K Pmk = K for all m.
Proof. If K/k is separably algebraic, then the conclusion follows from the elementary theory of finite algebraic extensions. Conversely, if K/k is finite algebraic but not separable, then the maximal separable extension of k in K cannot be all of K, and hence K Pk cannot be equal to K. Finally, if there exists an element t of K transcendental over k, then k(t 1 1Pm) has degree p m over k(t), and hence there exists a t such that t 1 1Pm does not lie in K. This proves our proposition. There is a class of extensions which behave particularly well from the point of view of changing the ground field, and are especially useful in algebraic geometry. We put some results together to deal with such exten sions. Let K be an extension of a field k, with algebraic closure K8• We claim that the following two conditions are equivalent : REG
1.
k is algebraically closed in K (i.e. every element of K algebraic over k lies in k), and K is separable over k.
REG 2. K is linearly disjoint from
ka over k.
We show the equivalence. Assume REG 2. By Proposition 4. 1, we know that K is separably generated over k. It is obvious that k must be algebraically closed in K. Hence REG 2 implies REG 1 . To prove the converse we need a lemma. Lemma 4. 10.
Let k be algebraically closed in extension K. Let x be some element of an extension of K, but algebraic over k. Then k(x) and K are linearly disjoint over k, and [k(x) : k] = [K(x) : K].
Proof Let f(X) be the irreducible polynomial for x over k. Then f remains irreducible over K ; otherwise, its factors would have coefficients algebraic over k, hence in k. Powers of x form a basis of k(x) over k, hence the same powers form a basis of K(x) over K. This proves the lemma. To prove REG 2 from REG 1 , we may assume without loss of generality that K is finitely generated over k, and it suffices to prove that K is linearly disjoint from an arbitrary finite algebraic extension L of k. If L is separable algebraic over k, then it can be generated by one primitive element, and we can apply Lemma 4. 1 0. More generally, let E be the maximal separable subfield of L containing k. By Proposition 3. 1 , we see that it suffices to prove that K E and L are linearly disjoint over E. Let (t) be a separating transcendence base for K over k. Then K is separably algebraic over k(t). Furthermore, (t) is also a separating transcendence base for KE over E, and KE is separable algebraic
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over E(t). Thus KE is separable over E, and by definition KE is linearly disjoint from L over K because L is purely inseparable over E. This proves that REG 1 implies REG 2. Thus we can define an extension K of k to be regular if it satisfies either one of the equivalent conditions REG 1 or REG 2. Proposition
4.1 1 .
(a) Let K be a regular extension of k, and let E be a sub.field of K containing k. Then E is regular over k. (b) Let E be a regular extension of k, and K a regular extension of E. Then K is a regular extension of k. (c) If k is algebraically closed, then every extension of k is regular.
Proof. Each assertion is immediate from the definition conditions REG 1 and REG 2. Theorem
Let K be a regular extension of k, let L be an arbitrary extension of k, both contained in some larger field, and assume that K, L are free over k. Then K, L are linearly disjoint over k. 4. 1 2.
Proof (Artin). Without loss of generality, we may assume that K is finitely generated over k. Let x 1 , , xn be elements of K linearly indepen dent over k. Suppose we have a relation of linear dependence •
•
•
+ · · · + X n Yn = 0 with Y i E L. Let cp be a kavalued place of L over k. Let (t) be a transcen dence base of K over k. By hypothesis, the elements of (t) remain alge braically independent over L, and hence cp can be extended to a place of KL which is identity on k(t). This place must then be an isomorphism of K on its image, because K is a finite algebraic extension of k(t) (remark at the end of Chapter VII, §3). After a suitable isomorphism, we may take a place equivalent to cp which is the identity on K. Say lfJ( Yi iYn) is finite for all i (use Proposition 3.4 of Chapter VII). We divide the relation of linear dependence by Yn and apply lfJ to get L X i lfJ( Y i iYn) = 0, which gives a linear relation among the xi with coefficients in ka, contradicting the linear disjointness. This proves the theorem. Xt Yt
Theorem
Let K be a regular extension of k, free from an extension L of k over k. Then KL is a regular extension of L. 4. 1 3.
Proof. From the hypothesis, we deduce that K is free from the algebraic closure La of L over k. By Theorem 4. 1 2, K is linearly disjoint from La over k. By Proposition 3. 1, K L is linearly disjoint from La over L, and hence KL is regular over L.
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Let K, L be regular extensions of k, free from each other over k. Then KL is a regular extension of k. 4. 1 4.
Proof. Use Corollary 4. 1 3 and Proposition 4. 1 1 (b). Theorem 4. 1 3 is one of the main reasons for emphasizing the class of regular extensions : they remain regular under arbitrary base change of the ground field k. Furthermore, Theorem 4. 1 2 in the background is important in the study of polynomial ideals as in the next section, and we add some remarks here on its implications. We now assume that the reader is acquainted with the most basic properties of the tensor product (Chapter XVI, § 1 and §2). Corollary
Let K = k(x) be a finitely generated regular extension, free from an extension L of k, and both contained in some larger field. Then the natural kalgebra homomorphism 4. 1 5.
L ®k k[x] + L[x] is an isomorphism. Proof By Theorem 4. 1 2 the homomorphism is injective, and it is obvi ously surjective, whence the corollary follows. Corollary
Let k(x) be a finitely generated regular extension, and let p be the prime ideal in k [X] vanishing on (x), that is, consisting of all polynomials f(X) E k [X] such that f(x) = 0. Let L be an extension of k, free from k(x) over k. Let P L be the prime ideal in L[X] vanishing on (x). Then P L = pL[X], that is P L is the ideal generated by p in L[X], and in particular, this ideal is prime. 4. 1 6.
Proof Consider the exact sequence 0 + p + k [X] + k [x] + 0. Since we are dealing with vector spaces over a field, the sequen�e remains exact when tensored with any kspace, so we get an exact sequence
0 + L ®k p + L[X] + L ®k k[x] + 0. By Corollary 4. 1 5 , we know that L ®k k[x] ::::::: L[x] , and the image of L ®k p in L[X] is pL[X], so the lemma is proved. Corollary 4. 1 6 shows another aspect whereby regular extensions behave well under extension of the base field, namely the way the prime ideal p remains prime under such extensions.
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D E R IVATI O N S
A derivation D of a ring R is a mapping D : R + R of R into itself which is l inear and satisfies the ordinary rule for derivatives, i.e., D(x + y) = D x + Dy
and
D(xy) = xDy + yDx.
As an example of derivations, consider the polynomial ring k[X] over a field k. For each variable Xi , the partial derivative iJjiJXi taken in the usual manner is a derivation of k [X]. Let R be an entire ring and let K be its quotient field. Let D: R + R be a derivation. Then D extends uniquely to a derivation of K, by defining (
D uI v
)
=
vDu  uDv v2
•
It is immediately verified that the expression on the righthand side is independent of the way we represent an element of K as ujv (u, v E R), and satisfies the conditions defining a derivation. Note. In this section, we shall discuss derivations of fields. For deriva
tions in the context of rings and modules, see Chapter XIX, §3. A derivation of a field K is trivial if Dx = 0 for all x E K. It is trivial over a subfield k of K if Dx = 0 for all x E k. A derivation is always trivial over the prime field : One sees that
D ( 1 ) = D( 1 · 1)
=
2D( 1 ),
whence D( 1) = 0. We now consider the problem of extending derivations. Let L = K(x) = K(x 1 ,
•
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, xn )
be a finitely generated extension. If f E K [X], we denote by iJfjiJxi the polynomials iJfjiJXi evaluated at (x). Given a derivation D on K, does there exist a derivation D * on L coinciding with D on K ? If f(X) E K [X] is a polynomial vanishing on (x), then any such D * must satisfy (1) where fD denotes the polynomial obtained by applying D to all coefficients of f. Note that if relation ( 1) is satisfied for every element in a finite set of generators of the ideal in K [X] vanishing on (x), then (1) is satisfied by every polynomial of this ideal. This is an immediate consequence of the rules for derivations. The preceding ideal will also be called the ideal determined by (x) in K [X].
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The above necessary condition for the existence of a D * turns out to be sufficient. Theorem 5.1. Let D be a derivation of a field K. Let =
(x) (x 1 , . . . , xn ) be a finite family of elements in an extension of K. Let { f:z(X) } be a set of generators for the ideal determined by (x) in K [X]. Then, if (u) is any set of elements of K(x) satisfying the equations o .t:zD (x) + L (of:zjoxi)ui, =
there is one and only one derivation D * of K(x) coinciding with D on K, and such that D * xi ui for every i. =
Proof The necessity has been shown above. Conversely, if g(x), h(x) are in K [x], and h(x) # 0, one verifies immediately that the mapping D * defined by the formulas
D * ( g/h)
=
hD * g  gD*h ' h2
is well defined and is a derivation of K (x). Consider the special case where (x) consists of one element x. Let D be a given derivation on K.
Case 1 . x is separable algebraic over K. Let f(X) be the irreducible polynomial satisfied by x over K. Then f'(x) # 0. We have 0
=
fD(x) + f'(x)u,
=
whence u fD(x)jf' (x). Hence D extends to K(x) uniquely. If D is trivial on K, then D is trivial on K (x).
Case 2. x is transcendental over K. Then D extends, and u can be selected arbitrarily in K (x). Case 3. x is purely inseparable over K, so x P  a 0, with a E K. Then D extends to K(x) if and only if Da 0. In particular if D is trivial on K, then u can be selected arbitrarily. =
Proposition
=
A finitely generated extension K (x) over K is separable algebraic if and only if every derivation D of K(x) which is trivial on K is trivial on K(x). 5.2.
Proof If K(x) is separable algebraic over K, this is Case 1 . Conversely, if it is not, we can make a tower of extensions between K and K(x), such
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that each step is covered by one of the three above cases. At least one step will be covered by Case 2 or 3. Taking the uppermost step of this latter type, one sees immediately how to construct a derivation trivial on the bottom and nontrivial on top of the tower. Proposition 5.3.
Given K and elements (x) = (x 1 , , xn ) in some extension field, assume that there exist n polynomials !;, E K [X] such that : (i) !;,(x) = 0, and (ii) det (o!;,joxi ) =F o. •
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Then (x) is separably algebraic over K. Proof. Let D be a derivation on K(x), trivial on K. Having 1"i(x) = 0 we must have D!;,(x) = 0, whence the Dx i satisfy n linear equations such that the coefficient matrix has nonzero determinant. Hence Dxi = 0, so D is trivial on K(x). Hence K(x) is separable algebraic over K by Proposition 5.2. The following proposition will follow directly from Cases 1 and 2. Proposition 5.4. Let K
k(x) be a finitely generated extension of k. An element z of K is in K Pk if and only if every derivation D of K over k is such that Dz = 0. Proof. If is in K P k, then it is obvious that every derivation D of K over k vanishes on z. Conversely, if z ¢ K Pk, then z is purely inseparable over K P k, and by Case 3 of the extension theorem, we can find a derivation D trivial on K P k such that Dz = 1 . This derivation is at first defined on the field K P k(z). One can extend it to K as follows. Suppose there is an element w E K such that w ¢ K Pk(z). Then w P E K P k, and D vanishes on w P. We can then again apply Case 3 to extend D from K P k(z) to K Pk(z, w). Proceeding stepwise, we finally reach K, thus proving our proposition. =
z
The derivations D of a field K form a vector space over K if we define zD for z E K by (zD) (x) = zDx. Let K be a finitely generated extension of k, of dimension r over k. We denote by i> the Kvector space of derivations D of K over k (derivations of K which are trivial on k). For each z E K, we have a pairing
(D, z) �+ Dz of ( i>, K) into K. Each element z of K gives therefore a Klinear functional of i>. This functional is denoted by dz. We have
d(yz) = y dz + z dy, d(y + z) = dy + dz. These linear functionals form a subspace define y dz by (D, y dz) �+ yDz.
tF
of the dual space of
D,
if we
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Proposition 5.5.
Assume that K is a separably generated and finitely generated extension of k of transcendence degree r. Then the vector space � (over K) of derivations of K over k has dimension r. Elements t 1 , , tr of K from a separating transcendence base of K over k if and only if dt , dtr form a basis of the dual space of � over K. Proof If t 1 , , tr is a separating transcendence base for K over k, then we can find derivations D 1 , . . . , Dr of K over k such that Di ti = � ii ' by Cases 1 and 2 of the extension theorem. Given D E � ' let wi = Dti. Then clearly D = L wiDi, and so the Di form a basis for � over K, and the dti form the dual basis. Conversely, if dt 1 , , dtr is a basis for ft over K, and if K is not separably generated over k( t), then by Cases 2 and 3 we can find a derivation D which is trivial on k(t) but nontrivial on K. If D 1 , , Dr is the dual basis of dt 1 , , dtr (so Di ti = � ii ) then D, D 1 , , Dr would be linearly independent over K, contradicting the first part of the theorem. .
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Corollary 5.6. Let K be a finitely generated and separably generated
extension of k. Let z be an element of K transcendental over k. Then K is separable over k(z) if and only if there exists a derivation D of K over k such that Dz =F 0. Proof If K is separable over k(z), then z can be completed to a separat ing base of K over k and we can apply the proposition. If Dz =F 0, then dz =F 0, and we can complete dz to a basis of � over K. Again from the proposition, it follows that K will be separable over k(z). Note.
Here we have discussed derivations of fields. For derivations in the context of rings and modules, see Chapter XVI. As an application, we prove : Theorem 5.7. (ZariskiMatsusaka).
Let K be a finitely generated sepa rable extension of a field k. Let y, z E K and z ¢ K Pk if the characteristic is p > 0. Let u be transcendental over K, and put ku = k( u), Ku = K( u) (a) For all except possibly one value of c E k, K is a separable extension of k( y + cz). Furthermore, Ku is separable over ku(Y + uz). (b) Assume that K is regular over k, and that its transcendence degree is at least 2. Then for all but a finite number of elements c E k, K is a regular extension of k(y + cz). Furthermore, Ku is regular over ku(Y + uz). Proof. We shall use throughout the fact that a subfield of a finitely generated extension is also finitely generated (see Exercise 4). If w is an element of K, and if there exists a derivation D of K over k such that Dw =F 0, then K is separable over k( w), by Corollary 5.6. Also by Corollary 5.6, there exists D such that Dz =F 0. Then for all elements c E k, except possibly one, we have D ( y + cz) = Dy + cDz =F 0. Also we may extend D to Ku over ku by putting Du = 0, and then one sees that .
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D(y + uz) = Dy + uDz =F 0, so K is separable over k(y + cz) except possibly for one value of c, and Ku is separable over ku( Y + uz). In what follows, we assume that the constants c 1 , c 2 , are different from the exceptional constant, and hence that K is separable over k(y + ciz) for i = 1 , 2. Assume next that K is regular over k and that the transcendence degree is at least 2. Let Ei k(y + ciz) (i 1 , 2) and let E; be the algebraic closure of Ei in K. We must show that E; Ei for all but a finite number of constants. Note that k(y, z) E 1 E 2 is the compositum of £ 1 and £ 2 , and that k(y, z) has transcendence degree 2 over k. Hence E� and E; are free over k. Being subfields of a regular extension of k, they are regular over k, and are therefore linearly disjoint by Theorem 4. 1 2. .
=
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K
/L� £ '1 (y, z)
E�(y, z)
"' k(y, z)/
E'1 ""' k(y + c 1 z)
k By construction, E� and £2 are finite separable algebraic extensions of E 1 and E 2 respectively. Let L be the separable algebraic closure of k(y, z) in K. There is only a finite number of intermediate fields between k(y, z) and Furthermore, by Proposition 3. 1 the fields E� (y, z) and E2(y, z) are linearly disjoint over k(y, z). Let c 1 range over the finite number of constants which will exhaust the intermediate extensions between and k(y, z) obtainable by lifting over k(y, z) a field of type E; . If c 2 is now chosen different from any one of these constants c 1 , then the only way in which the condition of linear disjointness mentioned above can be compatible with our choice of c 2 is that E;(y, z) = k(y, z), i.e. that E; = k(y + c 2 z). This means that k(y + c2 z) is algebraicall y closed in K, and hence that K is regular over k(y + c2 z). As for Ku , let u 1 , u 2 , be infinitely many elements algebraically indepen dent over K. Let k' = k(u 1 , u 2 , . . . ) and K' = K (u 1 , u 2 , . . . ) be the fields obtained by adjoining these elements to k and K respectively. By what has already been proved, we know that K' is regular over k' (u + uiz) for all but a finite number of integers i, say for i = 1 . Our assertion (a) is then a consequence of Corollary 4. 14. This concludes the proof of Theorem 5.7.
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Theorem 5.8.
Let K = k(x 1 , . . . , x,.) = k(x) be a finitely generated regular extension of a field k. Let u 1 , , u,. be algebraically independent over k(x). Let •
•
•
and let ku = k(u 1 , , u,. , u,. + 1 ). Then ku(x) is separable over ku, and if the transcendence degree of k(x) over k is > 2, then ku(x) is regular over ku. Proof By the separability of k(x) over k, some xi does not lie in K P k, say x,. ¢ K P k. Then we take •
•
•
and so that
u,. + 1
= y + u,. z, and we apply Theorem 5.7 to conclude the proof.
Remark.
In the geometric language of the next chapter, Theorem 5.8 asserts that the intersection of a kvariety with a generic hyperplane
u 1 X 1 + . . . + u,.X,.  un + 1 = 0 is a kuvariety, if the dimension of the kvariety is > 2. In any case, the extension ku(x) is separable over ku .
EX E R C I S ES 1 . Prove that the complex numbers have infinitely many automorphisms. [Hint : Use transcendence bases.] Describe all automorphisms and their cardinality.
2. A subfield k of a field K is said to be algebraically closed in K if every element of K which is algebraic over k is contained in k. Prove : If k is algebraically closed in K, and K , L are free over k, and L is separable over k or 1). is separable over k, then L is algebraically closed in KL.
3. Let k
c
E
c
K be extension fields. Show that
tr. deg. (K/k)
= tr. deg. (K/E) + tr. deg. (E/k).
If {xi } is a transcendence base of Ejk, and { Yi } is a transcendence base of K/E, t hen {xi, Yi } is a transcendence base of K/k. 4. Let K/k be a finitely generated extension, and let K Show that E/k is finitely generated.
:::::>
E
:::::>
k be a subextension.
5. Let k be a field a nd k(x 1 , . . . , x,.) = k(x) a finite separable extension. u 1 , . . . , u,. be algebraically independent over k. Let
Let ku
= k(u 1 ,
•
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)
u,. .
Show that ku(w)
= ku(x).
Let
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6. Let k(x) = k(x 1 , Let uii (i = 1 , . . . ,
•
, X11 ) be a separable extension of transcendence degree r ; j = 1 , . . . , n) be algebraically independent over k(x). Let •
•
Y.l
=
r
> 1.
II
� U lJ· ·X)· " � 1
j=
Let ku = k(u;j)an i,j· (a) Show that ku(x) is separable algebraic over k(y1 , , Yr ). (b) Show that there exists a polynomial P(u) e k [u] having the following prop erty. Let (c) = (c;j) be elements of k such that P (c) =F 0. Let •
Y� I
•
•
II
=
� · · X)· " � Cl}
j= 1
Then k(x) � separable algebraic over k(y' ). 7. Let k be a field and k [x 1 , , X11 ] = R a finitely generated entire ring over k with quotient field k(x). Let L be a finite extension of k(x). Let I be the integral closure of R in L. Show that I is a finite Rmodule. [Use Noether normalization, and deal with the inseparability problem and the separable case in two steps.] •
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8. Let D be a derivation of a field K. Then D 11 : K + K is a linear map. Let P,. = Ker D 11, so P,. is a n additive subgroup of K. An element x e K is called a logarithmic derivative (in K) if there exists y e K such that x = Dyjy. Prove : (a) An element x e K is the logarithmic derivative of an element y e P,. but y ¢ P,.  1 (n > 0) if and o nly if
(D + x)11 ( 1 )
=
0
and
(b) Assume that K = U P,., i.e. given x e K then x e P,. for some n > 0. Let F be a subfield of K s uch that DF c F. Prove that x is a logarithmic derivative in F if a nd o nly if x is a logarithmic derivative in K. [Hint : If x = Dyjy then 1 (D + x) = y  D a y and conversely.] 9. Let k be a field of characteristic 0, and let z 1 , . . . , zr be algebraically independent over k. Let (e;i ), i = 1 , . . . , m and j = 1 , . . . , r be a matrix of integers with r > m, a nd assume that this matrix has rank m. Let
i = 1 , . . . , m.
for
Show that w1 , . . . , wm are algebraically independent over k. [Hint : Consider the Khomomorphism m apping the Kspace of derivations of K/k into K 1 vanishing on (w) , and after taking an irreducible factor we may assume that this polynomial is irreducible , and so is a prime element in the factorial ring R [ W] . Let G(W) E R[W] vanish on (w) . To prove that P divides G , after selecting some irreducible factor of G vanishing on (w) if necessary , we may assume without loss of generality that G is a prime element in R [ W] . One of the variables W; occurs in P(W) , say Wm , so that wm is algebraic over k (w 1 , , wm _ 1 ) . Then (w 1 , , wm _ 1 ) are algebraically independent, and hence Wm also occurs in G . Furthermore , P(w1 , , wm  I , Wm ) is irreducible as a polynomial in k (w 1 , , wm  J )[Wm ] by the Gauss lemma as in Chapter IV, Theorem 2 . 3 . Hence there exists a polynomial H(Wm ) E k (w 1 , , wm  l )[Wm ] such that •
•
•
•
•
•
•
•
•
•
•
•
•
G(W)
=
•
•
H(Wm )P(W) .
Let R ' = R [ w 1 , , wm  d . Then P, G have content 1 as polynomials in R ' [Wm 1 · By Chapter IV Corollary 2 . 2 we conclude that H E R ' [Wm ] = R [ W] , which proves Theorem 2 . 4 . •
•
•
Next w e consider homogeneous ideals and projective space . A polynomial f(X ) E k[X] can be written as a linear combination with monomials M( v)(X ) = X�1 M( v) by
I vi
• • •
=
X�n and c( v)
deg M( v) = �
E
k. We denote the degree of
V; .
If in this expression for f the degrees of the monomials x< v> are all the same (whenever the coefficient c( v) is =I= 0) , then we say that / is a form , or also that f is a homogeneous (of that degree) . An arbitrary polynomial f(X ) in K[X] can also be written
f(X )
=
L t (2) Suppose we have an ascending sequence of submodules of M as above. Let N be the union of all the M i (i = 1 , 2, . . . ). Then N is finitely gen erated, say by elements xb . . . , xr , and each generator is in some M; . Hence there exists an index j such that
41 3
41 4
X, §1
NOETH E R IAN R I N G S AND MODU LES
Then whence equality holds and our implication is proved. (2) => (3) Let N 0 be an element of S. If N 0 is not maximal, it is properly contained in a submodule N 1 • If N 1 is not maximal, it is properly contained in a submodule N 2 • Inductively, if we have found N; which is not maximal, it is contained properly in a submodule N; + 1 • In this way we could construct an infinite chain, which is impossible. (3) => ( 1 ) Let N be a submodule of M. Let a0 E N. If N =F (a 0 ), then there exists an element a 1 E N which does not lie in (a 0 ). Proceeding induc tively, we can find an ascending sequence of submodules of N, namely where the inclusion each time is proper. The set of these submodules has a maximal element, say a submodule (a 0 , a b . . . , ar ), and it is then clear that this finitely generated submodule must be equal to N, as was to be shown. Proposition 1 . 1. Let M be a Noetherian Amodule. Then every submodule
and every factor module of M is Noetherian. Proof. Our assertion is clear for submodules (say from the first condi tion). For the factor module, let N be a submodule and f : M M/N the canonical homomorphism. Let M 1 c M 2 c · · · be an ascending chain of sub modules of M/N and let M; f  1 {M;). Then M 1 c M 2 c · · · is an ascending chain of submodules of M, which must have a maximal element, say M r , so that M; Mr for r > i . Then f(M;) M; and our assertion follows. __..
=
=
=
Proposition 1 .2.
Let M be a module, N a submodule . Assume that N and MjN are Noetherian. Then M is Noetherian. Proof. With every submodule L of M we associate the pair of modules L H (L
n
N, (L + N)/N).
We contend : If E c F are two submodules of M such that their associated pairs are equal, then E = F. To see this, let x E F. By the hypothesis that (E + N)/N = (F + N)/N there exist elements u, v E N and y E E such that y + u = x + v . Then x
y
Since y E E, it follows the x ascending sequence
=
u  vEF nN
E E
=
E n N.
and our contention is proved . If we have an
BAS IC CRITE RIA
X, § 1
41 5
then the associated pairs form an ascending sequence of submodules of N and M/N respectively, and these sequences must stop. Hence our sequence also stops, by our preceding contention. E 1 c E2 Propositions 1 . 1 and 1 .2 may be summarized by saying that in an exact s equence 0 __.. M' __.. M __.. M" + 0, M is Noetherian if and only if M' and M" are Noetherian. •
•
•
Let M be a module, and let N, N' be submodules. If M = N + N' and if both N, N' are Noetherian, then M is Noetherian. A finite direct sum of Noetherian modules is Noetheria n. Proof. We first observe that the direct product N x N ' is Noetherian since it contains N as a submodule whose factor module is isomorphic to N', and Proposition 1 .2 applies. We have a surjective homomorphism Corollary 1 .3.
N
X
N' __.. M
such that the pair (x, x') with x E N and x' E N' maps on x + x' . By Prop osition 1 . 1 , it follows that M is Noetherian . Finite products (or sums) follow by induction. A ring A is called Noetherian if it is Noetherian as a left module over itself. This means that every left ideal is finitely generated. Proposition 1 .4.
Let A be a N oetherian ring and let M be a .finitely generated module. Then M is Noetherian. Proof. Let x
1,
•
•
•
, xn be generators of M. There exists a homomorphism f : A A . . . X A __.. M X
X
of the product of A with itself n times such that This homomorphism is surjective. By the corollary of the preceding proposition, the product is Noetherian, and hence M is Noetherian by Proposition 1 . 1 .
Let A be a ring which is Noetherian, and let qJ : A __.. B be a surjective ringhomomorphism. Then B is Noetherian. c b" c · · · be an ascending chain of left ideals of B Proof. Let b c and let o; = qJ  1 ( b i). Then the o; form an ascending chain of left ideals of A which must stop, say at or . Since qJ{o;) = b i for all i, our proposition is proved. Proposition 1 .5. 1
· · ·
Proposition 1 .6.
Let A be a commutative Noetherian ring, and let S be a multiplicative subset of A. Then s  1 A is Noetherian. Proof. We leave the proof as an exercise.
41 6
X, §2
N OETH E R IAN R I N G S AND MODU LES
Examples .
In Chapter IV , we gave the fundamental examples of Noeth erian rings , namely polynomial rings and rings of power series . The above propositions show how to construct other examples from these , by taking factor rings or modules , or submodules. We have already mentioned that for applications to algebraic geometry , it is valuable to consider factor rings of type k[X]/a , where a is an arbitrary ideal . For this and similar reasons , it has been found that the foundations should be laid in terms of modules , not just ideals or factor rings . Notably , we shall first see that the prime ideal associated with an irreducible algebraic set has an analogue in terms of modules . We shall also see that the decomposition of an algebraic set into irreducibles has a natural formulation in terms of modules , namely by expressing a submodule as an intersection or primary modules . In §6 we shall apply some general notions to get the Hilbert polynomial of a module of finite length, and we shall make comments on how this can be interpreted in terms of geometric notions . Thus the present chapter is partly intended to provide a bridge between basic algebra and algebraic geometry .
§2. A S S O C I AT E D P R I M E S Throughout this section, we let A be a commutative ring. Modules and homo morphisms are Amodules and Ahomomorphisms unless otherwise specified. Proposition 2. 1 . Let S be a multiplicative subset of A, and assume that S
does not contain 0. Then there exists an ideal of A which is maximal in the set of ideals not intersecting S, and any such ideal is prime.
Proof. The existence of such an ideal p follows from Zorn's lemma (the set of ideals not meeting S is not empty, because it contains the zero ideal, and is clearly inductively ordered). Let p be maximal in the set. Let a, b E A, ab E p, but a $ p and b ¢ p. By hypothesis, the ideals (a, p) and { b, p) generated by a and p (or b and p respectively) meet S, and there exist therefore elements s, s' E S, c, c ' , x, x ' E A, p, p ' E p such that s
=
ca
+ xp and s'
=
c
'b + x'p'.
Multiplying these two expressions, we obtain
ss'
=
cc' ab +
p
"
with some p " E p, whence we see that ss' lies in p. This contradicts the fact that p does not intersect S, and proves that p is prime. An element a of A is said to be nilpotent if there exists an integer n that a" = 0.
> 1
such
ASSOC IATED PRIMES
X, §2
41 7
Corollary 2 . 2. An element a of A is nilpotent if and only if it lies in every
prime ideal of A. Proof. If an = 0, then a n E p for every prime p, and hence a E p . If an =F 0 for any positive integer n, we let S be the multiplicative subset of powers of a, namely { 1 , a, a 2 , }, and find a prime ideal as in the proposition to prove the converse. Let a be an ideal of A. The radical of a is the set of all a E A such that an E a for some integer n > 1 , (or equivalently, it is the set of elements a E A whose image in the factor ring A /a is nilpotent). We observe that the radical of a is an ideal, for if an = 0 and bm = 0 then (a + b)k = 0 if k is sufficiently large : In the binomial expansion, either a or b will appear with a power at least equal to n or m. •
•
•
Corollary 2.3 . An element a of A lies in the radical of an ideal a if and only
if it lies in every prime ideal containing a.
Proof. Corollary 2 . 3 is equivalent to Corollary 2 . 2 applied to the ring A/a. We shall extend Corollary 2 . 2 to modules . We first make some remarks on localization. Let S be a multiplicative subset of A. If M is a module, we can define s  1 M in the same way that we defined s  1 A. We consider equivalence classes of pairs (x, s) with x E M and s E S, two pairs (x, s) and (x', s') being equivalent if there exists s 1 E S such that s 1 (s'x  sx') = 0. We denote the equivalence class of (x, s) by xjs, and verify at once that the set of equivalence classes is an additive group (under the obvious operations). It is in fact an Amodule, under the operation
(a, xjs) 1+ axjs.
We shall denote this module of equivalence classes by s  1 M. (We note that s  1 M could also be viewed as an s  1 Amodule.) If p is a prime ideal of A, and S is the complement of p in A, then s  1 M is also denoted by M " . It follows trivially from the definitions that if N __.. M is an injective homo morphism, then we have a natural injection s  1 N __.. s  1 M. In other words, if N is a submodule of M, then s  1 N can be viewed as a submodule of s  1 M. If x E N and s E S, then the fraction xjs can be viewed as an element of s  1 N or s  1 M. If xjs = 0 in s  1 M, then there exists s 1 E s such that s 1 x = 0, and this means that xjs is also 0 in s  1 N. Thus if p is a prime ideal and N is a sub module of M, we have a natural inclusion of N" in M" . We shall in fact identify N" as a submodule of M" . In particular, we see that M" is the sum of its sub modules (Ax)" , for x E M (but of course not the direct sum). Let x E M. The annihilator a of x is the ideal consisting of all elements a E A such that ax = 0. We have an isomorphism (of modules) A/a � Ax
41 8
X, §2
NOETH E R IAN R I N G S AND MODULES
under the map a + ax. Lemma 2.4. Let x be an element of a module M, and let a be its annihilator. Let p be a prime ideal of A. Then ( Ax )" =F 0 if and only if p contains a.
Proof. The lemma is an immediate consequence of the definitions, and will be left to the reader. Let a be an element of A . Let M be a module. The homomorphism x t+
ax,
XEM
will be called the principal homomorphism associated with a, and will be de noted by aM . We shall say that aM is locally nilpotent if for each x E M there exists an integer n (x) > 1 such that an <x >x = 0. This condition implies that for every finitely generated submodule N of M, there exists an integer n > 1 such that an N = 0 : We take for n the largest power of a annihilating a finite set of generators of N. Therefore, if M is finitely generated, aM is locally nilpotent if a nd only if it is n ilpo ten t. Proposition 2 . 5 . L e t M be a module, a E A. Then aM is locally nilpotent if and only if a lies in every prime ideal p such that M " =F 0.
Proof. Assume that aM is locally nilpotent. Let p be a prime of A such that M" =F 0. Then there exists x E M such that ( A x )" =F 0. Let n be a positive integer such that anx = 0. Let a be the annihilator of x. Then an E a, and hence we can apply the lemma, and Corollary 4.3 to conclude that a lies in every prime p such that M" =F 0. Conversely , suppose aM is not locally nilpotent , so there exists x E M such that an x = 0 for all n > 0 . Let S = { 1 , a , a2 , . . . }, and using Proposition 2 . 1 let p be a prime not intersecting S . Then (Ax) P =I= 0 , so MP =I= 0 and a ¢. p , as desired . Let M be a module. A prime ideal p of A will be said to be associated with M if there exists an element x E M such that p is the annihilator of x. In par ticular, since p =F A, we must have x =F 0. Proposition 2.6. Let M be a module =F 0. Let p be a max im a l element in the set of ideals which are annihilators of elements x E M, x =F 0. Then p is prime.
Proof. Let p be· the annihilator of the element x =F 0. Then p =F A. Let a, b E A, ab E p, a $ p. Then ax =F 0. But the ideal (b, p) annihilates ax, and contains p. Since p is maximal, it follows that b E p, and hence p is prime. Corollary 2. 7.
If A is Noetherian and M is a module =I= 0, then there exists a prime associated with M. Proof. The set of ideals as in Proposition 2 . 6 is not empty since M =t= 0 , and has a maximal element because A is Noetherian .
X, §2
ASSOC IATED PRIMES
41 9
Corollary 2.8. Assume that both A and M are Noetherian, M =F 0. Then
there exists a sequence of submodules M
=
M 1 ::J M 2 ::J
• • •
::J Mr = 0
that each factor module M i/Mi + 1 is isomorphic to such . przme P i ·
A /P i
for some
Proof. Consider the set of submodules having the property described in the corollary. It is not empty, since there exists an associated prime p of M, and if p is the annihilator of x, then Ax � Ajp. Let N be a maximal element in the set. If N =F M, then by the preceding argument applied to M/N, there exists a submodule N ' of M containing N such that N '/N is isomorphic to A /p for some p, and this contradicts the maximality of N. Proposition 2. 9. Let A be Noetherian, and a E A. Let M be a module.
Then aM is injective if and only if a does not lie in any associated prime of M.
Proof. Assume that aM is not injective, so that ax = 0 for some x E M, x =I= 0 . By Corollary 2 . 7 , there exists an associated prime p of Ax, and a is an element of p. Conversely, if aM is injective, then a cannot lie in any associated prime because a does not annihilate any nonzero element of M. Proposition 2. 10. Let A be Noetherian, and let M be a module. Let a E A.
The following conditions are equivalent : (i) aM is locally nilpotent.
(ii) a lies in every associated prime of M. (iii ) a lies in every prime p such that M " =F 0. If p is a prime such that MP =I= 0, then p contains an associated prime of M.
Proof. The fact that (i) implies ( ii ) is obvious from the definitions, and does not need the hypothesis that A is Noetherian. Neither does the fact that (iii) implies (i) , which has been proved in Proposition 2. 5 . We must therefore prove that (ii) implies (iii) which is actually implied by the last statement. The latter is proved as follows . Let p be a prime such that Mp =I= 0 . Then there exists x E M such that (Ax)p =t= 0 . By Corollary 2 . 7 , there exists an associated prime q of (Ax)., in A . Hence there exists an element y Is of (Ax)., , with y E Ax, s ¢. p , and y Is =t= 0 , such that q is the annihilator of y Is. It follows that q c p, for otherwise , there exists b E q , b ¢. p , and 0 = by Is , whence y Is = 0 , contra , bn be generators for q. For each i , there exists s; E A , diction . Let b 1 , s; ¢. p , such that s; b;y = 0 because b;y l s = 0 . Let t = s 1 • • • sn . Then it is trivially verified that q is the annihilator of ty in A. Hence q c p, as desired. •
•
•
Let us define the support of M by supp(M) = set of primes p such that M" =F 0.
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X, §2
NOET H E R IAN R I N G S AN D MODU LES
We also have the annihilator of M, ann(M)
set of elements a E A such that aM
=
=
0.
We use the notation ass(M)
=
set of associated primes of M.
For any ideal a we have its radical, rad(a)
=
set of elements a E A such that a n E a for some integer n
> 1.
Then for finitely generated M, we can reformulate Proposition 2 . 1 0 by the following formula: rad(ann(M))
= p
n
e s upp( M )
p
= p
n p.
e ass( M )
Corollary 2. 1 1 . Let A be Noetherian, and let M be a module . The following
conditions are equivalent : (i) There exists only one associated prime of M. (i i) We have M =F 0, and for every a E A, the homomorphism aM is injective, or locally nilpotent. If these conditions are satisfied, then the set of elements a E A such that aM is locally nilpotent is equal to the associated prime of M. Proof. Immediate consequence of Propositions 2 . 9 and 2 . 1 0 . Proposition 2. 12. Let N be a submodule of M. Every associated prime of
N is associated with M also. An associated prime of M is associated with N or with MjN.
Proof. The first assertion is obvious. Let p be an associated prime of M, and say p is the annihilator of the element x =F 0. If Ax n N 0, then Ax is isomorphic to a submodule of M /N , and hence p is associated with M /N. Suppose Ax n N =I= 0 . Let y ax E N with a E A and y =t= 0 . Then p annihilates y. We claim p ann(y) . Let b E A and by 0 . Then ba E p but a ¢. p , so b E p . Hence p is the annihilator of y in A , and therefore is associated with N, as was to be shown . =
=
=
=
X, §3
§3.
PRIMARY D ECOM POS ITION
42 1
P R I MARY D ECOM POSITION
We continue to assume that A is a commutative ring, and that modules (resp. homomorphisms) are Amodules (resp. Ahomomorphisms), unless otherwise specified. Let M be a module. A submodule Q of M is said to be primary if Q =F M, and if given a E A, the homomorphism aM; Q is either injective or nilpotent. Viewing A as a module over itself, we see that an ideal q is primary if and only if it satisfies the following condition : Given a, b E A, ab E q and a ¢ q, then bn E q for some n > 1 . Let Q be primary. Let p be the ideal of elements a E A such that aM I Q is nilpotent. Then p is prime. Indeed, suppose that a, b E A, ab E p and a ¢ p. Then aM; Q is injective, and consequently a lt 1 Q is injective for all n > 1 . Since (ab)M/ Q is nilpotent, it follows that bM ; Q must be nilpotent, and hence that b E p, proving that p is prime. We shall call p the prime belonging to Q , and also say that Q is p primary . We note the corresponding property for a primary module Q with prime p: Let b
A and x E M be such that bx E Q. If x ¢. Q then b
p. Examples . Let m be a maximal ideal of A and let q be an ideal of A such that m k C q for some positive integer k. Then q is primary , and m belongs to q . We leave the proof to the reader. The above conclusion is not always true if m is replaced by some prime ideal p . For instance , let R be a factorial ring with a prime element t. Let A be the subring of polynomials /(X) E R[X] such that E
E
a0 + a 1 X + . . . with a 1 divisible by t. Let p = ( tX, X 2 ) . Then p is prime but p 2 = ( t zx z, tX 3, X 4 ) /(X)
=
is not primary, as one sees because X 2 ¢ p 2 but t k ¢ p 2 for all k > 1 , yet t2 X 2 E p 2 . Proposition 3. 1 . Let M be a trzodule, and Q 1 , • • • , Q r submodules which are pprimary for the same prime p. Then Q 1 n · · · n Q r is also pprimary. Proof. Let Q = Q 1 n · · · n Q r . Let a E p. Let n i be such that (aM 1Q )n i = 0
for each i = 1, . . . , r and let n be the maximum of n . , . . . , nr . Then a lt 1Q = 0, so that aM ; Q is nilpotent. Conversely, suppose a ¢ p. Let x E M, x ¢ Qi for some j. Then anx ¢ Q i for all positive integers n, and consequently aM /Q is injective. This proves our proposition.
422
NOETHER IAN R I N G S AN D MODU LES
X, §3
Let N be a submodule of M. When N is written as a finite intersection of primary submodules, say N = Q1 n
· · ·
n
Qr ,
we shall call this a primary decomposition of N. Using Proposition 3 . 1 , we see that by grouping the Q i according to their primes, we can always obtain from a given primary decomposition another one such that the primes belonging to the primary ideals are all distinct. A primary decomposition as above such that the prime ideals p 1 , • • • , Pr belonging to Q 1 , , Q r respectively are distinct, and such that N cannot be expressed as an intersection of a proper subfamily of the primary ideals { Q 1 , • . • , Qr } will be said to be reduced. By deleting some· of the primary modules appearing in a given decomposition, we see that if N admits some primar y decomposition, then it admits a reduced one. We shall prove a result giving certain uni q ueness properties of a reduced primary decomposition. Let N be a submodule of M and let x � i be the canonical homomorphism. Let Q be a submodule of M = M / N and let Q be its inverse image in M. Then directly from the definition , one sees that Q is primary if and only if Q is primary; and if they are primary , then the prime belonging to Q is also the prime belonging to Q . Furthermore , if N = Q 1 n . . . n Qr is a primary decomposition of N in M, then .
.
•
(0) = Q l n . . . n Q r is a primary decomposition of (0) in M, as the reader will verify at once from the definitions . In addition , the decomposition of N is reduced if and only if the decomposition of (0) is reduced since the primes belonging to one are the same as the primes belonging to the other. Let Q 1 n n Q r = N be a reduced primary decomposition, and let P i belong to Qi . If P ; does not contain p ; (j 1= i) then we say that P i is isolated . The isolated primes are therefore those prim es which are minimal in the set of primes belonging to the pri m ary modules Q i . ·
·
·
Theorem 3.2. Let N be a submodule of M, and let · · ·
N = Q1 n n Q r = Q'1 n n Q� be a reduced primary decomposition of N. Then r = s. The set of primes belonging to Q b . , Qr and Q '1 , . . . , Q; is the same. If {p h . . . , Pm } is the set of isolated primes belonging to these decompositions, then Qi = Q� for i = 1 , . , m, in other words, the primary modules corresponding to isolated primes are uniquely determined. .
.
· · ·
.
.
Proof. The uniqueness of the number of terms in a reduced decomposition and the uniqueness of the family of primes belonging to the primary components will be a consequence of Theorem 3 . 5 below .
X, §3
P R IMARY DECOMPOS ITION
423
There remains to prove the uniqueness of the primary module belonging to an isolated prime, say p 1 • By definition, for each j = 2, . . . , r there exists a i E p i and a i ¢ p 1 • Let a = a 2 · · · ar be the product. Then a E pi for all j > 1, but a ¢ p 1 • We can find an integer n > 1 such that alt 1 Qi = 0 for j = 2, . . . , r. Let N 1 = set of x E M such that a"x E N. We contend that Q 1 = N 1 . This will prove the desired uniqueness . Let x E Q 1 • Then a"x E Q 1 n · · · n Q r = N, so x E N 1 • Conversely, let x E N 1 , so that a"x E N, and in particular a" x E Q 1 • Since a ¢ p 1 , we know by definition that aM ; Q 1 is injective. Hence x E Q 1 , thereby proving our theorem. Theorem 3.3. Let M be a Noetherian module. Let N be a submodu le of
M. Then N admits a primary decomposition.
Proof. We consider the set of submodules of M which do not admit a primary decomposition. If this set is not empty, then it has a maximal element because M is Noetherian. Let N be this maximal element. Then N is not primary, and there exists a E A such that aM ; N is neither injective nor nilpotent. The increasing sequence of modules Ker aM ; N c Ker ait 1N c Ker a �;N c · · · stops, say at a� ; N · Let lfJ : MjN __.. M/N be the endomorphism lfJ = a�;N · Then Ker qJ 2 = Ker lfJ · Hence 0 = Ker lfJ n Im lfJ in MjN , and neither the kernel nor the image of cp is 0 . Taking the inverse image in M, we see that N is the intersection of two submodules of M, unequal to N. We conclude from the maximality of N that each one of these submodules admits a primary de composition, and therefore that N admits one also, contradiction. We shall conclude our discussion by relating the primes belonging to a primary decomposition with the associated primes discussed in the previous section. Proposition 3.4. Let A and M be Noetherian. A submodule Q of M is
primary if and only if M/Q has exactly one associated prime p, and in that case, p belongs to Q, i.e. Q is pprimary. Proof. Immediate consequence of the definitions, and Corollary 2 . 1 1 .
Theorem 3.5. Let A and M be Noetherian. The associated primes of M
are precisely the primes which belong to the primary modules in a reduced primary decomposition of 0 in M. In particular, the set of associated primes of M is finite. Proof. Let
0
=
Q1 n
···
n
Qr
424
X , §4
NOETHERIAN RINGS AND MODULES
be a reduced primary decomposition of 0 in M. We have an injective homo morphism r M + EBM/Q; . i= 1
By Proposition 2 . 1 2 and Proposition 3 . 4 , we conclude that every associated prime of M belongs to some Q; . Conversely, let N = Q2 n · · · n Q r . Then N =F 0 because our decomposition is reduced. We have
Hence N is isomorphic to a submodule of M/Q 1 , and consequently has an associated prime which can be none other than the prime p 1 belonging to Q 1 • This proves our theorem. Theorem 3.6. Let A be a Noetherian ring. Then the set of divisors of zero
in A is the settheoretic union of all primes belonging to primary ideals in a reduced primary decomposition of 0 . Proof. An element of a E A is a divisor of 0 if and only if aA is not injective. According to Proposition 2. 9, this is equivalent to a lying in some associated prime of A (viewed as module over itself) . Applying Theorem 3 . 5 concludes the proof.
§4.
N A KAY A M A ' S L E M M A
We let A denote a commutative ring, but not necessarily Noetherian . When dealing with modules over a ring , many properties can be obtained first by localizing , thus reducing problems to modules over local rings . In practice , as in the present section , such modules will be finitely generated . This section shows that some aspects can be reduced to vector spaces over a field by reducing modulo the maximal ideal of the local ring . Over a field , a module always has a basis . We extend this property as far as we can to modules finite over a local ring. The first three statements which follow are known as Nakayama's lemma . Lemma 4 . 1 . Let a be an ideal ofA which is contained in every maximal ideal
of A . Let E = {0}.
E
be a finitely generated Amodule. Suppose that a E
= E.
Then
NAKAYAMA'S LEMMA
X, §4
425
, xs be Proof. Induction on the number of generators of E. Let x generators of E. By hypothesis, there exist elements a 1 , , as E a such that 1,
•
•
•
•
•
•
so there is an element a (namely as) in a such that ( 1 + a)xs lies in the module generated by the first s  1 generators. Furthermore 1 + a is a unit in A, otherwise 1 + a is contained in some maximal ideal, and since a lies in all maximal ideals, we conclude that 1 lies in a maximal ideal, which is not possible. Hence xs itself lies in the module generated by s  1 generators, and the proof is complete by induction. a
Lemma 4 . 1 applies in particular to the case when A is a local ring, and = m is its maximal ideal. Lemma 4.2. Let A be a local ring, let E be a .finitely generated Amodule, and F a submodule. If E = F + mE, then E = F.
Proof. Apply Lemma 4 . 1 to E/F. Lemma 4.3. If X ] ,
.
.
•
Let A be a local ring. Let E be a finitely generated A module. , Xn a re generators for E mod m E then they are g en era to rs fo r E. ,
Proof. Take F to be the submodule generated by x 1 ,
•
•
•
, xn .
Theorem 4.4. Let A be a local ring and E a finite projective Amodule.
Then E is free. In fact, if x 1 , , xn are elements of E whose residue classes .X 1 , , xn are a basis of E/mE over Ajm , then x 1 , , xn are a basis of E over A. If x 1 , . . . , xr are such that .X 1 , , xr are linearly independent over A/m, then they can be completed to a basis of E over A. •
•
•
•
•
.
•
.
.
•
•
•
P ro of I am indebted to George Bergman for the following proof of the first statement. Let F be a free module with basis e 1 , , en, and let f: f., � E be the homomorphism mapping e ; to x;. We want to prove that f is an isomor phism. By Lemma 4.3 , f is surjective. Since E is projective, it follows that f splits, i.e. we can write F = Po EB Ph where Po = Ker f and P1 is mapped isomorphically onto E by f Now the linear independence of x1 , , Xn mod mE shows that •
•
•
•
Po
c
mE
=
mPo
c
mP 1
•
•
•
Hence P0 C mPo Also, as a direct summand in a finitely generated module, Po is finitely generated. So by Lemma 4.3, Po = (0) and f is an isomorphism, as was to be proved. As to the second statement, it is immediate since we can complete a given
426
NOETH E R IAN R I NG S AND MODULES
X, §5
sequence x b . . . , Xr with x . , . . . , Xr linearly independent over A im , to a sequence x1 , • • • , Xn with x 1 , • • • , Xn lineary independent over Aim , and then we can apply the first part of the proof. This concludes the proof of the theorem . Let E be a module over a local ring A with maximal ideal m. We let E(m) = E/mE. I f f : E __.. F is a homomorphism, then f induces a homo morphism fc m > : E(m) __.. F(m). If f is surjective, then it follows trivially that fc m > is surjective. Proposition 4.5. Let f : E
__..
F be a homomorphism of modules, finite over a
local ring A. Then : (i) If h m ) is surjective, so is f. (ii) Assume f is injective. If h m> is surjective, then f is an isomorphism. (iii) Assume that E, F are free. Iffc m > is injective (resp. an isomorphism) then f is injective (resp. an isomorphism). Proof. The proofs are immediate consequences of Nakayama's lemma and will be left to the reader. For instance, in the first statement, consider the exact sequence E
__..
F
__..
F/ Im f
__..
0
and apply Nakayama to the term on the right . In (iii) , use the lifting of bases as in Theorem 4 . 4 .
§5.
F I LT E R E D A N D G R A D E D M O D U L E S
Let A be a commutative ring and E a module. By a filtration of E one means a sequence of submodules E
=
E0 ::J E 1 ::J E 2 ::J • • • ::J En ::J • • •
Strictly speaking, this should be called a descending filtration. We don't consider any other. Example.
Let a be an ideal of a ring A, and E an Amodule. Let
Then the sequence of submodules {En } is a filtration. More generally, let {En } be any filtration of a module E. We say that it is an afiltration if aEn c En + 1 for all n. The preceding example is an afiltration.
FILTE RED AND G RADED MODULES
X , §5
427
We say that an afiltration is astable, or stable if we have aEn = En + 1 for all n sufficiently large. Proposition 5. 1 . Let {En } and {£�} be stable afiltrations of E. Then there
exists a positive integer d such that for all n > 0.
Proof. It suffices to prove the proposition when E� = an E. Since a En c En + for all n, we have an £ c En . By the stability hypothesis, there exists d such that 1
which proves the proposition. A ring A is called graded (by the natural numbers) if one can write A as a direct sum (as abelian group) ,
m
such that for all integers m, n > 0 we have A n A c A n + m · It follows in par ticular that A 0 is a subring, and that each component A n is an A 0 module. Let A be a graded ring. A module E is called a graded module if E can be expressed as a direct sum (as abelian group)
such that A n Em c En + m · In particular, E, is an A 0 module. Elements of En are then called homogeneous of degree n. By definition, any element of E can be written uniquely as a finite sum of homogeneous elements. Example.
Let k be a field, and let X 0 , . . . , Xr be independent variables. The polynomial ring A = k[ X 0 , , Xr] is a graded algebra, with k = A0 • The homogeneous elements of degree n are the polynomials generated by the monomials in X 0 , Xr of degree n, that is •
.
•
•
•
•
,
X� · · · X�.. with
r
L d i = n.
i=O
An ideal I of A is called homogeneous if it is graded, as an Amodule. If this is the case, then the factor ring A/I is also a graded ring. Proposition 5 . 2 . Let A be a graded ring. Then A is Noetherian if and only
if A 0 is Noetherian, and A is finitely generated as A 0 algebra.
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NOETH E R IAN R I N G S AND MODULES
X, §5
Proof. A finitely generated algebra over a Noetherian ring is Noetherian , because it is a homomorphic image of the polynomial ring in finitely many variables, and we can apply Hilbert's theorem. Conversely, suppose that A is Noetherian. The sum
is an ideal of A, whose residue class ring is A 0 , which is thus a homomorphic image of A, and is therefore Noetherian. Furthermore, A + has a finite number of generators x 1 , , xs by hypothesis. Expressing each generator as a sum of homogeneous elements, we may assume without loss of generality that these generators are homogeneous, say of degrees d 1 , , ds respectively, with all d; > 0. Let B be the subring of A generated over A 0 by x 1 , , X5 • We claim that A n c B for all n. This is certainly true for n = 0. Let n > 0. Let x be homogeneous of degree n. Then there exist elements a; E A n  d, such that s x = L a; x; . 1 •
.
.
•
•
•
•
•
•
i=
Since d; > 0 by induction, each a; is in A 0 [x 1 , also, and concludes the proof.
.
•
.
, X5] = B, so this shows x E B
We shall now see two ways of constructing graded rings from filtrations. First, let A be a ring and a an ideal. We view A as a filtered ring, by the powers an . We define the first associated graded ring to be 00
Sa(A) = S = EB
an .
n=O
Similarly, if E is an Amodule, and E is filtered by an afiltration, we define
Then it is immediately verified that Es is a graded Smodule. Observe that if A is Noetherian, and a is generated by elements x 1 , , xs then S is generated as an A algebra also by x 1 , , xs , and is therefore also Noetherian. •
•
•
•
•
•
Lemma 5 .3. Let A be a Noetherian ring, and E a finitely generated module,
with an afiltration. Then Es is finite over S if and only if the filtration of E is astable. Proof. Let
n
Fn = ffi E; , i=O
FILTERED AND G RADED MODULES
X , §5
429
and let
Then G n is an Ssubmodule of E5 , and is finite over S since Fn is finite over A. We have Since S is Noetherian, we get : E5 is finite over S E5 = G N for some N E N + m = a m E N for all m > 0
the filtration of E is astable. This proves the lemma. Theorem 5.4. (ArtinRees).
Let A be a Noetherian ring, a an ideal, E a .finite Amodule with a stable afiltration. Let F be a submodule, and let Fn = F n En . Then { Fn } is a stable afiltration of F. Proof. We have a(F
n
E,.)
c
aF
n
aE,
c
F
n
En + I '
so {F, } is an afiltration of F. We can then form the associated graded Smodule fS , which is a submodule of E5 , and I S finite over S since S is Noetherian. We apply Lemma 5 . 3 to conclude the proof. We reformulate the ArtinRees theorem in its original form as follows. Corollary 5.5. Let A be a Noetherian ring, E a finite Amodule, and F a
submodule. Let a be an ideal. There exists an integer s such that for all integers n > s we have a" E n F = a"  s(a5E n F). Proof. Special case of Theorem 5 . 4 and the definitions . Theorem 5.6. (Krull).
Let A be a Noetherian ring, and let a be an ideal contained in every maxitnal ideal of A. Let E be a finite A module. Then 00
n a" E
n= l
Proof. Let F proof.
=
=
0.
n a" E and apply Nakayama' s lemma to conclude the
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NOETH E R IAN R I N G S AND MODU LES
Corollary 5. 7. Let o be a local Noetherian ring with maximal ideal m . Then 00
n
n= l
m"
=
0.
Proof. Special case of Theorem 5 .6 when E
=
A.
The second way of forming a graded ring or module is done as follows. Let A be a ring and a an ideal of A. We define the second associated graded ring 00
EB a"fa" + 1 • n=O Multiplication is defined in the obvious way. Let a E a" and let a denote its residue class mod a n + 1 . Let b E am and let 5 denote its residue class !TIOd a m + 1 . We define the product a5 to be the residue class of ab mod a m + n + 1 • It is easily verified that this definition is independent of the choices of representatives and defines a multiplication on gr0(A) which makes gr0(A) into a graded ring. Let E be a filtered Amodule. We define g r a(A)
=
00
EB E,./En + 1 · n=O If the filtration is an afiltration, then gr(E) is a graded gr0(A)module. gr( E )
=
Proposition 5. 8. Assume that A is Noetherian, and let a be an ideal of A.
Then gr0(A) is Noetherian. If E is a finite Amodule with a stable afiltration, then gr(E) is a finite gr0(A) module . Proof·. Let x 1 , in aja 2 • Then
•
•
•
, xs be generators of a. Let x i be the residue class of x i
is Noetherian, thus proving the first assertion. For the second assertion, we have for some d, for all m > 0. Hence gr(E) is generated by the finite direct sum gr(£) 0 (!)
· · ·
(f) gr(E)d .
But each gr(E),. = E,./E,. + 1 is finitely generated over A, and annihilated by a, so is a finite A/amodule. Hence the above finite direct sum is a finite A/a module, so gr(E) is a finite gr0(A)module, thus concluding the proof of the proposition.
TH E H I LBERT POLYNOM IAL
X, §6
431
§6. T H E H I L B E R T P O L Y N O M I A L The main point of this section is to study the lengths of certain filtered modules over local rings , and to show that they are polynomials in appropriate cas es. However, we first look at graded modules, and then relate filtered modules to graded ones by using the construction at the end of the preceding section. We start with a graded Noetherian ring together with a finite graded Amodule E, so oc
and E
=
ffi
n= O
En .
We have seen in Proposition 5 . 2 that A 0 is Noetherian, and that A is a finitely generated A 0 algebra. The same type of argument shows that E has a finite number of homogeneous generators , and En is a finite A 0 module for all n > 0 . Let cp be an EulerPoincare Zvalued function on the class of all finite A 0 modules , as in Chapter III , §8 . We define the Poincare series with respect to cp to be the power series PqJ(E,
t)
00
=
L qJ (En )t
n
n=O
E
Z[[t]].
We write P(E, t) instead of PqJ(E, t) for simplicity. Theorem 6. 1 . ( Hi l bert  Serre ).
A 0 algebra. Then
P(E,
t) is
a
Let s be the number of generators of A as rational function of type
P(E,
t)
=
•
f(t)
n (1
i= 1

td i)
with suitable positive integers d i , and f(t) E Z[t]. Proof. Induction on s. For s = 0 the assertion is trivially true . Let s > 1 . , x5] , deg . X; = d; > 1 . Multiplication by X5 on E gives rise Let A = A 0 [x 1 , to an exact sequence •
Let
•
•
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NOETH E R IAN R I NGS AN D MODULES
Then K, L are finite Amodules (being submodules and factor modules of E) , and are annihilated by X5, so are in fact graded A 0 [x 1 , , x5_ dmodules . By definition of an EulerPoincare function , we get .
•
.
Multiplying by tn + ds and summing over n, we get ( 1  t ds)P(E, t)
=
P(L, t)  t d sP(K, t) + g(t),
where g(t) is a polynomial in Z[t]. The theorem follows by induction.
. . ' xsJ then d; deg as shown . in the proof. The next result shows what happens when all the degrees are Remark . In Theorem 6. 1 ' if A
=
A o [X 1 '
=
X;
equal to 1 . Theorem 6.2. Assume that A is generated as an A0algebra by homogeneous
elements of degree 1 . Let d be the order of the pole of P( E, t) at t 1 . Then for all sufficiently large n, ({J (En ) is a polynomial in n of degree d  1 . (For this statement, the zero polynomial is assumed to have degree  1 .) Proof. By Theorem 6. 1 , cp(En ) is the coefficient of t n in the rational function =
P(E, t)
f{t)/( 1  t)s .
 t, we write P(E, t)
Cancelling powers of 1 h(t) E Z[t]. Let
=
h(t)
For convenience we let

t
)d
(_�)
=
=
h(t)/( 1  t)d, and h( 1 )
I=
=
L ak tk . k=O
(
)
� d+k 1 k t. � d 1
k O

0 for n > 0 and
(_�)
=
l for n
=
 1 . We
then get for all n > m. The sum on the righthand side is a polynomial in n with leading term nd  1 0. (ii) For all sufficiently large n, this length is a polynomial g(n) of degree < s, where s is the least number of generators of q. ( iii) The degree and leading coefficient of g(n) depend only on E and q, but not on the chosen filtration. Proof. Let G
=
grq(A)
=
EB q njqn + 1 .
Then gr (E) = EB En/E n + 1 is a graded Gmodule, and G 0 = Ajq. By Proposition 5.8, G is Noetherian and gr(E) is a finite Gmodule. By the remarks preceding the theorem, E/En has finite length, and if qJ denotes the length, then n qJ(E/En ) = L qJ(Ej  1 /Ej). j= 1 If x 1 , , X5 generate q , then the images .X 1 , , xs in q jq 2 generate G as A /q algebra, and each x i has degree 1. By Theorem 6.2 we see that •
•
•
is a polynomial in n of degree < s
•

•
•
1 for sufficiently large n. Since
it follows by Lemma 6 .4 below that cp (E / En ) is a polynomial g(n) of degree < s for all large n . The last statement concerning the independence of the degree
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NOETH E R IAN R I NG S AND MODULES
X, §6
of g and its leading coefficient from the chosen filtration follows immediately from Proposition 5 . 1 , and will be left to the reader. This concludes the proof. From the theorem, we see that there is a poly n omial X.E , q such that for all sufficiently large n. If E = A, the n XA . q is usually called the characteristic polynomial of q In particular, we see that 0
for all sufficiently large no For a continuation of these topics into dimension theory , see [AtM 69] and [Mat 80] . We shall now study a particularly important special case having to do with polynomial ideals . Let k be a field , and let A = k[X0 , . . . , XN]
be the polynomial ring in N + 1 variable . Then A is graded , the elements of degree n being the homogeneous polynomials of degree n . We let a be a homo geneous ideal of A , and for an integer n > 0 we define: cp(n) =
dim k A n cp(n , a ) = dim k a n x (n , a ) = dim k A n/an = dim k A n  dimk a n = cp(n)  cp(n, a ) . As earlier in this section , A n denotes the kspace of homogeneous elements of degree n in A , and similarly for a n . Then we have cp(n) =
(N ) . + n
N
We shall consider the binomial polynomial (1)
( 7)) d
T(T  1 ) =
o
o
(T  d + 1 ) T d = + lower terms . d! d! o
Iff is a function , we define the difference function llf by llf(T) = f(T + 1 )  f(T) . Then one verifies directly that (2)
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435
Lemma 6.4. Let P E Q[T] be a polynomial of degree d with rational
coefficients.
(a) If P(n) E Z for all sufficiently large integers n, then there exist integers c0 , . . . , cd such that
In particular, P(n) E Z for all integers n . (b) Iff: Z � Z is any function, and if there exists a polynomial Q(T) E Q[T] such that Q(Z) C Z and df(n) = Q(n) for all n sufficiently large, then there exists a polynomial P as in (a) such thatf(n) = P(n)for all n sufficiently large. Proof. We prove (a) by induction . If the degree of P is 0 , then the assertion is obvious . Suppose deg P > 1 . By ( 1 ) there exist rational numbers c0 , . . . , cd such that P(D has the expression given in (a) . But dP has degree strictly smaller than deg P . Using (2) and induction , we conclude that c0, . . . , cd  I must be integers . Finally cd is an integer because P(n) E Z for n sufficiently large . This proves (a) . As for (b) , using (a) , we can write Q(T) = c0
(d 1 ) T

+ .. . +
cd  I
with integers c0, . . . , cd  I · Let P I be the "integral" of Q , that is
Then d ( f  P I )(n) = 0 for all n sufficiently large . Hence (f  P I )(n) is equal to a constant cd for all n sufficiently large , so we let P = P I + cd to conclude the proof. Proposition 6.5 .
Let a , b be homogeneous ideals in A . Then
cp(n, a + b ) = cp(n , a ) x (n, a + b ) = x (n, a )
+ +
cp(n, b )  cp(n , x (n, b )  x (n,
a n b)
a n b).
Proof. The first is immediate , and the second follows from the definition of X ·
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Theorem 6.6. Let F be a homogeneous polynomial of degree d. Assume that
F is not a divisor of zero mod a, that is : if G
x N then En = En + 1 • Example 1 . If k is a field, A is a kalgebra, and E is a finitedimen sio nal vector space over k which is also an Amodule, then E is Artinian as well as Noetherian. Example 2.
Let A be a commutative Noetherian local ring with maximal ideal m, and let q be an mprimary ideal. Then for every positive integer n, A /q n is Artinian. Indeed, A jq n has a JordanHolder filtration in which each factor is a finite dimensional vector space over the field A/m, and is a module of finite length. See Proposition 7 . 2 . Conversely, suppose that A is a local ring which is both Noetherian and Artinian. Let m be the maximal ideal. Then there exists some positive integer n such that mn = 0. Indeed, the descending sequence m n stabilizes, and Nakayama's lemma implies our assertion. It then also follows that every primary ideal is nilpotent. As with Noetherian rings and modules, it is easy to verify the following statements : Proposition 7 . 1 . Let A be a ring, and let 0 + E'
+
E
+
E"
+
0
be an exact sequence of Amodules. Then E is Artinian if and only if E' and E" are Artinian. We leave the proof to the reader. The proof is the same as in the Noetherian case, reversing the inclusion relations between modules. Proposition 7 . 2 . A module E has a finite simple filtration if and only if E
is both Noetherian and Art in ian.
Proof. A simple module is generated by one element, and so is Noetherian . Since it contains no proper submodule =I= 0 , it is also Artinian . Proposition 7 . 2 is then immediate from Proposition 7 . 1 . A module E is called decomposable if E can be written as a direct sum
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NOET H E R IAN R I N G S AND MODULES
X, §7
with E 1 =F E and E 2 =F E. Otherwise, E is called indecomposable . If E is decomposable as above, let e 1 be the projection on the first factor, and e 2 = 1 e 1 the projection on the second factor. Then e b e 2 are idempotents such that 
Conversely, if such idempotents exist in End( E) for some module E, then E is decomposable, and e; is the projection on the submodule e; E. Let u : E + E be an endomorphism of some module E. We can form the descending sequence 2 Im u => Im u ::J Im u 3 ::J • • • If E is Artinian, this sequence stabilizes, and we have for all sufficiently large n. We call this submodule u00(E), or Im U 00 • Similarly, we have an ascending sequence Ker u c Ker u 2 c Ker u 3 c . . . which stabilizes if E is Noetherian, and in this case we write Ker uoo
=
Ker un
for n sufficiently large.
Proposition 7 .3. (Fitting's Lemma). Assume that E is Noetherian and
Artinian. Let u E End(E). Then E has a direct sum decomposition
Furthermore, the restriction ofu to Im U00 is an automorphism, and the restric tion of u to Ker U00 is nilpotent. Proof. Choose n such that Im u 00 Im u n and Ker u00 Ker u n . We have =
=
Im
U 00
n Ker
U 00
=
{0},
for if x lies in the intersection, then x = u n(y) for some y E E, and then 0 = un (x ) = u 2 n (y). So y E Ker u 2 n = Ker un , whence X = u n(y) = 0. Secondly, let x E E. Then for some y E un( E ) we have
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441
Then we can write which shows that E = Im U 00 + Ker U 00 • Combined with the first step of the proof, this shows that E is a direct sum as stated. The final assertion is immediate, since the restriction of u to Im u00 is sur jective, and its kernel is 0 by the first part of the proof. The restriction of u to Ker U 00 is nilpotent because Ker U 00 = Ker un . This concludes the proof of the pro position. We now generalize the notion of a local ring to a noncommutative ring. A ring A is called local if the set of nonunits is a twosided ideal. Proposition 7 .4. Let E be an indecomposable module over the ring A. Assume
E Noetherian and Art in ian. Any endomorphism of E is either nilpotent or an automorphism. Furthermore End(£) is local. Proof. By Fitting's lemma, we know that for any endomorphism u, we have E = lm u 00 or E = Ker u 00 • So we have to prove that End(£) is local. Let u be an endomorphism which is not a unit, so u is nilpotent. For any endomorphism v it follows that uv and vu are not surjective or injective respec tively, so are not automorphisms. Let u 1 , u 2 be endomorphisms which are not units. We have to show u 1 + u 2 is not a unit. If it is a unit in End(£), let vi = u i (u 1 + u 2 )  1 • Then v 1 + v 2 = 1. Furthermore, v 1 = 1 v 2 is invertible by the geometric series since v 2 is nilpotent. But v 1 is not a unit by the first part of the proof, contradiction. This concludes the proof. 
Theorem 7.5. (KrullRemakSchmidt).
Let E =F 0 be a module which is both Noetherian and Artinian. Then E is a .finite direct sum of indecomposable modules. Up to a permutation, the indecomposable components in such a direct sum are uniquely determined up to isomorphism. Proof. The existence of a direct sum decomposition into indecomposable modules follows from the Artinian condition. If first E = E Et> E 2 , then either £ 1 , E 2 are indecomposable, and we are done ; or, say , £ 1 is decomposable. Repeating the argument, we see that we cannot continue this decomposition indefinitely without contradicting the Artinian assumption. There remains to prove uniqueness. Suppose 1
where E b F.i are indecomposable. We have to show that r = s and after some permutation, E i � F i . Let e i be the projection of E on E i , and let ui be the projection of E on Fi , relative to the above direct sum decompositions. Let :
vJ. = e 1 uJ. and WJ· = uJ. e 1 •
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NOETH E R IAN R I NG S AND MODULES
Then L ui = idE implies that
X, §7
s
L vi wi i E 1 = idE . · j 1 ==
By Proposition 7 . 4 , End{£ 1 ) is local, and therefore some vi wi is an automor phism of E 1 • After renumbering, we may assume that v 1 w 1 is an automorphis m of E 1 • We claim that v 1 and w 1 induce isomorphisms between E 1 and F . , This follows from a lemma. Lemma 7 6 Let M, N be modules, and assume N indecomposable. Let u : M � N and v : N � M b e such that vu is an automorphism. Then u, v .
.
are isomorphisms. Proof. Let e = u(vu)  1 v. Then e 2 = e is an idempotent, lying in End(N), and therefore equal to 0 or 1 since N is assumed indecomposable. But e =F 0 because idM =F 0 and So e = idN . Then u is injective because vu is an automorphism ; v is injective because e = idN is injective ; u is surjective because e = idN ; and v is surjective because vu is an automorphism. This concludes the proof of the lemma. Returning to the theorem, we now see that
Indeed, e 1 induces an isomorphism from F 1 to E 1 , and since the kernel of e 1 is E 2 (f) · · · (f) E r it follows that But also, F 1 = E 1 (mod E 2 (f) · · · (f) Er), so E is the sum ofF 1 and E 2 (f) · · · (f) Er , whence E is the direct sum, as claimed. But then The proof is then completed by induction. We apply the preceding results to a commutative ring A. We note that an idempotent in A as a ring is the same thing as an idempotent as an element of End( A), viewing A as module over itself. Furthermore End(A) � A. Therefore, we· find the special cases : Theorem 7. 7. Let A be a Noetherian and Artin ian commutative ring.
X, Ex
443
EXER CISES
(i) If A is indecomposable as a ring, then A is local. (ii) In general, A is a direct product of local rings, which are Art in ian and Noetherian. Another way of deriving this theorem will be given in the exercises.
EX E R CI S ES 1 . Let A be a commutative ring. Let M be a module, and N a submodule. Let N = Q 1 n . · · n Q, be a primary decomposition of N. Let Q; = QJN. Show t hat 0 = Q 1 n . . n Q, is a primary decomposition of 0 in M/N. State and prove the converse. ·
2. Let p be a prime ideal, and a, b ideals of A. If ab
c
p, show that a
c
p or b
c
p.
3 . Let q be a primary ideal. Let a, b be ideals, and assume ab c q. Assume that b is finitely generated. Show that a c q or there exists some positive integer n such that b" c q. 4. Let A be Noetherian, and let q be a pprimary ideal. Show that there exists some n > 1 such that p " c q. 5. Let A be an arbitrary commutative ring and let S be a multiplicative subset. Let p be a prime ideal and let q be a pprimary ideal. Then p intersects S if and only if q intersects S. Furthermore, if q does not intersect S, then s  1q is s  1 pprimary in 1 s  A.
6. If a is an ideal of A, let as = s 1 a. If lfJs : A + s  A is the canonical map, abbreviate 1 (/Js (as) by as n A, even though lfJs is not injective. Show that there is a bijection between the prime ideals of A which do not intersect S and the prime ideals of S  1 A, given by 1
Prove a similar statement for primary ideals instead of prime ideals. 7. Let a = q 1 n · · · n q, be a reduced primary decomposition of an ideal. Assume that q1, q ; do not intersect S, but that q i intersects S for j > i. Show that •
•
•
,
is a reduced primary decomposition of as . 8. Let A be a local ring. Show that any idempotent :1: 0 in A is necessarily the unit element. (An idempotent is an element e E A such that e2 = e.) 9. Let A be an Artinian commutative ring. Prove :
(a) All prime ideals are maximal. [Hint : Given a prime ideal p, let x e A, x(p) = 0. Consider the descending chain (x) ::::> (x 2 ) ::::> (x 3 ) ::::> ] •
•
•
•
444
X, Ex
NOETH E R IAN R I NG S AN D MODULES
number of pnme, or m a x i ma l , Ide al s . [H in t : Am on g a l l finttc I n tersect tons of maximal I de al s ptck a m i n i m a l one.] (c) The tdeal N o f ni lpotent elements In A is n i l poten t , t h at I S there ex t sts a po s i t i ve I n tege r k �uch t hat N " = (0). [Hint : L e t k be such t h at N " = N" 1 • Let a = N " . L e t b be a m t n i ma l ide al i= 0 such t h a t ba i= 0. Then b • � pnnc1pal and ba = b. ] (d ) A t s Noetherian. ( c ) There exists an I nteger r such that
(b)
Th ere IS
o nl y
a fin ite
,
�
A = n A/n{ where t he prod uct ts taken over a l l
m a x i ma l ideals.
(f) We h a ve
where aga tn
1 0. Let A, B be
the prod uct IS taken over a l l
prime ideals p.
maximal ideals m A , m 8 , respec t i vely. Let f : A + B be a homomorph tsm. We s a y t hat f I S local 1 f f  1 ( m8 ) = m A . Suppose this is the case. Assu me A , B N o et h eria n , and assume that : 1 . A /ntA
2. m A
�
local nngs with
�
Bjtn n ts an tsomorph 1 sm ,
mn/nt � I S surjective :
3. B t s a fi n i t e A module, via f. Prove t h at r I S
surjective. [Hint :
Apply Nakayama
tW ICe.]
For an ideal a , recall from Chapter IX , §5 that ?1 (a) is the set of primes containing a. 1 1 . Let A be a commutative ring and M an A module. Define the support of M by
supp(M)
=
{ p E spec( A) : M P i= 0 }.
If M is finite over A, show that supp(M) = ?1 (ann(M)) , where ann(M) is the annihilator of M in A , that is the set of elements a E A such that aM = 0 . 1 2 . Let A be a Noetherian ring and M a finite Amodule . Let I be an ideal of A such that supp(M) C ?! (/ ) . Then l "M = 0 for some n > 0 . 1 3 . Let A be any commutative ring, and M, N modules over A . I f M is finitely presented, and S is a multiplicative subset of A , show that
This is usually applied when A is Noetherian and M finitely generated, in which case M is also finitely presented since the module of relations is a submod ule of a finitely generated free module. 14. (a) Prove Proposit ion 6. 7 (b) . (b) Prove that the degree of the polynomial P in Theorem 6.9 is exactly
r.
Locally constant dimensions 15. Let A be a Noetherian local ring. Let E be a finite Amodule. Assume that A has no n i l potent elements. For each pnme Ideal p of A , le t k( p) be the residue class field. If d t m k ( p ) EP/p E P is co n st an t for all p, show th at E t s free. [Hint : Let x 1 , , xr E A be •
•
•
X, Ex
EXERCISES
445
such that the residue classes mod the maximal ideal form a basis for EjmE over k(m). We get a surjective h omomorphism
Let J be the kernel. Show that JP
c
mP A; for all p so J
c
p
for all p and J = 0.]
1 6. Let A be a Noetherian local ring without nilpotent elements. Letf : E + F be a homo morph ism of Amod ules, and suppose E, F are finite free. For each prime p of A let fcp) : E /p E P
P
+
F p/pF P
be the correspondtng k(p )homomorph ism, where k(p) = A p/pA P is the residue class field at p. Assume that is constant. (a) Prove that Fjim f and Im f are free, and that there is an isomorphism F
[H int :
�
l m f e> (F/Im f ) .
Use Exercise 15.]
(b) Prove that Ker j is free and E projective is free.]
�
(Ker f) e> (lm f ). [Hint :
Use that finite
The next exercises depend on the notion of a complex, which we have not yet formally defined . A (finite) complex E is a sequence of homomorphisms of modules o
dl
dn
d 0 � £0 � £1 � • • • � En � 0
and homorphisms d ; : Ei � Ei+ 1 such that di + 1 o di = 0 for all i . Thus Im(di) C Ker (d i + 1 ) . The homology Hi of the complex is defined to be
Hi = Ker(di + 1 ) / l m (d i) . By definition , H0 = E0 and H n = E n/l m (d n) . You may want to look at the first section of Chapter XX , because all we use here is the basic notion , and the following property , which you can easily prove . Let E, F be two complexes . By a homomorphism /: E � F we mean a sequence of homomorphisms
/; : Ei � F ; making the diagram commutative for all i :
Fi
+
d}
pi+ l
Show that such a homomorphism / induces a homomorphism H( f ) : H(E) ")> H(F) on the homology ; that is , for each i we have an induced homomorphism
446
NOETH E R IAN R I N G S AND MODULES
X, Ex
The following exercises are inspired from applications to algebraic geometry , as for instance in Hartshorne , AIgebraic Geometry , Chapter III , Theorem 1 2 . 8 . See also Chapter XXI , § 1 to see how one can construct complexes such as those considered in the next exercises in order to compute the homology with respect to less tractable complexes .
Reduction of a complex mod p 1 7. Let 0 + K0 + K 1 + · · · + K" + 0 be a complex of finite free modules over a local Noetherian ring A without nilpotent elements. For each prime p of A and module E, let E(p) = EP/ pEP , and similarly let K(p) be the complex localized and reduced mod p . For a given integer i, assume that
is constant, where Hi is the ith homology of the reduced complex. Show that H i(K) is free and that we have a natural isomorphism
[Hint :
First write d :P > for the map induced by d i on K i(p ). Write dimk(p) Ker d:P)
=
dimkcp> K i(p)

dimk
Im d:p> ·
Then show that the dimensions dimk C p) Im d:P > and dimk
Im Then apply Exercise 1 2.]
d:;> 1
must be constant.
Comparison of homology at the special point 1 8. Let A be a Noetherian local ring. Let K be a finite complex, as follows : 0
+
K0
+
···
+
K"
+
0,
such that K; is finite free for all i. For some index i assume that
is surjective. Prove : {a) This map is an isomorphism. (b) The following exact sequences split :
(c) Ever y term in these sequences is free.
1 9. Let A be a Noetherian local ring. Let K be a complex as in the previous exercise. For some i assume that
is surjective (or equivalently is an isomorphism by the previous exercise). Prove that
X, Ex
EXERCISES
447
the following conditions are equivalent : (a) H;  1 (K)(m) ... Hi  1 (K(m)) is surjective. (b) Hi  1 (K )(m) � Hi  1 (K(m)) is an isomorph ism. (c) Hi( K ) is free. [Hint : Lift bases until you are blue in the face.] (d) If these conditions hold, then each one of the two inclusions
splits, and each one of these modules is free. Reducing mod corresponding inclusions
m
yields the
and induce the isomorphism on cohomology as stated in (b). [Hint : the preceding exercise.]
Apply
C H A PT E R
XI
R ea l F i e l d s
§1 .
O R D E R ED FI E LDS
Let K be a field. An ordering of K is a subset P of K having the following properties : ORD 1 .
Given x E K, we have either x E P, or x = 0, or  x E P, and these three possibilities are mutually exclusive. In other words, K is the disjoint union of P, {0}, and  P.
ORD 2.
If x, y E P, then x + y and xy E P.
We shall also say that K is ordered by P, and we call P the set of positive elements .
Let us assume that K is ordered by P. Since 1 =F 0 and 1 = 1 2 = (  1) 2 we see that 1 E P. By ORD 2, it follows that 1 + · · · + 1 E P, whence K has characteristic 0. If x E P, and x =F 0, then xx  1 = 1 E P implies that x  1 E P. Let x, y E K. We define x < y (or y > x) to mean that y  x E P. If x < 0 we say that x is negative. This means that  x is positive. One verifies trivially the usual relations for inequalities, for instance : x < y and y < z
implies
X < Z,
x < y and z > O
implies
xz < yz, 1 1  < . y X
x < y and
X, y
>0
implies
We define x < y to mean x < y or x = y. Then x < y and y < x imply x = y. If K is ordered and x E K, x =F 0, then x 2 is positive because x 2 = (  x) 2 and either x E P or  x E P. Thus a sum of squares is positive, or 0.
Let E be a field. Then a product of sums of squares in E is a sum of squares. If a, b E E are sums of squares and b =F 0 then a/b is a sum of squares. 449
450
REAL FI ELDS
XI, §1
The first assertion is obvious, and the second also, from the expression a/b = ab(b  1 ) 2 • If E has characteristic =F 2, and  1 is a sum of squares in E, then every element a E E is a sum of squares, because 4a = ( 1 + a) 2  ( 1  a) 2 • If K is a field with an ordering P, and F is a subfield, then obviously, P n F defines an ordering of F, which is called the induced ordering. We observe that our two axioms ORD 1 and ORD 2 apply to a ring. If A is an ordered ring, with 1 =F 0, then clearly A cannot have divisors of 0, and one can extend the ordering of A to the quotient field in the obvious way : A faction is called positive if it can be written in the form a/b with a, b E A and a, b > 0. One verifies trivially that this defines an ordering on the quotient field. Example.
We define an ordering on the polynomial ring R[t] over the real numbers. A polynomial
with an =F 0 is defined to be positive if an > 0. The two axioms are then trivially verified. We note that t > a for all a E R. Thus t is infinitely large with respect to R. The existence of infinitely large (or infinitely small) elements in an ordered field is the main aspect in which such a field differs from a subfield of the real numbers. We shall now make some comment on this behavior, i.e. the existence of infinitely large elements. Let K be an ordered field and let F be a subfield with the induced ordering. As usual, we put l x l = x ifx > O and l x l =  x if x < O. We say that an element rx in K is infinitely large over F if I rx I > x for all x E F. We say that it is infinitely small over F if O < l rx l < l x l for all x E F, x =F 0. We see that rx is infinitely large if and only if rx  1 is infinitely small. We say that K is archimedean over F if K has no elements which are infinitely large over F. An intermediate field F h K ::J F1 ::J F, is maximal archimedean over F in K if it is archimedean over F, and no other intermediate field containing F1 is archimedean over F. If F1 is archimedean over F and F2 is archimedean over F1 then F2 is archimedean over F. Hence by Zorn ' s lemma there always exists a maximal archimedean subfield F1 of K over F. We say that F is maximal archimedean in K if it is maximal archimedean over itself in K . Let K be an ordered field and F a sub field. Let o be the set of elements of K which are not infinitely large over F. Then it is clear that o is a ring, and that for any rx E K, we have rx or rx  1 E o. Hence o is what is called a valuation ring, containing F. Let m be the ideal of all rx E K which are infinitely small over F. Then m is the unique maximal ideal of o , because any element in o which is not in m has an inverse in o. We call o the valuation ring determined by the ordering of K/F.
R EAL FI ELDS
XI, §2
45 1
Let K be an ordered field and F a subfield. Let o be the valuation ring determined by the ordering of K/F, and let m be its maximal ideal. Then ojm is a real field.
Proposition 1 . 1 .
Proof Otherwise, we could write  1 = L rx f + a with rx; E o and a E m. Since L rxf is positive and a is infinitely small, such a relation is clearly impossible.
§ 2.
R EA L F I E L D S
A field K is said to be real if  1 is not a sum of squares in K. A field K is said to be real closed if it is real, and if any algebraic extension of K which is real must be equal to K. In other words, K is maximal with respect to the property of reality in an algebraic closure. Proposition 2.1 .
Let K be a real field. (i) If a E K, then K(Ja) or K( �) is real. If a is a sum of squares in K, then K(''v'a) is real. If K(Ya) is not real, then  a is a sum of squares in K. (ii) Iff is an irreducible polynomial of odd degree n in K [X] and if rx is a root off, then K(rx) is real. Proof Let a E K. If a is a square in K, then K( Ja) = K and hence is real by assumption. Assume that a is not a square in K. If K( Ja) is not real, then there exist b i , c i E K such that  1 = L (b; + c i Ja ) 2 = L (b f + 2c; b; Ja + c f a).
Since Ja is of degree 2 over K, it follows that  1 = L bf + a L cf .
If a is a sum of squares in K, this yields a contradiction. In any case, we con clude that f t + L b a = L C l2· is a quotient of sums of squares, and by a previous remark, that  a is a sum of squares . Hence K(Va) is real , thereby proving our first assertion.
452
XI, §2
R EAL FI ELDS
As to the second, suppose K(rt ) is not real. Then we can write with polynomials g; in K[ X ] of degree < n  1 . There exists a polynomial h in K[ X ] such that  1 = L g i(X) 2 + h(X)f( X ). The sum of g;( X ) 2 has even degree, and this degree must be > 0, otherwise  1 is a sum of squares in K. This degree is < 2n  2. Since f has odd degree n, it follows that h has odd degree < n  2. If p is a root of h then we see that  1 is a sum of squares in K(p). Since deg h < deg f, our proof is finished by induction. Let K be a real field. By a real closure we shall mean a real closed field L which is algebraic over K. Theorem 2.2. Let K be a real field. Then there exists a real closure of K.
If R is real closed, then R has a unique ordering. The positive elements are the squares of R. Every positive element is a square, and every polynomial of odd degree in R[X] has a root in R . We have Ra R(v=i). =
Proof By Zorn 's lemma, our field K is contained in some real closed field algebraic over K. Now let R be a real closed field. Let P be the set of nonzero elements of R which are sums of squares. Then P is closed under addition and multiplication. By Proposition 2. 1, every element of P is a square in R, and given a E R, a =F 0, we must have a E P or  a E P. Thus P defines an ordering. Again by Proposition 2. 1 , every polynomial of odd degree over R has a root in R . Our assertion follows by Example 5 of Chapter VI , §2 . Corollary 2.3.
Let K be a real field and a an element of K which is not a sum of squares. Then there exists an ordering of K in which a is negative.
Proof The field K( �) is real by Proposition 1 . 1 and hence has an ordering as a subfield of a real closure. In this ordering,  a > 0 and hence a is negative. Proposition 2.4.
Let R be afield such that R =F Ra but Ra R(j=T). Then R is real and hence real closed. Proof Let P be the set of elements of R which are squares and =F 0. We contend that P is an ordering of R. Let a E R, a =F 0. Suppose that a is not a square in R. Let rt be a root of X 2  a 0. Then R { rt) R(j=T), and hence there exist c, d E R such that rt c + dj=T. Then rt 2 c 2 + 2cdj=T  d 2 • =
=
=
=
=
X I , §2
R EAL FI ELD S
453
Since 1 , J=l are linearly independent over R, it follows that c = 0 (because a ¢ R 2 ), and hence  a is a square. We shall now prove that a sum of squares is a square. For simplicity, write i = j"=t. Since R(i) is algebraically closed, given a, b E R we can find c, d E R such that (c + di) 2 = a + bi. Then a = c 2  d 2 and b = 2cd. Hence a 2 + b 2 = (c 2 + d 2 ) 2 , as was to be shown. If a E R, a =F 0, then not both a and  a can be squares in R. Hence P is an ordering and our proposition is proved. Theorem 2.5.
Let R be a real closed field, and f(X) a polynomial in R [X] . Let a, b E R and assume that f(a) < 0 and f(b) > 0. Then there exists c between a and b such that f(c) = 0. Proof Since R( FJ ) is algebraically closed, it follows that f splits into a product of irreducible factors of degree 1 or 2. If X 2 + rxX + p is irreducible ( rx, P E R) then it is a sum of squares, namely
and we must have 4P > rx 2 since our factor is assumed irreducible. Hence the change of sign of f must be due to the change of sign of a linear factor, which is trivially verified to be a root lying between a and b. Lemma 2.6. Let K be a subfield of an ordered field E. Let rx E E be algebraic
over K, and a root of the polynomial
with coefficients in K. Then I rx I < 1 + I an  t l + · · · + I a o 1 Proof If I rx I < 1, the assertion is obvious. If I rx I > 1 , we express I rx I n in terms of the terms of lower degree, divide by 1 rx I n  1 , and get a proof for our lemma. Note that the lemma implies that an element which is algebraic over an ordered field cannot be infinitely large with respect to that field. Let f(X) be a polynomial with coefficients in a real closed field R, and assume that f has no multiple roots. Let u < v be elements of R. By a Sturm sequence for f over the interval [u, v] we shall mean a sequence of polynomials having the following properties :
454
XI, §2
R EAL F I E LDS
ST 1 . The last polynomial fm is a nonzero constant. ST 2. There is no point x E [u, v] such that jj(x) = fi + 1 (x) value 0 < j < m  1. ST 3. ST 4.
If x E [u, v] and h{x) = 0 for some j and h + 1 (x) have opposite signs. We have jj(u)
1=
=
0 and h{v) =F 0 for all j
1, . . . , m =

=
0 for any
1 , then fi  1 (x)
0, . . . , m.
For any x E [u, v] which is not a root of any polynomial /; we denote by J.Vs(x) the number of sign changes in the sequence
{f(x), ft (x),
· · · ,
fm(x)},
and call W8(x) the variation of signs in the sequence. Theorem2.7. (Sturm's Theorem).
The number of roots off between u and v is equal to W8(u)  W8(v)for any Sturm sequence S.
Proof. We observe that if a 1 < a 2 < · · · < exr is the ordered sequence of roots of the polynomials jj in [u, v] U = 0, . . . , m  1), then W8(x) is constant on the open intervals between these roots, by Theorem 2.5. Hence it will suffice to prove that if there is precisely one element ex such that u < a < v and ex is a root of some fi , then W8(u)  W8(v) = 1 if ex is a root of f, and 0 otherwise. Suppose that a is a root of some jj , for 1 < j < m  1. Then .fj_ (C(), fJ + 1 (ex) have opposite signs by ST 3, and these signs do not change when we replace ex by u or v. Hence the variation of signs in { jj  1 ( u ), h{ u ), jj + 1 ( u) } and { .fj _ 1 ( v), jj( v ), jj + 1 ( v) } is the same, namely equal to 2. If ex is not a root of f, we conclude that 1
WS(u)
=
W8(v).
If ex is a root of f, then f(u) and f(v) have opposite signs, but f ' (u) and f '(v) have the same sign, namely, the sign of /' ( ex) . Hence in this case,
Ws(u)
=
W5(v) + 1 .
This proves our theorem. It is easy to construct a Sturm sequence for a polynomial without multiple roots. We use the Euclidean algorithm, writing f = g 1 f '  !2 ,
!2
=
fm  2
=
g 2 !1  !3 ' . g m  1 /m  1  fm ,
R EAL FI ELD S
XI , §2
455
using f' = f1 • Since f, f ' have no common factor, the last term of this sequence is nonzero constant. The other properties of a Sturm sequence are triviall y verified, because if two successive polynomials of the sequence have a com mon zero, then they must all be 0, contradicting the fact that the last one is not.
Let K be an ordered field, f a n irreducible polynomial of degree > 1 over K . The number of roots off in two real closures of K inducing the given ordering on K is the same.
Corollary 2.8.
Proof We can take v sufficiently large positive and u sufficiently large negative in K so that all roots of f and all roots of the polynomials in the Sturm sequence lie between u and v, using Lemma 2.6. Then J.Vs(u)  W8(v) is the total number of roots of f in any real closure of K ind ucing the given orde ring. Theorem 2.9. Let K be an ordered field, and let R, R' be real closures of K,
whose orderings induce the given ordering on K. Then there exists a unique isomorphism a : R R' over K, and this isomorphism is orderpreserving. Proof We first show that given a finite subextension E of R over K, there exists an embedding of E into R' over K. Let E = K(a), and let +
f(X) = Irr(a, K, X). Then f(a) = 0 and the corollary of Sturm ' s Theorem (Corollary 2.8) shows that f has a root p in R ' . Thus there exists an isomorphism of K (a ) on K ( P) over K, mapping a on p. Let a t , . . . , an be the distinct roots off in R, and let P t , . . . , Pm be the distinct roots of f in R ' . Say a t < · · · < an in the ordering of R, P t < · · · < Pm in the ordering of R'. We contend that m = n and that we can select an embedding a of K ( ab . . . , an ) into R' such that aai = pi for i = 1 , . . , n. Indeed, let '}'; be an element of R such that .
yf = a; + t  a; for .
i
= 1,
...,n 
1
.
and let E t = K( a t , . . . , an , y b . . , Y n t ) By what we have seen, there exists an embedding a of E t into R', and then aa; + t  aa; is a square in R'. Hence _
This proves that m > n. By symmetry, it follows that m = n. Furthermore, the condition that aa; = f3i for i = 1 , . . . , n determines the effect of a on
456
XI, §2
R EAL F I E LDS
K(rx 1 ' . . . ' rx ) . We contend that (J is orderpreserving. 'Let y E K(rx . . ' rxn ) and 0 < y. Let y E R be such that y 2 = y. There exists an embedding of n
b
K( rx 1 , .
.
.
, rxn , Y 1 ,
· · ·
,Y
n 
1,
.
Y)
into R' over K which must induce a on K(rx . , . . . , rx ) and is such that ay is a square, hence > 0, as contended. Using Zorn's lemma, it is now clear that we get an isomorphism of R onto R' over K. This isomorphism is orderpreserving because it maps squares on squares, thereby proving our theorem. n
Proposition 2. 10.
no relation
Let K be an ordered field, K' an extension such that there is 
1
=
n
� ' a · rx �
i= 1
I
I
with a i E K, a i > 0, and rx; E K ' . Let L be the .field obtainedfrom K' by adjoining the square roots of all positive elements of K. Then L is real. Proof If not, there exists a relation of type 1
=
n
' i
� rx a· � =
1
I
I
with a; E K, a i > 0, and rx; E L. (We can take a; = 1.) Let r be the smallest integer s � ch that we can write such a relation with rx i in a subfield of L, of type with bi E K, bi > 0. Write (X 1. =
X r· + Y 1v' · f[;
with X ; , Y i E K' ( y'/;;, . . . , �). Then
1
= =
Ur
L a;(X ; + Y i fi.) 2 L a i (x f + 2x i y i fi. + y f br) .
By hypothesis, fir is not in K'(b . , . . . , �). Hence contradicting the minimality of r. Theorem 2.1 1 .
Let K be an ordered field. There exists a real closure R of K inducing the given ordering on K.
XI, §3
R EAL ZER OS AND HOMOMORPH ISMS
457
Proof Take K' K in Proposition 2. 1 0. Then L is real, and is contained in a real closure. Our assertion is clear. =
Corollary 2.12.
Let K be an ordered field, and K' an extension field. In order that there exist an ordering on K' inducing the given ordering of K, it is necessary and sufficient that there is no relation of type "n } 1 � a . rx =
i= 1
I
l
with a i E K, a i > 0, and rx i E K'. Proof. If there is no such relation, then Proposition 2. 1 0 states that L is contained in a real closure, whose ordering induces an ordering on K', and the given ordering on K, as desired. The converse is clear. Example. Let Qa be the field of algebraic numbers. One sees at once that Q admits only one ordering, the ordinary one. Hence any two real closures of Q in Qa are isomorphic, by means of a unique isomorphism. The real closures of Q in Qa are precisely those subfields of Qa which are of finite degree under Q8 • Let K be a finite real extension of Q, contained in Q8• An element rx of K is a sum of squares in K if and only if every conjugate of rx in the real numbers is positive, or equivalently, if and only if every conjugate of rx in one of the real closures of Q in Qa is positive. Note.
The theory developed in this and the preceding section is due to Artin Schreier. See the bibliography at the end of the chapter.
§ 3.
R EA L Z E R O S AN D H O M O M O R P H I S M S
Just as we developed a theory of extension of homomorphisms into an algebraically closed field, and Hilbert 's Nullstellensatz for zeros in an alge braically closed field, we wish to develop the theory for values in a real closed field. One of the main theorems is the following : Theorem
Let k be a field, K k(x . . . , xn ) a finitely generated extension. Assume that K is ordered. Let Rk be a real closure of k inducing the same ordering on k as K. Then there exists a homomorphism over k.
3. 1 .
=
b
458
X I , §3
R EAL FI ELDS
As applications of Theorem 3. 1 , one gets : Corollary
3.2.
assume
Notation being as in the theorem, let y 1 , • • • , Y m E k[x] and
Y1 < Y 2 < · · · < Ym is the given ordering of K . Then one can choose qJ such that lfJY1 < · · · < lfJYm · Proof Let Yi E K a be such that yf = Yi + 1  Yi · Then K (y 1 , . . . , Y n  1 ) given ordering on K . We apply the theorem to the has . an ordering inducing the r1 ng k[ X 1 ' · • • ' Xn ' Y 1 ' · · · ' Y m 1 b Y · · · ' Ym  1 ] ·  1
h
( Art i n) . Let k be a real field admitting only one ordering. Let f(X 1 ' . . . ' Xn ) E k(X) be a rational function having the property that for all (a) = (a b . . . , an ) E Rin ) such that f(a) is defined, we have f(a) > 0. Then j (X) is a sum of squares in k(X). Proof Assume that our conclusion is false. By Corollary 2.3, there exists an ordering of k(X) in which f is negative. Apply Corollary 3.2 to the ring k[ X 1 , . . . , X , h( X)  1 ] Corollary 3.3.
n
where h(X) is a polynomial denominator for f(X). We can find a homo morphism qJ of this ring into R k (inducing the identity on k) such that lfJ(f) < 0. But contradiction. We let a i = qJ(X i) to conclude the proof. Corollary 3 . 3 was a Hilbert problem. The proof which we shall describe for Theorem 3 . 1 differs from Artin' s proof of the corollary in several technical aspects . We shall first see how one can reduce Theorem 3. 1 to the case when K has transcendence degree 1 over k, and k is real closed. Lemma
Let R be a real closed field and let R 0 be a subfield which is algebraically closed in R (i.e. such that every element of R not in R 0 is tran scendental over R 0 ). Then R 0 is real closed. 3.4.
Proof Let f(X) be an irreducible polynomial over R 0 . It splits in R into linear and quadratic factors. Its coefficients in R are algebraic over R 0 , and hence must lie in R 0 • Hence f(X) is linear itself, or quadratic irreducible already over R 0 • By the intermediate value theorem, we may assume that f is positive
X I , §3
R EAL ZEROS AND HOMOMORPH ISMS
459
definite, i.e. f(a) > 0 for all a E R0 . Without loss of generality, we may assume that f(X) = X 2 + b 2 for some b E R0 • Any root of this polynomia l will bring J=l with it and therefore the only algebraic extension of R0 is R0( J=}) . This proves that R0 is real closed. Let Rx be a real closure of K inducing the given ordering on K. Let R0 be the algebraic closure of k in Rx . By the lemma, R0 is real closed. We consider the field R0(x 1 , , xn ). If we can prove our theorem for the ring R0[x 1 , , xn], and find a homomorphism •
•
•
•
•
•
then we let a : R0 __.. Rx be an isomorphism over k (it exists by Theorem 2.9), and we let qJ = a o ljJ to solve our problem over k. This reduces our theorem to the case when k is real closed. Next, let F be an intermediate field, K => F => k, such that K is of tran scendence degree 1 over F. Again let RK be a real closure of K preserving the ordering, and let RF be the real closure of F contained in Rx . If we know our theorem for extensions of dimension 1, then we can find a homomorphism
We note that the field k(l/Jx b . . . , l/Jx n ) has transcendence degree < n 1 , and is real, because it is contained in RF . Thus we are reduced inductively to the case when K has dimension 1, and as we saw above, when k is real closed. One can interpret our statement geometrically as follows. We can write K = R(x, y) with x transcendental over R, and {x, y) satisfying some irreducible polynomial f(X, Y) = 0 in R [X, Y]. What we essentially want to prove is that there are infinitely many points on the curve f(X, Y) = 0, with coordinates lying in R, i.e. infinitely many real points. The main idea is that we find some point (a, b) E R < 2 > such that f(a, b) = 0 but D 2 f(a, b) =F 0. We can then use the intermediate value theorem. We see that f(a, b + h) changes sign as h changes from a small positive to a small negative element of R. If we take a' E R close to a, then f(a', b + h) also changes sign for small h, and hence f(a', Y) has a zero in R for all a' sufficiently close to a . In this way we get infinitely many zeros. To find our point, we consider the polynomial f(x, Y) as a polynomial in one variable Y with coefficients in R(x). Without loss of generality we may assume that this polynomial has leading coefficient 1 . We construct a Sturm sequence for this polynomial, say 
{f(x, Y) , !1 (x, Y), . . . , fm (x, Y)} . Let d = deg f. If we denote by A(x) = (a d  1 (x), . . . , a0(x)) the coefficients of f(x, Y), then from the Euclidean alogrithm, we see that the coefficients of the
460
XI, §3
R EAL FI ELDS
polynomials in the Sturm sequence can be expressed as rational functions
in terms of ad  t (x), . . . , a0 (x). Let 1 + ad  t (x) + · · · + a 0 (x) + s, where s is a positive integer, and the signs are selected so that each term in this sum gives a positive contribution. We let u(x) =  v(x ), and select s so that neither u nor v is a root of any polynomial in the Sturm sequence for f. Now we need a lemma.
v(x)
Lemma
=
Let R be a real closed field, and { h i (x)} a finite set of rational functions in one variable with coefficients in R. Suppose the rational field R(x) ordered in some way, so that each hi (x) has a sign attached to it. Then there exist infinitely many special values c of x in R such that h i( c) is defined and has the same sign as h i(x), for all i. Proof. Considering the numerators and denominators of the rational functions, we may assume without loss of generality that the h i are polynomials. We then write 3.5.
where the first product is extended over all roots A. of h i in R, and the second product is over positive definite quadratic factors over R. For any � E R, p( �) is positive. It suffices therefore to show that the signs of (x  A.) can be preserved for all A. by substituting infinitely many values tX for x. We order all values of A. and of x and obtain ...
0. 2. Let F be a finite extension of Q. Let lfJ : F + Q be a Qlinear functional such that cp(x2) > 0 for all x E F, x "# 0. Let a E F, ex "# 0. If cp{exx2) > 0 for all x E F, show that ex is a sum of squares in F, and that F is totally real, i.e. every embedding of F in the complex numbers is contained in the real numbers. [Hint : Use the fact that the trace gives an identification of F with its dual space over Q, and use the approximation theorem of Chapter X II, § 1 .] I.
462
XI, Ex
R EAL FI ELDS
3 . Let cx < t < p be a real interval, and let f(t) be a real polynomial which is positive on this I nterval. Show that f ( t) can be written in the form
where Q2 denotes a square, and c > 0. Hint : Split the polynomial , and use the I dentity :
(t  a ) ({J  t) =
( t  a)2({J  t) + ( t  a) ({J  t)2 a p  (l
.
The above seemingly innocuous result is a key step in developing the spectral theorem for bounded hermitian operators on H ilbert space . See the appendix of [La 72] and also [La 85] .
Remark .
4 . Show that the field of real numbers has only the identity automorphism. [Hint : Show that an automorphism preserves the ordering.]
Real places For the next exercise s , cf. Krull [Kr 32] and Lang [La 5 3 ] . These exercises form a connected sequence , and solutions will be found in [La 53] . 5 . Let K be a field and suppose that there exists a real place of K; that is , a place cp with values in a real field L. Show that K is real . 6. Let K be an ordered real field and let F be a subfield which is maximal archimedean in K. Show that the canonical place of K with respect to F is algebraic over F ( i . e . i f o i s the valuation ring of elements of K which are not infinitely large over F , and m is its maximal ideal , then o/m is algebraic over F) . 7 . Let K be an ordered field and let F be a subfield which is maximal archimedean in K. Let K' be the real closure of K (preserving the ordering) , and let F ' be the real closure of F contained in K'. Let cp be the canonical place of K' with respect to F '. Show that cp(K') is F ' valued , and that the restriction of cp to K is equivalent to the canonical place of K over F . 8 . Define a real field K to be quadratically closed if for all a E K either Va or � lies in K. The ordering of a quadratically closed real field K is then uniquely determined , and so is the real closure of such a field , up to an isomorphism over K. Suppose that K is quadratically closed . Let F be a subfield of K and suppose that F is maximal archimedean in K. Let cp be a place of K over F , with values in a field which is algebraic over F . Show that cp is equivalent to the canonical place of K over F . 9 . Let K be a quadratically closed real field . Let cp be a real place of K, taking its values in a real closed field R . Let F be a maximal subfield of K such that cp is an isomorphism on F, and identify F with cp(F) . Show that such F exists and is maximal archimedean in K. Show that the image of cp is algebraic over F , and that cp is induced by the canonical place of K over F.
1 0. Let K be a real field and let cp be a real place of K, taking its values in a real closed field R . Show that there is an extension of cp to an Rvalued place of a real closure of K. [Hint : first extend cp to a quadratic closure of K. Then use Exercise 5 . ]
EXE RC ISES
XI, Ex
463
1 1 . Let K C K 1 C K2 be real closed fields . Suppose that K is maximal archimedean in K 1 and K 1 is maximal archimedean in K2 Show that K is maximal archimedean in Kz . 1 2 . Let K be a real closed field . Show that there exists a real closed field R containing •
K and having arbitrarily large transcendence degree over K, and such that K is maximal archimedean in R . , fr be homogeneous polyno mials of odd 1 3 . Let R be a real closed field . Let /1 , degrees in n variables over R . If n > r, show that these polynomials hav e a non trivial common zero in R . (Comments : If the forms are generic (in the sense of Chapter IX) , and n = r + 1 , it is a theorem of Bezout that in the algebraic closure Ra the forms have exactly d 1 • • • dm common zeros , where d; is the degree of /;. You may •
•
•
assume this to prove the result as stated . If you want to see this worked out , see [La 53] , Theorem 1 5 . Compare with Exercise 3 of Chapter IX . )
Bibliography [Ar 24] [Ar 27] [ArS 27] [Kr 32] [La 53] [La 72] [La 85]
E . ARTIN , Kennzeichnung des Korpers der reellen algebraischen Zahlen , Abh. Math . Sem . Hansischen Univ . 3 ( 1 924) , pp . 3 1 9323 E. ARTIN , U ber die Zerlegung definiter Funktionen in Quadrate , Abh . Math . Sem . Hansischen Univ. S ( 1 927) , pp . 1 00 1 1 5 E. ARTIN and E. ScHREIER , Algebraische Konstruktion reeller Korper, A bh . Math . Sem . Hansischen Univ . 5 ( 1 927) , pp . 8599 W. KRULL , Allgemeine Bewertungstheorie , J. reine angew. Math. ( 1 932) , pp . 1 69 1 96 S. LANG , The theory of real places , Ann . Math . 57 No . 2 ( 1 953) , pp. 378
39 1
S . LANG , Differential manifolds , AddisonWesley , 1 972; reprinted by Springer Verlag, 1 985; superceded by [La 99a]. S. LANG, Real and functional analysis . Third edition , Springer Verlag,
1 993
[La 99a]
S. LANG, Fundamentals of D ifferential Geometry, Springer Verlag, 1 999
C H A PT E R
XII
A bso lute Va lues
§1 .
D E FI N ITI O N S , D E P E N D E N C E, AN D I N DEPEN DENCE
Let K be a field. An absolute value v on K is a realvalued function x 1+ I x lv on K satisfying the following three properties : A V 1.
We have I x l v > 0 for all x E K, and I x l v
AV
2.
For all x, y E K, we have l xy l v
AV
3.
For all x, y E K, we have I x + Y lv < I x lv + I Y l v .
=
0 if and only if x
=
0.
l x lv i Y i v ·
=
If instead of A V 3 the absolute value satisfies the stronger condition then we shall say that it is a valuation, or that it is nonarchimedean. The absolute value which is such that I x l v = I for all x =F 0 is called trivial . We shall write I x I instead of I x l v if we deal with just one fixed absolute value. We also refer to v as the absolute value. An absolute value of K defines a metric. The distance between two elements x, y of K in this metric is I x  y 1 . Thus an absolute value defines a topology on K. Two absolute values are called dependent if they define the same topology. If they do not, they are called independent. We observe that I l l = 1 1 2 1 = I (  1 ) 2 1 = 1 1 1 2 whence
Ill Also, 1  x l
=
=
Ill
l x l for all x E K, and l x  1 1
=
=
1.
l x l  1 for x i= 0. 465
466
ABSOLUTE VALU ES
XII, §1
Let I I t and I b be nontrivial absolute values on a field K. They are dependent if and only if the relation
Proposition 1.1.
lxlt < 1
implies I x b < 1 . If they are dependent, then there exists a number A > 0 such that l x l t l x l� for all x E K . Proof If the two absolute values are dependent, then our condition is satisfied, because the set of x E K such that I x I t < 1 is the same as the set such that lim x n 0 for n + oo . Conversely, assume the condition satisfied. Then I x I t > 1 implies I x b > 1 since I x  t I t < 1 . By hypothesis, there exists an element x0 E K such that l xo l t > 1 . Let a = l x o l t and b = l x0 1 2 . Let log b . A log a =
=
=
Let x E K, x 1= 0. Then I x I t = I x0 I� for some number rx. If m, n are integers such that m/n > rx and n > 0, we have I X I t > I X o 1 71 n whence
and thus
I xnjx� l 2 < 1 . This impli es that I x b < I x 0 1 �1n. Hence
Similarly , one proves the reverse inequality, and thus one gets
for all x E K, x 1= 0. The assertion of the proposition is now obvious, i.e. l x b = l x l�. We shall give some examples of absolute values. Consider first the rational numbers. We have the ordinary absolute value such that I m I = m for any positive integer m. For each prime number p, we have the padic absolute value vP , defined by the formula
XII, § 1
D E F I N ITIONS, DEPENDENCE, AN D IND EPENDENC E
467
where r is an integer, and m, n are integers =1= 0, not divisible by p. One sees at once that the padic absolute value is nonarchimedean. One can give a similar definition of a valuation for any field K which is the quotient field of a principal ring. For instance, let K = k(t) where k is a field and t is a variable over k. We have a valuation vP for each irreducible polynomial p(t) in k[t], defined as for the rational numbers, but there is no way of normalizing it in a natural way. Thus we select a number c with 0 < c < 1 and for any rational function p) /g where f, g are polynomials not divisible by p, we define
The various choices of the constant c give rise to dependent valuations. Any subfield of the complex numbers (or real numbers) has an absolute value, induced by the ordinary absolute value on the complex numbers. We shall see later how to obtain absolute values on certain fields by embedding them into others which are already endowed with natural absolute values.
Suppose that we have an absolute value on a field which is bounded on the prime ring (i.e. the integers Z if the characteristic is 0, or the integers mod p if the characteristic is p ). Then the absolute value is necessarily nonarchimedean. Proof For any elements x, y and any positive integer n, we have
Taking nth roots and letting n go to infinity proves our assertion. We note that this is always the case in characteristic > 0 because the prime ring is finite ! If the absolute value is archimedean, then we refer the reader to any other book in which there is a discussion of absolute values for a proof of the fact that it is dependent on the ordinary absolute value. This fact is essentially useless (and is never used in the sequel), because we always start with a concretely given set of absolute values on fields which interest us. In Proposition 1 . 1 we derived a strong condition on dependent absolute values. We shall now derive a condition on independent ones.
Let K be a field and I 1 1 , , I Is nontrivial pairwise independent absolute values on K. Let x 1 , xs be elements of K, and l > 0. Then there exists x E K such that Theorem 1.2. (Approximation Theorem). (ArtinWhaples). •
•
for all i.
•
•
,
•
•
468
XII, §2
ABSOLUTE VALU ES
Proof Consider first two of our absolute values, say Vt and v 2 • By hypo thesis we can find rx E K such that I rx I t < 1 and I rx I s > 1 . Similarly, we can find P e K such that i P i t > 1 and 1 P i s < 1 . Put y = P/rx. Then l y l t > l and i Y i s < 1 . We shall now prove that there exists z E K such that I z I t > 1 and I l i < 1 for j = 2, . . . , s. We prove this by induction, the case s = 2 having just been proved. Suppose we have found z E K satisfying z
for j = 2, . . . , s  1 . l z l t > 1 and l z l i < 1 If I z I s < 1 then the element zny for large n will satisfy our requirements. If I z I s > 1, then the sequence
tends to 1 at v t and vs , and tends to O at vi U = 2, . . . , s  1). For large n, it is then clear that t n y satisfies our requirements. Using the element z that we have just constructed, we see that the sequence zn/( 1 + zn ) tends to 1 at v t and to 0 at vi for j = 2, . . . , s. For each i = 1 , . . . , s we can therefore construct an element Z; which is very close to 1 at v; and very close to 0 at vi (j =I= i ) . The element then satisfies the requirement of the theorem.
§2.
CO M P LETI O N S
Let K be a field with a nontrivial absolute value v, which will remain fixed throughout this section. One can then define in the usual manner the notion of a Cauchy sequence. It is a sequence {x n } of elements in K such that, given £ > 0, there exists an integer N such that for all n, m > N we have
We say that K is complete if every Cauchy sequence converges. Proposition
There exists a pair (Kv , i) consisting of afield Kv , complete under an absolute value, and an embedding i : K + Kv such that the absolute value on K is induced by that of Kv (i.e. ! x l v = I ix I for x E K), and such that iK is dense in Kv . If (K� , i' ) is another such pair, then there exists a unique 2. 1 .
COM PLETIONS
X I I , §2
469
+
isomorphism qJ : Kv K� preserving the absolute values, and making the following diagram commutative :
\KI Proof The uniqueness is obvious. One proves the existence in the well known manner, which we shall now recall briefly, leaving the details to the reader. The Cauchy sequences form a ring, addition and multiplication being taken componentwise. One defines a null sequence to be a sequence {x n } such that lim x n 0. The null sequences form an ideal in the ring of Cauchy sequences, and in fact form a maximal ideal. (If a Cauchy sequence is not a null sequence, then it stays away from 0 for all n sufficiently large, and one can then take the inverse of almost all its terms. Up to a finite number of terms, one then gets again a Cauchy sequence.) The residue class field of Cauchy sequences modulo null sequences is the field K v . We em bed K in K v " on the diagonal ", i.e. send x E K on the sequence (X , X, X , ). We extend the absolute value of K to Kv by continuity. If {x n } is a Cauchy sequence, representing an element � in Kv , we define I � I lim I Xn 1 . It is easily proved that this yields an absolute value (independent of the choice of repre sentative sequence {xn } for �), and this absolute value induces the given one on K. Finally , one proves that Kv is complete. Let { � n } be a· Cauchy sequence in Kv . For each n, we can find an element X n E K such that I �n  Xn I < ljn. Then one verifies immediately that { x n } is a Cauchy sequence in K. We let � be its limit in K v . By a three£ argument, one sees that { �n } converges to � ' thus proving the completeness. =
.
.
•
=
A pair (Kv , i) as in Proposition 2. 1 may be called a completion of K. The standard pair obtained by the preceding construction could be called the completion of K. Let K have a nontrivial archimedean absolute value v. If one knows that the restriction of v to the rationals is dependent on the ordinary absolute value, then the completion Kv is a complete field, containing the completion of Q as a closed subfield, i.e. containing the real numbers R as a closed subfield. It will be worthwhile to state the theorem of GelfandMazur concerning the structure of such fields. First we define the notion of normed vector space. Let K be a field with a nontrivial absolute value, and let E be a vector space over K. By a norm on E (compatible with the absolute value of K) we shall mean a function � + I � I of E into the real numbers such that : NO 1 .
I � I > 0 for all � E E, and
=
0 if and only if �
=
0.
470
XII, §2
ABSOLUTE VALU ES
NO
For all x
2.
e
K and
If �' � ' E E then
NO 3.
I�
g
e
E we have lxgl  lxl lgl. + �' I < I � I + I �' I ·
Two no rms I 1 1 and I 1 2 are called equivalent if there exist numbers C 1 , C 2 > 0 such that for all � E E we have
Suppose that E is finite dimensional, and let over K. If we write an element
w b . . . , w"
be a basis of E
in terms of this basis, with X ; E K , then we can define a norm by putting 1�1
=
max l xd . i
The three properties defining a norm are trivially satisfied. Proposition
Let K be a complete field under a nontrivial absolute value, and let E be a finitedimensional space over K. Then any two norms on E (compatible with the given absolute value on K ) are equivalent. 2.2.
Proof We shall first prove that the topology on E is that of a product space, i.e. if w 1 , , w" is a basis of E over K, then a sequence •
•
•
":,�(v)
=
1 xw 1
+ ... +
xw n
"'
x�v) E K I
'
is a Cauchy sequence in E only if each one of the n sequences x�v> is a Cauchy sequence in K. We do this by induction on n. It is obvious for n = 1 . Assume n > 2. We consider a sequence as above, and without loss of generality, we may assume that it converges to 0. (If necessary, consider �  �<Jt> for v , J1 __.. oo . ) We must then show that the sequences of the coefficients converge to 0 also. If this is not the case, then there exists a number a > 0 such · that we have for some j, say j = 1 ,
I x I > a
for arbitrarily large v . Thus for a subsequence of ( v ), �;x\v> converges to 0, and we can write
We let Yf be the righthand side of this equation. Then the subsequence Yf converges (according to the lefthand side of our equation). By induction, we
XII, §2
COM PLETIONS
47 1
conclude that its coefficients in terms of w 2 , • • • , wn also co nverge in K, say to y 2 , • • • , yn . Taking the limit, we get
contradicting the linear independence of the wi . We must finally see that two norms inducing the same topology are equivalent. Let I I t and I l z be these norms. There exists a number C > 0 such that for any � E E we have I � I t < C implies I � l z < 1 . Let a E K be such that 0 < I a I < 1 . For every � E E there exists a unique integer s such that Hence I as � 1 z < 1 whence we get at once
The other inequality follows by symmetry, with a similar constant. (GelfandMazur) . Let A be a commutative algebra over the real numbers, and assume that A contains an element j such that j 2 =  1 . Let C = R + Rj. Assume that A is normed (as a vector space over R), and that j xy j < l x l l Y l for all x, y E A . Given x0 E A, x0 =I= 0, there exists an element c E C such that x0  c is not invertible in A . Theorem 2.3.
Proof (Tornheim). Assume that x0  z is invertible for all z E C. Consider the mapping f : C __.. A defined by
It is easily verified (as usual) that taking inverses is a continuous operation. Hence f is continuous, and for z =F 0 we have
1 Xo 
z
1
From this we see that f ( z ) approaches 0 when z goes to infinity (in C). Hence the map z �+ I f(z) I is a continuous map of C into the real numbers > 0, is bounded, and is small outside some large circle. Hence it has a maximum, say M. Let D
472
XII, §2
ABSOLUTE VALU ES
be the set of elements z E C such that lf(z)l = M. Then D is not empty; D is bounded and closed. We shall prove that D is open , hence a contradiction. Let c0 be a point of D, which, after a translation, we may assume to be the origin. We shall see that if r is real > 0 and small, then all points on the circle of radius r lie in D. Indeed, consider the sum S(n)
n 1 L
=

n k= 1
1
Xo
 OJk r
where w is a primitive nth root of unity. Taking formally the logarithmic n n n derivative of x  r = 0 (X  wk r) shows that k= 1 n
L k= 1 X
1 
wk r '
and hence, dividing by n, and by x n  1 , and substituting x 0 for X, we obtain S( n)
1 =
x0  r( rIx0 t 
• .
If r is small (say I r/x0 I < 1 ), then we see that lim I S(n) I
n +
=
oo
__!__ Xo
=
M.
Suppose that there exists a complex number A. of absolute value 1 such that 1
M. x0  �t.r < �
Then there exists an interval on the unit circle near A., and there exists £ > 0 such that for all roots of unity ( lying in this interval, we have 1
[ S(n) n 2":1
r
x 0  �.;,r
< M  e
]
(This is true by continuity.) Let us take n very large. Let bn be the number of nth roots of unity lying in our interval. Then bn /n is approximately equal to the length of the interval (times 2n) : We can express S(n) as a sum =
1
x
0
1 
wk r
+ L: n x 0
1 
wkr
,
X I I , §2
COM PLETIONS
473
the first sum L1 being taken over those roots of unity wk lying in our interval, and the second sum being taken over the others. Each term in the second sum has norm < M because M is a maximum. Hence we obtain the estimate
This contradicts the fact that the limit of I S(n) I is equal to M. Corollary
Let K be a field, which is an extension of R, and has an absolute value extending the ordinary absolute value on R. Then K = R or K = C. 2.4.
Proof Assume first that K contains C. Then the assumption that K is a field and Theorem 2.3 imply that K = C. If K does not contain C, in other words, does not contain a square root of  1 , we let L = K(j) wherej 2 =  1 . We define a norm on L (as an Rspace) by putting I X + yj I
=
IXI + IyI
for x, y E K. This clearly makes L into a normed Rspace. Furthermore, if z = x + yj and z ' = x ' + y'j are in L, then
l zz' l
=
< < <
con verges to z 1 w 1 + + zn wn . Hence E is complete. Furthermore, since any two extensions of v to E are equivalent, we can apply Proposition 1 . 1 , and we see that we must have A. = 1 , since the extensions induce the same absolute value v on K. This proves what we want. +
· · ·
· · ·
From the uniqueness we can get an explicit determination of the absolute value on an algebraic extension of K. Observe first that if E is a normal extension of K, and a is an automorphism of E over K, then the function
x �+ I ax I is an absolute value on E extending that of K. Hence we must have
l ax l
=
lxl
for all x E E. If E is algebraic over K, and a is an embedding of E over K in K8, then the same conclusion remains valid, as one sees immediately by embedding E in a normal extension of K. In particular, if rx is algebraic over K, of degree n, and if rx b . . . , rxn are its conjugates (counting multiplicities, equal to the degree of inseparability), then all the absolute values I rx; I are equal. Denoting by N the norm from K(rx) to K, we see that
and taking the nth root, we get :
Let K be complete with respect to a nontrivial absolute value. Let rx be algebraic over K, and let N be the norm from K(rx) to K. Let n = [ K(rx) : K] . Then
Proposition
2.6.
In the special case of the complex numbers over the real numbers, we can write rx = a + bi with a , b E R, and we see that the formula of Proposition 2.6 is a generalization of the formula for the absolute value of a complex number, rx = (a 2 + b 2 ) 1 ; 2 , since a 2 + b 2 is none other than the norm of rx from C to R.
COM PLETIONS
XII, §2
475
Comments and examples .
The process of completion is widespread in mathematics. The first example occurs in getting the real numbers from the rational numbers , with the added property of ordering . I carry this process out in full in [La 90a] , Chapter IX , § 3 . In all other examples I know , the ordering property does not intervene . We have seen examples of completions of fields in this chapter , especially with the padic absolute values which are far away from ordering the field . But the real numbers are nevertheless needed as the range of values of absolute values , or more generally norms . In analysis , one completes various spaces with various norms . Let V be a vector space over the complex numbers , say . For many applications , one must also deal with a seminorm, which satisfies the same conditions except that in NO 1 we require only that II � II > 0. We allow II � II = 0 even if � * 0 . One may then form the space of Cauchy sequenc es , the subspace of null sequences , and the factor space V. The seminorm can be extended to a seminorm on V by continuity , and this extension actually turns out to be a norm . It is a general fact that V is then complete under this extension . A Banach space is a complete normed vector space . Example. Let V be the vector space of step functions on R , a step function being a complex valued function which is a finite sum of characteristic functions of intervals ( closed , open , or semiclosed , i . e . the intervals may or may no t contain their endpoints). For f E V we define the L 1 seminorm by
11 /1/ t
==
J l f(x) I
R
dx .
The completion of V with respect to this semi norm is defined to be L I (R) . One then wants to get a better idea of what elements of L I (R) look like . It is a simple lemma that given an L I Cauchy sequence in V, and given e > 0 , there exists a subsequence which converges uniformly except on a set of measure less than e . Thus elements of L I (R) can be identified with pointwise limits of L I Cauchy sequences in V. The reader will find details carried out in [La 85] . Analysts use other norms or seminorms , of course , and other spaces , such as the space of coo functions on R with compact support, and norms which may bound the derivatives . There is no end to the possible variations. Theorem 2. 3 and Corollary 2.4 are also used in the theory of Banach algebras , representing a certain type of Banach algebra as the algebra of continuous func tions on a compact space , with the GelfandMazur and GelfandNaimark theo rems . Cf. [Ri 60] and [Ru 73] . Arithmetic example .
For padic Banach spaces in connection with the number theoretic work of Dwork , see for instance Serre [Se 62] , or also [La 90b] , Chapter 1 5 . In this book we limit ourselves to complete fields and their finite extensions .
476
XII, §3
ABSOLUTE VALU ES
Bibliography [La 85] [ La 90a] [La 90b] [Ri 60] [Ru 73] [Se 62]
§3.
S . LANG , Real and Functional Analysis , Springer Verlag , 1 993 S . LANG , Undergraduate Algebra , Second Edition , Springer Verlag , 1 990 S . LANG , Cyclotomic Fields I and II , Springer Verlag 1 990 (combined from the first editions, 1 97 8 and 1 980) C . RICKART , Banach Algebras , Van Nostrand ( 1 960) , Theorems 1 . 7 . 1 and 4.2.2. W. RUDIN , Functional Analy sis , McGraw Hill ( 1 973) Theorems 1 0 . 1 4 and 1 1 . 18. J. P. S ERRE , Endomorphismes completement continus des espaces de Banach padiques , Pub . Math. IHES 12 ( 1 962) , pp . 6985
FI N IT E EXTE N S I O N S
Throughout this section we shall deal with a field K having a nontrivial absolute value v . We wish to describe how this absolute value extends to finite extensions of K. If E is an extension of K and w is an absolute value on E extending v , then we shall write w I v. If we let K v be the completion, we know that v can be extended to K v , and then uniquely to its algebraic closure K� . If E is a finite extension of K, or even an algebraic one, then we can extend v to E by embedding E in K� by an iso morphism over K, and taking the induced absolute value on E. We shall now prove that every extension of v can be obtained in this manner. •
Proposition 3. 1 . Let E be a .finite extension of K. Let w be an absolute value
on E extending v, and let Ew be the completion. Let Kw be the closure of K in Ew and identify E in Ew. Then Ew = EKw (the composite field) .
Proof We observe that Kw is a completion of K, and that the composite field E K w is algebraic over K w and therefore complete by Proposition 2.5. Since it contains E, it follows that E is dense in it, and hence that E w = E K w . If we start with an embedding a : E K� (always assumed to be over K), then we know again by Proposition 2.5 that aE Kv is complete. Thus this construction and the construction of the proposition are essentially the same, up to an isomorphism. In the future, we take the embedding point of view. We must now determine when two em beddings give us the same absolute value on E. Given two em beddings a, r : E K� , we shall say that they are conjugate over Kv if there exists an automorphism A of K� over Kv such that a = A r. We see that actually A is determined by its effect on rE , or rE K v . +
·
+
·
FINITE EXTEN SIONS
XII, §3
477
Proposition 3.2. Let E be an algebraic extension of K. Two embeddings a, r : E + K� give rise to the same absolute value on E if and only if they are conjugate over K v .
Proof Suppose they are conjugate over Kv . Then the uniqueness of the ex tension of the absolute value from Kv to K� guarantees that the induced absolute values on E are equal. Conversely, suppose this is the case. Let A. : r£ aE be an isomorphism over K. We shall prove that A. extends to an is omorphism of rE Kv onto aE · Kv over Kv . Since tE is dense in tE · Kv , an element x E rE · Kv can be written +
·
X =
lim tXn
with xn E E . Since the absolute values induced by a and t on E coincide, it follows that the sequence A.r x = ax n converges to an element of aE Kv which we denote by .A.x. One then verifies immediately that A.x is independent of the particular sequence tx n used, and that the map A. : tE · Kv + aE K v is an iso morphism, which clearly leaves K v fixed. This proves our proposition. In view of the previous two propositions, if w is an extension of v to a finite extension E of K, then we may identify E w and a composite extension EK v of E a nd K v . If N = [ E : K] is finite, then we shall call n
·
·
the local degree. Proposition 3.3.
Let E be a.finite separable extension of K, ofdegree N. Then
Proof We can write E = K ( rt) for a single element rt. Let f(X) be its irreducible polynomial over K. Then over Kv , we have a decomposition f(X)
=
ft (X) · · · fr(X)
into irreducible factors f'i(X). They all appear with multiplicity 1 according to our hypothesis of separability. The em beddings of E into K� correspond to the maps of rt onto the roots of the f'i . Two em beddings are conjugate if and only if they map rt onto roots of the same polynomial f'i . On the other hand, it is clear that the local degree in each case is precisely the degree of f'i . This proves our proposition. Proposition 3.4.
Let E be a finite extension of K. Then L [E w : Kv]
1 . We consider the nature of the solutions of a congruence of type •
•
•
•
•
•
•
•
•
(*)
This congruence is equivalent with the linear congruence ( ** )
If some coefficient c i (i = 1 , . . . , n) is not = 0 (mod p ) , then the set of solutions is not empty, and has the usual structure of a solution of one inhomogeneous linear equation over the field ofp. In particular, it has dimension n 1 . A congruence ( * ) or ( ** ) with some ci ¢ 0 (mod p) will be called a proper 
congruence.
As a matter of notation, we write D i f for the formal partial derivative of f with respect to X i . We write grad f(X)
=
(D 1 f(X), . . . , Dn f(X)).
Proposition 7.2. Let f(X) E o[X]. Let r be an integer > 1 and let A E o be
such that
f (A )
=
D i f (A)
=
D i f(A)
=!
0 (mod p 2 '  1 ), 0 (mod p'  1 ),
for all
i
=
1,
0 (mod p'),
for some i
=
1 , . . . , n.
. . . , n,
Let be an integer > 0 and let B E o < n > be such that v
B
=
A (mod p') and f(B)
=
0 (mod p 2 '  1 + v).
A vector Y E o < n > satisfies Y
=
B (mod p r + v) and f( Y)
=
0 (mod p 2 r+ v )
X I I , §7
ZE ROS OF POLYNOM IALS IN COM PLETE F I E LDS
if and only if Y can be written in the form Y satisfying the proper congruence
f(B) + r{ + v grad f(B) · C
=
=
B
+
n' + v c ,
493
with some C E o
0 (mod p 2 r + v ).
Proof The proof is shorter than the statement of the proposition. Write Y = B + nr + v c. By Taylor ' s expansion, f(B
+
n' + v c ) =
f(B)
+
n' + v
grad f(B) · C (mod p 2 ' + 2 v ) .
'
To solve this last congruence mod p 2 + v , we obtain a proper congruence by hypothesis, because grad f(B) = grad f (A) = 0 (mod p '  1 ) . Corollary
Assumptions being as in Proposition 7.2, there exists a zero off in o< n > which is congruent to A mod p '. Proof We can write this zero as a convergent sum 7.3.
solving for C 1 , C 2 ,
•
•
•
inductively as in the proposition.
Let f be a polynomial in one variable in o [X], and let a E o be such that f(a) 0 (mod lJ) bu t f ' (a) =I= 0 (mod lJ). Th en there exists b E o, b a (mod p ) such that f(b) = 0. Proof Take n = 1 and r = 1 in the proposition, and apply Corollary 7.3.
Corollary
7.4 .
=
=
Corollary 7.5. Let m be a positive integer not divisible by the characteristic
of K. There exists an integer r such that for any a E o, a equation xm  a = 0 has a root in K. Proof Apply the proposition.
=
1 (mod p '), the
Example .
In the 2adic field Q 2 , there exists a square root of  7, i.e. � E Q 2 , because  7 = 1  8. When the absolute value is not discrete, it is still possible to formulate a criterion for a polynomial to have a zero by Newton approximation . (Cf. my paper, "On quasialgebraic closure , " Annals of Math . ( 1 952) pp . 373390. Proposition
Let K be a complete under a nonarchimedean absolute value (nontrivial). Let o be the valuation ring and let f(X) E o [X] be a poly nomial in one variable. Let cx 0 E o be such that I f( cx o ) I < I f' (cxo ) 2 1 (here f' denotes the formal derivative of f). Then the sequence 7.6.
cx i + l
=
f( cx i) cx i f '( cxi )
494
XII, §7
ABSOLUTE VALUES
converges to a root rx off in o, and we have o) < f(rx < 1. I rx  OCo I = f' ( oc o ) 2 Proof Let c = I f (rx 0 )/f' (rx 0 ) 2 1 < 1 . We show inductively that : 1 . I rxi I < 1 , 2. I rx i  rxo I < c, 3.
f (rx ;) < c2 i. = f '(rx i ) 2
These three conditions obviously imply our proposition. If i = 0, they are hypotheses. By induction, assume them for i. Then : 1 . I f (rx i)/ f' (rx i ) 2 1 < c 2 i gives I rxi + 1  rx; l < c 2 i < 1 , whence I rxi + 1 I < 1 . 2 . I rxi + 1  rxo I < max { I rxi + 1  rx; l , I rxi  rxo I } = c. 3. By Taylor ' s expansion, we have f (rx i ) !( rxi + 1 ) = !( rx ;)  ! ' ( rx ;) f
'( rxi)
+
( )
f (rxi) 2 P f '(rx i)
for some p E o, and this is less than or equal to f (rx ;) 2 f '( rx i)
in absolute value. Using Taylor' s expansion on f '(rxi + 1 ) we conclude that From this we get f (rxi + 1 ) i+1 < c 2 f '(rxi + 1 ) 2 =
as desired. The technique of the proposition is also useful when dealing with rings, say a local ring o with maximal ideal m such that m r = 0 for some integer r > 0. If one has a polynomial f in o[X] and an approximate root rx0 such that f '(rx 0) =I= 0 mod
m,
then the Newton approximation sequence shows how to refine rx0 to a root of f. Example in several variables . Let K be complete under a nonarchimedean absolute value. Let f(X 1 , . . . , Xn + 1 ) E K[X] be a polynomial with coefficients in K. Let (a 1 , , an, b) E K n + 1 . Assume that f(a, b) = 0. Let Dn + 1 be the •
.
.
X I I , Ex
EXE RC ISES
495
partial derivative with respect to the (n + 1 )th variable, and assume that Dn + I f( a, b) * 0. Let (a) E K n be sufficiently close to (a) . Then there exists an element 5 of K close to b such that f (a, b) = 0. This statement is an immediate corollary of Proposition 7 . 6 . By multiplying all a;, b by a suitable nonzero element of K one can change them to elements of o . Changing the variables accordingly , one may assume without loss of gen erality that a; , b E o , and the condition on the partial derivative not vanishing is preserved . Hence Proposition 7 . 6 may be applied . After perturbing (a) to (a) , the element b becomes an approximate solution off( a, X) . As (a) approaches (a) , f(a, b) approaches 0 and Dn + tf(a, b) approaches Dn+ tf(a , b) =F 0. Hence for (a) sufficiently close to (a ), the conditions of Proposition 7 . 6 are satisfied, and one may refine b to a root of f(a, X) , thus proving the assertion . The result was used in a key way in my paper "On Quasi Algebraic Closure" . It is the analogue of Theorem 3 . 6 of Chapter XI , for real fields . In the language of algebraic geometry (which we now assume) , the result can be reformulated as follows . Let V be a variety defined over K. Let P be a simple point of V in K. Then there is a whole neighborhood of simple points of V in K. Especially , suppose that V is defined by a finite number of polynomial equations over a finitely generated field k over the prime field . After a suitable projection , one may assume that the variety is affine , and defined by one equation f(X b . . . , Xn + I ) = 0 as in the above statement , and that the point is P == (a I , . . . , a n , b) as above . One can then select a; = X; close to a; but such that (x i , . . . , xn ) are algebraically independent over k . Let y b� the refinement of b such that f(x, y) = 0. Then (x, y) is a generic point of V over k, and the coordinates of (x, y) lie in K. In geometric terms , this means that the function field of the variety can be embedded in K over k, just as Theorem 3 . 6 of Chapter XI gave the similar result for an embedding in a real closed field , e . g . the real numbers .
EX E R C I S ES 1. (a) Let K be a field with a valuation. If
is a polynomial in K [X], define I j I to be the max on the values l ai l (i = 0, . . . , n). Show that this defines an extension of the valuation to K[X], and also that the valuation can be extended to the rational field K(X). How is Gauss' lemma a special case of the above statement ? Generalize to polynomials in several variables. (b) Let f be a polynomial with complex coefficients. Define I f I to be the maximum of the absolute values of the coefficients. Let d be an integer > 1 . Show that
496
XII, Ex
ABSOLUTE VALU ES
there exist constants C 1 , C (depending only on d) such that, if I, g are polynomials 2 in C[X] of degrees < d, then
Induction on the number of factors of degree 1 . Note that the right inequality is trivial.]
[Hint :
2. Let M0 be the set of absolute values consisting of the ordinary absolute value and all padic absolute values vP on the field of rational numbers Q. Show that for any rational number a E Q, a =F 0, we have
n l a lv =
1.
V E MQ
If K is a finite extension of Q, and M K denotes the set of absolute values on K extending those of M0 , and for each w E MK we let N w be the local degree [Kw : Qv], show that for ex E K, ex =F 0, we have
n l ex l�w =
l.
w e Mx
3. Show that the padic numbers Q P have no automorphisms other than the identity. [Hint : Show that such automorphisms are continuous for the padic topology. Use Corollary 7.5 as an algebraic characterization of elements close to 1 .] 4.
Let A be a principal entire ring, and let K be its quotient field. Let o be a valuation ring of K containing A , and assume o =F K. Show that o is the local ring A < P> for some prime element p. [This applies both to the ring Z and to a polynomial ring k[X] over a field k.]
5. Let A be an entire ring, and let K be its quotient field. Assume that every finitely generated ideal of A is principal. Let o be a discrete valuation ring of K containing A . Show that o = A < P> for some element p of A , and that p is a generator of the maximal ideal of o .
6. Let QP be a padic field. Show that QP contains infinitely many quadratic fields of type Q(�), where m is a positive integer.
7. Show that the ring of padic integers ZP is compact. Show that the group of units in ZP is compact. 8. If K is a field complete with respect to a discrete valuation, with finite residue class field, and if o is the ring of elements of K whose orders are > 0, show that o is compact. Show that the group of units of o is closed in o and is compact.
9. Let K be a field complete with respect to a discrete valuation, let o be the ring of integers of K, and assume that o is compact. Let /1 , f , be a sequence of polynomials in n 2 variables, with coefficients in o . Assume that all these polynomials have degree < d, and that they converge to a polynomial f (i.e. that I f  /; I + 0 as i + oo ) . If each /; has a zero in o , show that f has a zero in o . If the polynomials /; are homogeneous of degree d, and if each li has a nontrivial zero in o, show that I has a nontrivial zero in o. [Hint : Use the compactness of o and of the units of o for the homogeneous case.] (For applications of this exercise , and also of Proposition 7 . 6 , cf. my paper "On quasialgebraic closure ," Annals of Math . , 55 ( 1 952) , pp . 4 1 2444 . ) •
•
•
EXERC ISES
XII, Ex
497
1 0. Show that if p, p' are two distinct prime numbers, then the fields Q P and Q P . are not isomorphic. 1 1 . Prove that the field Q P contains all (p  1 )th roots of unity. [Hint : Use Proposition 7.6, applied to the polynomial x p  1  1 which splits into factors of degree 1 in the residue class field.] Show that two distinct (p  1 )th roots of unity cannot be congruent mod p.
1 2 . (a) Let f(X) be a polynomial of degree I in Z[X] . Show that the values f(a) for a E Z are divisible by infinitely many primes . (b) Let F be a finite extension of Q . Show that there are infinitely many primes p such that all conjugates of F (in an algebraic closure of QP ) actually are contained in QP . [Hint : Use the irreducible polynomial of a generator for a Galois extension of Q containing F . ]
1 3 . Let K be a field of characteristic 0, complete with respect to a nonarchimedean absolute value. Show that the series exp(x) log( 1 + x)
=
x 2 x3 1 +x +++
=
x 2 x3 x+
2!
2

3
3!

· · ·
· · ·
converge in some neighborhood ofO. (The main problem arises when the characteristic of the residue class field is p > 0, so that p divides the denominators n ! and n. Get an expression which determines the power of p occurring in n !.) Prove that the exp and log give mappings inverse to each other, from a neighborhood of 0 to a neighborhood of 1 . 14. Let K be as in the preceding exercise, of characteristic 0, complete with respect to a non archimedean absolute value. For every integer n > 0, show that the usual binomial expansion for ( 1 + x) 1 1" converges in some neighborhood of O. Do this first assuming that the characteristic of the residue class field does not divide n, in which case the asser tion is much simpler to prove. 1 5 . Let F be a complete field with respect to a discrete valuation, let o be the valuation ring, n a prime element, and assume that o/(n) = k. Prove that if a, b E o and a = b (mod n') with r > 0 then a P" = b P" (mod nr + " ) for all integers n > 0. 1 6 . Let F be as above. Show that there exists a system of representatives R for o/( n) in o such that R P = R and that this system is unique (Teichmiiller). [Hint : Let ex be a residue class in k. For each v > 0 let av be a representative in o of a P v and show that the sequence a� v converges for v + oo , and in fact converges to a representative a of ex, independent of the choices of av . ] Show that the system of representatives R thus obtained is closed under multiplication, and that if F has characteristic p, then R is closed under addition, and is isomorphic to k. 1 7 . (a) (Witt vectors again) . Let � be a perfect field of characteristic p. We use the Witt vectors as described in the exercises of Chapter VI . One can define an absolute value on W(k) , namely l x l = p r if xr is the first nonzero component of x. Show that this is an absolute value , obviously discrete , defined on the ring , and which can be extended at once to the quotient field . Show that this quotient field is complete , and note that W(k) is the valuation ring . The maximal ideal consists of those x such that x0 = 0, i . e . is equal to pW(k) .
498
XI I , Ex
ABSOLUTE VALU ES
(b) Assume that F has characteristic 0. M ap each vector x E W(k) on the element
where � i is a representative of x i in the special system of Exercise 1 5. Show that this map is an embedding of W(k) Into o .
1 8 . (Local uniformization). Let k be a field, K a finitely generated extension of transcendence degree 1 , and o a discrete valuation ring of K over k, with maximal ideal m. Assume that o/m = k. Let x be a generator ofm, and assume that K is separable over k(x). Show that there exists an element y E o such that K = k(x, y), and also having the following property. Let qJ be the place on K determined by o. Let a = qJ(x), b = ({J(y) (of course a = 0). Let f(X, Y) be the irreducible polynomial in k[X, Y] such that f(x, y) = 0. Then D2 f (a, b) # 0. [Hint : Write first K = k(x, z ) where z is integral over k[x]. Let z = z . , . . , z" (n > 2) be the conjugates of z over k(x), and extend o to a valuation ring .0 of k(x, z 1 , . . , z" ). Let .
.
be the power series expansion of z with a i E k, and let P,(x) = a0 + · · · + a, x r. For i = 1 , . . . , n let Yi =
Z;
 P,(x)

x'
Taking r large enough, show that y1 has no pole at .0 but y2 , , Yn have poles at .0. The elements y 1 , , Yn are conjugate over k(x). Let f(X, Y) be the irreducible poly nomial of (x, y) over k. Then f(x, Y) = 1/Jn(x) Y" + + I/J0(x) with 1/J i(x)k[x]. We may also assume 1/J i(O) # 0 (since f is irreducible). Write f(x, Y) in the form •
•
•
•
•
•
·
·
·
Show that I/Jn(x)y2 · Yn = u does not have a pole at .0. If w E .0, let w denote its residue class modulo the maximal ideal of .0. Then •
•
o
=�: f( x , Y)
1
= (  1 )"  u( Y  y 1 ).
Let y = y. , y = b. We find that D2 f(a, b) = (  1 ) "

1
u # 0.]
1 9 . Prove the converse of Exercise 1 7, i.e. if K = k(x, y), f(X, Y) is the irreducible poly nomial of (x, y) over k, and if a, b E k are such that f(a, b) = 0, but D2 f(a, b) # 0, then there exists a unique valuation ring o of K with maximal ideal m such that x = a and y = b (mod m). Furthermore, ojm = k, and x  a is a generator of m. [Hint : If g(x, y) E k[x, y] is such that g(a, b) = 0, show that g(x, y) = (x  a)A(x, y)/B(x, y) where A , B are polynomials such that B(a, b) # 0. If A(a, b) = 0 repeat the process. Show that the process cannot be repeated indefinitely, and leads to a proof of the desired assertion.] 20 . (lss'saHironaka Ann . of Math 83 ( 1 966) , pp . 3446) . This exercise requires a good working knowledge of complex variables . Let K be the field of merom orphic functions on the complex plane C . Let .0 be a discrete valuation ring of K (containing the
EXERC ISES
X I I , Ex
499
constants C ) . Show that the function z is in n . [Hint: Let a 1 , a2, be a discrete sequence of complex numbers tending to infinity, for instance the positive integers . Let v 1 , v 2 , , be a sequence of integers , 0 < V; < p I , for some prime number p , such that � v; p ; is not the padic expansion of a rational number. Letfbe an entire function having a zero of order v; p ; at a; for each i and no other zero . If z i s not in o, consider the quotient •
•
•
•
•
•

g(z)
=
f (z)
_" ____
n (z
i= 1

ai)Vipi
From the Weierstrass factorization of an entire function, show that g(z) = h(z)P" + 1 for some entire function h(z ). Now analyze the zero of g at the discrete valuation of o in terms of that of f and n (z a iripi to get a contradictio n.] If U is a noncompact Riemann surface, and L is the field of meromorphic functions on U, and if o is a discrete valuation ring of L containing the constants, show that every holomorphic function ({J on U lies in o . [Hint : Map ({J : U + C, and get a discrete valuation of K by composing qJ with merom orphic functions on C. Apply the first part of the exercise.] Show that the valuation ring is the one associated with a complex number. [Further hint : If you don't know about Riemann surfaces, do it for the complex plane. For each z E U, let fz be a function holomorphic on U and having only a zero of order 1 at z. If for some z0 the function fz o has order > 1 at o , then show that o is the valuation ring associated with z 0 • Otherwise, every function fz has order 0 at o . Conclude that the valuation of o is trivial on any holomorphic function by a limit trick analogous to that of the first part of the exercise.] 
Pa rt
Th ree
L I N EAR ALG E B RA and R E P R E S E N TATI O N S
We shall be concerned with modules and vector spaces, going into their structure under various points of view. The main theme here is to study a pair, consisting of a module, and an endomorphism, or a ring of endomorphisms, and try to decompose this pair into a direct sum of components whose structure can then be described explicitly. The direct sum theme recurs in every chapter. Sometimes, we use a duality to obtain our direct sum decomposition relative to a pairing, and sometimes we get our decomposition directly. If a module refuses to decompose into a direct sum of simple components, then there is no choice but to apply the Grothendieck construction and see what can be ob tained from it. The extension theme occurs only once, in Witt's theorem, in a brief counter point to the decomposition theme.
501
C H A PTE R
XI l l
M atrices an d Lin ear M aps
Presumably readers of this chapter will have had some basic acquaintance with linear algebra in elementary courses . We go beyond such courses by pointing out that a lot of results hold for free modules over a commutative ring . This is useful when one wants to deal with families of linear maps , and reduction modulo an ideal . Note that §8 and §9 give examples of group theory in the context of linear groups . Throughout this chapter, we let R be a commutative ring, and we let E, F be R modules . We suppress the prefix R in front of linear maps and modules .
§1 .
M AT R I C ES
By an m x n matrix in R one means a doubly indexed family of elements of R, (a ii), (i = 1 , . . . , m and j = 1 , . . . , n), usually written in the form
We call the elements a;i the coefficients or components of the matrix . A 1 x n matrix is called a row vector (of dimension , or size, n) and a mx 1 matrix is called a column vector (of dimension, or size, m) . In general, we say that (m, n ) is the size of the matrix, or also m x n. We define addition for matrices of the same size by components. If A = (a;i) and B = (b;i) are matrices of the same size, we define A + B to be the matrix whose ijcomponent is a ii + b ii . Addition is obviously associative. We define the multiplication of a matrix A by an element c E R to be the matrix (ca ii), 503
504
XI I I , § 1
MATR ICES AND LIN EAR MAPS
whose ijcomponent is ca ii . Then the set of m x n matrices in R is a module (i.e. an Rmodule). We define the product AB of two matrices only under certain conditions. Namely, when A has size (m, n) and B has size (n, r), i.e. only when the size of the rows of A is the same as the size of the columns of B. If that is the case, let A = (a ii) and let B = (bik ). We define AB to be the m x r matrix whose ik component is n L a ii bik · 1 j=
If A, B, C are matrices such that AB is defined and BC is defined, then so is (AB)C and A(BC) and we have (AB)C
=
A(BC) .
This is trivial to prove. If C = (ck 1), then the reader will see at once that the ifcomponent of either of the above products is equal to
L L a ii bik ck l · j
k
An m x n matrix is said to be a square matrix if m = n. For example, a 1 x 1 matrix is a square matrix, and will sometimes be identified with the element of R occurring as its single component.
For a given integer n
>
1 the set of square n
x
n matrices forms a ring.
This is again trivially verified and will be left to the reader. The unit element of the ring of n x n matrices is the matrix 0 ... 0 1 In = 0
0 0 1
whose components are equal to 0 except on the diagonal, in which case they are equal to 1 . We sometimes write I instead of In . If A = (a i i) is a square matrix, we define in general its diagonal components to be the elements a ii . We have a natural ringhomomorphism of R into the ring of n x n matrices, given by Thus ci is the square n x n matrix having all its components equal to 0 except " the diagonal components, which are equal to c. Let us denote the ring of n x n
XIII, § 1
MATR ICES
505
matrices in R by Mat n(R). Then Mat n(R) is an algebra over R (with respect to the above homomorphism). Let A = (a ii ) be an m x n matrix . We define its transpose 1A to be the matrix (ai; ) (j = 1 , . . . , n and i = 1 , . . . , m) . Then 1A is an n x m matrix . The reader will verify at once that if A , B are of the same size , then
If c E R then r(c A) have
=
c !4 . If A, B can be multiplied, then r B � is defined and we
We note the operations on matrices commute with homomorphisms. More precisely, let q> : R R' be a ringhomomorphism. If A, B are matrices in R, we define q>A to be the matrix obtained by applying q> to all the components of A. Then +
q>(A + B)
=
q>(AB)
q>A + q>B,
q>(' A)
=
=
(q>A)(q>B),
q>(cA)
=
q>(c)q>A,
'q>(A).
A similar remark will hold throughout our discussion of matrices (for instance in the next section). Let A = (a ii) be a square n x n matrix in a commutative ring R. We define the trace of A to be tr(A)
n
=
L ai i ; i= 1
in other words, the trace is the sum of the diagonal elements.
If A, B are n
x
n matrices, then tr(AB)
Indeed ' if A
=
(a · .) and B l)
=
=
tr(BA).
(b · ·) then
tr(AB)
IJ
=
L L aiv b vi i v
=
tr(BA).
As an application, we observe that if B is an invertible n tr(B  1 AB) Indeed, tr(B  1 AB)
=
tr(ABB  1 )
=
tr(A).
=
tr(A).
x
n matrix, then
506
MATR ICES AND LINEAR MAPS
§2.
T H E R A N K O F A M AT R IX
XI I I , §2
Let k be a field and let A be an m x n matrix in k. By the row rank of A we shall mean the maximum number of linearly independent rows of A, and by the column rank of A we shall mean the maximum number of linearly independen t columns of A. Thus these ranks are the dimensions of the vector spaces gen erated respectively by the rows of A and the columns of A. We contend that these ranks are equal to the same number, and we define the rank of A to be that number. Let A 1 , . . . , A " be the columns of A, and let A 1 , , A m be the rows of A. Let rx = ( x b . . . ' Xm ) have components Xj E k. We have a linear map .
x � x 1A 1 +
· ··
.
•
+ xm A m
of k< m> onto the space generated by the row vectors. Let W be its kernel. Then W is a subspace of k< ml and dim W + row rank =
m.
If Y is a column vector of dimension m, then the map
is a bilinear map into k, if we view the 1 x 1 matrix r X Y as an element of k. We observe that W is the orthogonal space to the column vectors A 1 , , A " , i.e. it is the space of all X such that X · Ai = 0 for all j = 1, . . . , n . By the duality theorem of Chapter III, we know that k<m > is its own dual under the pairing •
.
•
(X, Y) r+ X · Y and that k<m >jW is dual to the space generated by A 1 ,
•
•
•
, A " . Hence
dim k< m>/W = column rank, or dim W + column rank =
m.
From this we conclude that column rank = row rank, as desired. We note that W may be viewed as the space of solutions of the system of n linear equations
507
MATRICES AND LINEAR MAPS
X I I I , §3
in m unknowns x 1 , . . . , x m . Indeed, if we write out the preceding vector equation in terms of all the coordinates, we get the usual system of n linear equations. We let the reader do this if he or she wishes.
§3.
M AT R I C ES A N D LI N EA R M A PS
Let E be a module, and assume that there exists a basis CB = { e 1 , , en } for E over R . This means that every element of E has a unique expression as a linear combination •
•
•
with X; E R. We call (x 1 , , xn ) the components of x with respect to the basis. We may view this ntuple as a row vector. We shall denote by X the transpose of the row vector (x 1 , , x") . We call X the column vector of x with respect to •
•
•
•
•
•
the basis.
We observe that if { e '1 , , e� } is another basis of E over R, then m = n. Indeed, let p be a maximal ideal of R . Then E/pE is a vector space over the field RjpR, and it is immediately clear that if we denote by �; the residue class of ei mod pE , then { � 1 , , �"} is a basis for EjpE over R jp R . Hence n is also the dimension of this vector space, and we know the in variance of the cardinality for bases of vector spaces over fields. Thus m = n. We shall call n the dimension of the module E over R. We shall view R as the module of column vectors of size n. It is a free module of dimension n over R . It has a basis consisting of the unit vectors e 1 , , e" such that •
.
•
•
.
•
•
•
•
rei
=
( 0, . . . , 0, 1, 0, . . . , 0)
has components 0 except for its ith component, which is equal to 1 . An m x n matrix A gives rise to a linear map
by the rule X t+ AX .
Namely, we have A ( X vectors X, Y and c E R.
+
Y)
= AX
+ A Y and A(eX)
= cAX
for column
508
Xl l l , §3
MATR ICES AND LINEAR MAPS
The above considerations can be extended to a slightly more general context, which can be very useful. Let E be an abelian group and assume that R is a commutative subring of Endz (E) = Homz(E, E). Then E is an Rmodule. Furthermore, if A is an m a linear map
x
n matrix in R, then we get
defined by a rule similar to the above, namely X H AX. However, this has to be interpreted in the obvious way. If A = (a ii) and X is a column vector of elements of E, then Yt
AX = where Y I·
n
=
" �
j= 1
. .
'
...
Ym
a I) x J • ·
If A, B are matrices in R whose product is defined, then for any c E R we have LAB = LA LB and
LeA = c LA .
Thus we have associativity, namely
A(BX) = (AB)X. An arbitrary commutative ring R may be viewed as a module over itself. In this way we recover the special case of our map from R into R<m>. Further more, if E is a module over R, then R may be viewed as a ring of endomorphisms of E. Proposition 3. 1. Let E be a free module over R, and let {x 1 , . . . , Xn} be a basis. Let y 1 , . • • , Yn be elements of E. L et A be the matrix in R such that Yt Yn
Then { y b . . . , Yn} is a basis of E if and only if A is invertible. Proof. Let X, Y be the column vectors of our elements. Then AX = Y. Suppose Y is a basis. Then there exists a matrix C in R such that C Y = X.
MATRICES AND LINEAR MAPS
X I I I , §3
509
Then CAX = X, whence CA = I and A i s invertible. Conversely, assume that A i s invertible. Then X = A  1 Y and hence x 1 , , xn are in the module generated by y 1 , , Yn · Suppo s e that we have a relation •
.
.
•
•
•
with b i E R. Let B be the row vector (b 1 ,
•
•
•
, b n). Then
BY = 0 and hence BAX = 0. But {x 1 , , x n } is a basis. Hence BA = 0, and hence BAA  1 = B = 0. This proves that the components of Y are linearly indepen dent over R, and proves our proposition. •
.
•
We return to our situation of modules over an arbitrary commutative ring R . Let E, F be modules . We shall see how we can associate a matrix with a linear map whenever bases of E and F are given. We assume that E, F are free . We let CB = { �1 , , �n} and CB ' = {�� , . . . , �;, } be bases of E and F res p ec tive ly Let •
•
.
•
be a linear map. There exist unique elements a ii E R such that
or in other words, J< e j)
m
= L a ij �� 1 i=
(Observe that the sum is over the first index.) We define If X = X 1 e 1 + . . . + Xn �n i s expressed in terms of the basi s let us denote the column vector X of components of x by M. Let f be an nmultilinear map. If we take two indices i, j and i # j then fixing all the variables except the ith and jth variable, we can view f as a bilinear map on E, x Ei . Assume that E 1 = · · · = En = E. We say that the multilinear map f is alternating iff (x 1 , , xn) = 0 whenever there exists an index i, 1 < i < n  1, such that x i = x1 + 1 (in other words, when two adjacent elements are equal). •
•
•
Proposition 4.1 . Let f be an nmultilinear alternating map o n X 1 , . . . , X n E E. Then f ( · · · , X; , Xi + b
· · ·) =
 f (·
· ·
E.
Let
, X i + b Xi , · · · ) .
In other words, when we interchange two adjacent arguments off, the value , X n) = 0 . off changes by a sign. If x i = xi for i # j then f (x 1 , •
•
•
51 2
j y
=
E
MATR ICES AND LINEAR MAPS
XI I I , §4
Proof.
Restricting our attention to the factors in the ith and jth place, with i + 1 , we may assume f is bilinear for the first statement. Then for all x, E we have 0
=
f(x + y, x
+
y)
=
f(x, y)
+
f(y, x).
This proves what we want, namely f(y, x) =  f(x, y). For the second asser tion, we can interchange successively adjacent arguments of f until we obtain an ntuple of elements of E having two equal adjacent arguments. This shows that when x i = xi , i # j, then f(x b . . . , xn ) = 0. Corollary 4.2. Let f be an nmultilinear alternating map on E. Let
x 1 , , Xn E E. Let i # j and let a e R. Then the va lue off on (x 1 , , xn ) does not change if we replace xi by xi + a xi and leave all other components fixed. •
•
•
.
•
.
Proof. Obvious. A multilinear alternating map taking its value in R is called a multilinear alternating form. On repeated occasions we shall evaluate multilinear alternating maps on linear combinations of elements of E. Let
Let f be nmultilinear alternating on E. Then We expand this by multilinearity, and get a sum of terms of type
where u ranges over arbitrary maps of { 1 , . . . , n} into itself. If u is not a bijection (i.e. a permutation), then two arguments Va< i> and Vau> are equal for i =I= j, and the term is equal to 0. Hence we may restrict our sum to permutations u. Shuffling back the elements (va < t > ' . . . , Va ) to their standard ordering and using Proposition 4. 1, we see that we have obtained the following expansion : Lemma 4.3. If w 1 ,
•
•
•
,
wn are as above, then
L £( u)a l , a( l ) . . . an, a (n) f(v b . . . ' Vn ) a where the sum is taken over all permutations u of { 1 , . . . , n} and £(u) is the sign of the permutation. f(w l , . . . ' wn)
=
DETE RMI NANTS
X I I I , §4
51 3
For determinants, I shall follow Artin 's treatment in Galois Theory. By an n x n determinant we shall mean a mapping also written
D : Mat,.(R)
+
R
which, when viewed as a function of the column vectors A 1 , , A " of a matrix A, is multilinear alternating, and such that D(J) = 1 . In this chapter, we use mostly the letter D to denote determinants. We shall prove later that determinants exist. For the moment, we derive p roperties. •
Theorem 4.4 . (Cramer's Rule). Let A 1 , sion n. Let x 1 , , X n E R be such that •
•
•
•
, A" be column vectors of dimen
.
•
•
.
X1A 1
+
· · ·
+
n Xn A =
B
for some column vector B. Then for each i we have xi D( A 1 , , A") = D( A 1 , , B, . . . , A "), •
•
•
•
•
•
where B in this last line occurs in the ith place. Proof. Say i = 1 . We expand D(B, A 2 ,
n
•
•
•
, A ") = L xi D(A i, A 2 , j= 1
•
•
•
, A "),
and use Proposition 4. 1 to get what we want (all terms on the right are equal to 0 except the one having x 1 in it). Corollary 4.5. Assume that R is a field. Then A 1 ,
dependent if and only if D(A 1 ,
•
•
•
, A ")
=
0.
•
•
•
, A " are linearly
Proof. Assume we have a relation with xi E R. Then xi D(A) = 0 for all i. If some x i =F 0 then D(A) = 0. Con versely, assume that A 1 , , A" are linearly independent. Then we can express the unit vectors e 1 , e" as linear combinations •
•
.
.
,
•
•
51 4
XI I I , §4
MATR ICES AND LINEAR MAPS
with bii E R. But 1
=
D( e 1 ,
•
•
.
, e").
Using a previous lemma, we know that this can be expanded into a sum of terms involving D(A 1 , . . . , A " ), and hence D(A) cannot be 0. Proposition 4.6. If determinants exist, they are unique. If A 1 ,
the column vectors of dimension n, of the matrix A = (aii), then D(A 1 ' . ' A") = L £( u)aa ( 1 ) , 1 . . . aa (n) , n ' .
.
•
.
•
, A " are
0'
where the sum is taken over all permutations a of { 1 , . . . , n}, and £(a) is the sign of the permutation. Proof. Let e 1 , e" be the unit vectors as usual. We can write •
•
•
,
Therefore
D(A 1 ' . . . ' A ")
=
L i ( a)aa( l ) , l . . . aa ( n) , n 0'
by the lemma. This proves that the value of the determinant is uniquely deter mined and is given by the expected formula. Corollary 4.7. Let q> : R
+
R' be a ringhomomorphism into a commutative ring. If A is a square matrix in R, define q>A to be the matrix obtained by applying qJ to each component of A . Then
q>(D(A))
=
D(q>A).
Proof. Apply qJ to the expression of Proposition 4.6. Proposition 4.8. If A is a square matrix in R then
D(A)
=
D('A).
Proof. In a product a a ( 1 ) , l . . . aa ( n ) , n
each integer k from 1 to n occurs precisely once among the integers a( 1 ), . . . , u(n ). Hence we can rewrite this product in the form
D ETERMI NANTS
X I I I , §4
51 5
Since £(a) = £ ( a  1 ) , we can rewrite the sum in Proposition 4.6 in the form
L £ (u 1 )a 1 , a 
a

1( 1)
. . . a,., a
l (n) ·
In this sum, each term corresponds to a permutation a. However, as a ranges over all permutations, so does a  1 . Hence our sum is equal to
L £(a)a 1 , a< 1 > • • • an , a , a
which is none other than D('A), as was to be shown. Corollary
The determinant is multilinear and alternating with respect to the rows of a matrix. 4.9.
We shall now prove existence, and prove simultaneously one additional important property of determinants. When n = 1 , we define D(a) = a for any a E R. Assume that we have proved the existence of determinants for all integers < n (n > 2). Let A be an n x n matrix in R, A = (a ii). We let A ii be the (n  1 ) x (n  1) matrix obtained from A by deleting the ith row and jth column. Let i be a fixed integer, 1 < i < n. We define inductively i D(A) = (  1 ) i + 1 a i l D(A il ) + · · · + (  1 ) + na in D(A in). ( This is known as the expansion of D according to the ith row.) We shall prove that D satisfies the definition of a determinant. Consider D as a function of the kth column, and consider any term (  l ) i + ia ii D(A ii). Ifj =F k then a ii does not depend on the kth column, and D(A ii) depends linearly on the kth column. If j = k, then a ii depends linearly on the kth column, and D(A ii) does not depend on the kth column. In any case our term depends linearly on the kth column. Since D(A) is a sum of such terms, it depends linearly on the kth column, and thus D is multilinear. Next, suppose that two adjacent columns of A are equal, say A k = A k + 1 . Let j be an index =1= k and =F k + 1 . Then the matrix A ii has two adjacent equal columns, and hence its determinant is equal to 0. Thus the term corresponding to an index j =1 k or k + 1 gives a zero contribution to D(A). The other two terms can be written
The two matrices A ik and A i , k + 1 are equal because of our assumption that the kth column of A is equal to the ( k + 1 )th column. Similarly, a ik = ai , k + 1 •
51 6
XI I I , §4
MATR ICES AND LIN EAR MAPS
Hence these two terms cancel since they occur with opposite signs. This proves that our form is alternating, and gives : Proposition
4. 10.
Determinants ex ist and satisfy the rule of expansion
according to rows and columns.
(For columns, we use the fact that D(A) = DCA).) Example.
We mention explicity one of the most important determinants . Let x i , . . . , xn be elements of a commutative ring . The Vandermonde deter minant V = V(x I , . . . , xn ) of these elements is defined to be
1
V=
xi X nt
I
1
... ...
x 2n  I
...
x2
1
Xn X nn
I
whose value can be determined explicitly to be
V= n
l" E L :(E) can then be written in a unique way as c� 1 , with some c E R, namely c = q>(v 1 , , v" ). This proves what we wanted. Any two bases of L:(E) in the preceding corollary differ by a unit in R. In other words, if � is a basis of L:(E), then � = c� 1 = �c for some c E R , and c must be a unit. Our � 1 depends of course on the choice of a basis for E. When we consider R o j· < n > , E' X · · · x E
'
J( n )
q>
: E � F we can a�sociate the (/)
E x ··· x E�F
where f is the product of f with itself n times. The map L:( J ) : L: (E, F) � L:(E', F) given by (/) � (/) o
f < n> ,
is obviously a linear map, which defines our functor. We shall sometimes write f * instead of L; ( J) . In particular, consider the case when E = E' and F = R. We get an induced map
Let E be a free module over R, ofdimension n. Let {d} be a basis of L:(E). Let f : E � E be an endomorphism of E. Then f * � = D(f) � . Proposition 4. 19.
Proof This is an immediate consequence of Theorem 4. 1 1 . Namely, we let {v 1 , . . . , vn } be a basis of E, and then take A ( or � ) to be a matrix of f relative to this basis. By definition, f * � (v l , Vn ) = �(f(v l ) , , f(vn)), ·
·
·
,
·
·
·
DETERMI NANTS
XIII , §4
52 1
and by Theorem 4. 1 1 , this is equal to D(A) �( v b . . . , vn). By Corollary 4. 1 2, we conclude that f * � = D(A)� since both of these forms take o n the same value o n ( v 1 , , vn). The above considerations have dealt with the de terminant as a function on •
•
•
all endomorphisms of a free module . One can also view it multiplicatively , as a homomorphism.
det : G Ln (R) + R * from the group of invertible n x n matrices over R into the group of units of R . The kernel of this homomorphism, consisting of those matrices with deter minant 1 , is called the special linear group, and is denoted by SL n (R). We now give an application of determinants to the situation of a free module and a submodule considered in Chapter III , Theorem 7 . 8 . Proposition 4.20. Let R be a principal entire ring. Let F be a free module over R and let M be a finitely generated submodule . Let {e 1 , , em, . . . } be •
•
•
a basis of F such that there exist nonzero elements a 1 , . . . , am E R such that: (i) The elements a 1 e 1 , , am em form a basis of M over R . (ii) We have a; l a;+ 1 for i = l , . . . , m  1 . •
•
•
Let L� be the set of all smultilinear alternating forms on F. Let Js be the ideal generated by all elements f(y 1 , , y5), with f E L� and Yt , . . . , Ys E M. Then •
•
Proof. We first show that J s written in the form
c
•
(a 1
· · ·
as ) .
Indeed, an element y E M can be
Hence if y 1 , • • • , Ys E M , andfis multilinear a l te rnati ng on F, th en f(Y t , . . . , Ys) is e q u al to a sum in terms of type
This is nonzero only when e i 1 , , e is are distinct, in which case the product a1 a s di v ide s this term, and hence J s is contained in the stated ideal. Conversely, we show that there exists an smultilinear alternating form which gives precisely this product. We deduce this from determinants. We can write F as a direct s um •
· · ·
•
•
522
X I I I , §5
MATR ICES AND LIN EAR MAPS
with some submodule Fr . Let /; (i = 1, . . . , r) be the linear map F + R such t hat /; (ei) = �ii ' and such that/; has value 0 on Fr . For v 1 , , V5 E F we define •
•
•
Then f is multilinear alternating and takes on the value as well as the value This proves the proposition . Th e uniqu e n es s of Chapt er III , Th eorem 7 . 8 is now obvious, since first (a 1 ) is unique , then (a 1 a 2 ) is unique and the quotient (a 2 ) is unique , and so forth by induction . Remark. Compare the above theorem with Theorem 2 . 9 of Chapter XIX , in the theory of Fitting ideals , which gives a fancier context for th e res u l t.
§5.
D U A LITY
Let R be a commutative ring, and let E, F be modules over R. An R bilinear form o n E x F is a map
f: E
X
F + R
having the following properties : For each x E E, t he map
y t+ f(x, y) is Rlinear, and for each y E F, the map
x 1+ f(x, y) is Rlinear. We shall omit the p refi x R i n the rest of this section, and write (x, y) 1 or (x, y) instead o f f(x, y). If x E F, we write x j_ y if (x, y) = 0. Similarly, if S is a subset of F, we define x j_ S if x j_ y fo r all y E S. We then say that x is perpendicular to S. We let SJ.. consist of all elements o f E which are perpendicular to S. It is obviously a submodule of E. We define perpendicu larity on the other side in the same way. We define the kernel o ff on the left to be Fl. and the kernel on the right to be EJ.. . We say that f is nondegenerate on the left if its kernel on the left is 0. We say that f is nondegenerate on the right if its kernel on the right is 0. If E0 is the kernel o ff on the left, then we
DUALITY
XI I I , §5
523
get an induced bilinear map E/E0
x
F�R
which is nondegenerate on the left, as one verifies trivial ly from the definitions. Sim ilarly, if F 0 is the kernel off on the rig ht, we get an induced bilinear map E/E0
x
F/F 0 � R
which is nondegenerate on either side. This map arises from the fact that the value (x, y) depends only on the coset of x modulo E 0 and the coset of y modulo F0 • We shall denote by L2(E, F ; R) the set of all bilinear maps of E x F into R. It is clear that this set is a module (i.e. an R module) addition of maps being the usual one, and also multiplication of maps by elements of R. The form f gives rise to a homomorphism ,
q> 1 :
E � HomR(F, R)
such that q>1
(x )(y )
=
f(x, y)
=
(x, y),
for all x e Eand y e F. We shall call HomR(F, R) the dual module ofF, and denote it by F v . We have an isomorphism
given by f t+ q>1 , its inverse being defined in the obvious way : If q>
: E � HomR(F, R)
is a homomorphism, we let f be such that
f(x, y)
=
q>(
x) (y)
.
We shall say that I is nonsingular on the left if q>1 is an isomorphism, in other words if our form can be used to identify E with the dual module of F. We define nonsingular on the right in a similar way, and say that I is non singular if it is nonsingular on the left and on the right. Warning :
Nondegeneracy does not necessarily imply nonsingularity.
We shall now obtain an isomorphism
depending on a fixed nonsingular bilinear map f : E
x
F � R.
524
MATR ICES AN D LINEAR MAPS
X I I I , §5
Let A E EndR(E) be a linear map of E into itself. Then the map
(x, y) � (Ax, y) = ( A x , y) 1 is bilinear, and in this way, we associate linearly with each A E EndR(E) a bilinear map in L 2 (E, F ; R). Conversely, let h : E x F + R be bilinear. Given x E E, the map hx : F + R such that hx( Y) = h(x, y) is linear, and is in the dual space F v . By assumption, there exists a unique element x' E E such that for all y E F we have
h(x, y)
=
(x ' , y).
It is clear that the association x � x ' is a linear map of E into itself. Thus with each bilinear map E x F + R we have associated a linear map E + E. It is immediate that the mappings described in the last two paragraphs are inverse isomorphisms between EndR(E) and L 2 (E, F ; R). We emphasize of course that they depend on our form f. Of course, we could also have worked on the right, and thus we have a similar isomorphism L 2 (E, F ; R) � EndR(F)
depending also on our fixed nonsingular form f. As an application , let A : E � E b e linear, and let (x, y) � (Ax, y) be its associated bilinear map . There exists a uniqu e linear map such that
(Ax, y)
=
(x, rAy)
for all x E E and y E F. We call t A the transpose of A with respect to f It is immediately clear that if, A, B are linear maps of E into itself, then for C E R, r(cA)
=
c'A,
t(A + B)
=
rA
+
'B, and t(AB)
=
r Br A .
More g e nerall y , let E, F be modules with nonsingular bilinear forms denoted by ( , )£ and ( , )F respectively . Let A : E � F be a linear map . Then by the nonsingularity of ( , )£ the re exists a uniqu e linear map 1A : F � E such that (Ax, y)F
=
(x, 1Ay)E for all x E E and y E F .
We also call 1A the transpose with respect to these forms . Examples. For a nice classical example of a transpose , see Exe rci se 3 3 . For the systematic study when a linear map is equal to its transpose , see th e
DUALITY
XI I I , §5
525
spec tral theorems of Cha pter XV . Next I give anot her ex am p l e of a transpose from analysis as follows . Let E be the (infinite dimensional) vector space of ex functions on R , having compact support , i . e . equal to 0 outside some finite interval . We define the scalar prod uc t
J f(x)g(x)dx . X
(!, g) =
X
Let D : E � E be the derivative . Then one has th e fo rm ula (Df, g) =

(J Dg) . ,
Thus one says that 1D = D , even though the scalar produc t is not "nonsingular", but much of the formalism of nonsingular forms goes over. Also in analysis , one puts various norms on the spaces and one ex te nd s the bilinear form by continuity to t he completions , thus leaving the domain of algebra to enter the domain of estimates ( anal ys i s ) . Then the spectral theorems become mo re com plicated in such anal ytic contexts .
Let us assume that E = F. Let f : E x E + R be bilinear. By an auto morphism of the pair (E,/), o r simply off, we shall mean a linear automorphism A : E + E such that
(Ax, Ay) = (x, y) for all x, y E E. The group of automorphisms of f is denoted by Aut(/).
Let f: E x E + R be a nonsingular bilinear form. Let A : E + E be a linear map. Then A is an automorphism of f if and on ly if t AA = id, and A is invertible.
Proposition
5. 1 .
Proof From the equality (x, y) = (Ax, Ay) = (x, tAAy) holding for all x, y E E, we conclude that t A A = id if A is an automorphism of f. The converse is equally clear. Note. If E is free and finite dimensional, then the condition 'A A
implies that A is invertible.
= id
Let f : E x E + R be a bilinear form. We say that f is symmetric if f(x, y) = f(y, x) for all x, y E E. The set of symmetric bilinear forms on E will be denoted by L;(E). Let us take a fixed symmetric nonsingular bilinear form f on E, denoted by (x, y) � (x, y). An endomorphism A : E + E will be said to be symmetric with respect to f if 'A = A . It is clear that the set of sym metric endomorphisms o f E is a module, which we shall denote by Sym( E ).
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XI I I , §5
Depending on our fixed symmetric nonsingular f, we ha ve an isomorphism
which we describe as follows. If g is symmetric bilinear on E, then there exists a unique linear map A such that g(x, y) = (Ax, y) for all x, y E E. Using the fact that both f, g ar e symmetric, we obtain
(Ax, y)
=
(Ay, x)
=
( y,
'Ax)
=
('Ax, y).
Hence A = 'A. The association g � A gives us a homomorphism from L;(E) into Sym(E). Conversely, given a symmetric endomorphism A of E, we can define a symmetric form by the rule (x, y) � ( Ax, y), and the association of this form to A clearly gives a homomorphism of Sym(E) into L; (E) which is inverse to the preceding homomorphism. Hence Sym(E) and L; (E) are iso morphic. We recall that a bilinear form g : E x E � R is said to be alternating if g(x, x) = 0 for all x E E, and consequently g(x, y) =  g(y, x) for all x, y E E. The set of bilinear alternating forms on E is a module, denoted by L;(E). Let f be a fixed symmetric nonsingular bilinear form on E. An endo morphism A : E � E will be said to be skewsymmetric or alternating with respect to f if 'A =  A , and also (Ax, x) = 0 for all x E E. If for all a E R, 2a = 0 implies a = 0, then this second condition (Ax, x) = 0 is redundant, because (Ax, x) =  (Ax, x) implies (Ax, x) = 0. It is clear that the set of alternating endomorphisms of E is a module, denoted by Alt(E). Depending on our fixed symmetric nonsingular form f, we have an isomorphism L� (E) � Alt(E) described as usual. If g is an alternating bilinear form on E, its corresponding linear map A is the one such that
g(x, y)
=
(Ax, y)
for all x, y E E. One verifies trivially in a manner similar to the one used in the symmetric case that the correspondence g � A gives us our desired iso morphism. Examples . Let k be a field and let E be a finitedimensional vector space over k. Let f : E x E + E be a bilinear map , denoted b y (x , y) � xy . To each
X I I I , §6
MATR ICES AND BILINEAR FO RMS
527
x E E, we associate the linear map A x : E � E such that Then the map obtained by taking the trace, namely
is a bilinear form on E. If x y = yx, then this bilinear form is symmetric. Next, let E be the space of continuous functio ns on the interval [0, 1]. Let K(s, t) be a continuous function of two real variable s defined on the square 0 < s < 1 and 0 < t < 1 . For q>, t/1 E E we define ( ((), 1/1)
=
JJ qJ(s)K(s, t)l/l(t) ds dt,
the double integral being taken on the square. Then we obtain a bilinear form on E. If K(s, t) = K(t, s), then the bilinear form is symmetric. When we discuss matrices and bilinear forms in the next section, the reader will note the similarity between the preceding formula and the bilinear form defined by a matrix. Thirdly, let U be an open subset of a real Banach space E (or a finitedimen sional Euclidean space, if the reader insists), and let f : U � R be a map which is twice continuously differentiable. For each x E U, the derivative Df(x) : E � R is a continuous linear map, and the second derivative D 2f(x) can be viewed as a continuous symmetric bilinear map of E x E into R.
§6.
M AT R I C ES A N D B I LI N EA R FO R M S
We shall investigate the relation between the concepts introduced above and matrices. Let f : E x F � R be bilinear. Assume that E, F are free over R. Let CB = {v 1 , . . , vm } be a basis for E over R, and let CB' = {w b . . . , w,} be a basis for F over R. Let g ii = < vi , wi > · If .
and are elements of E and F respectively, with coordinates x i , Yi E R, then m ( x, y ) = L L U iiXi Yi · i= 1 j= 1 n
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X I I I , §6
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Let X, Y be the column vectors of coordinates for x, y respectively, with respect to our bases. Then
(x, y)
=
rxG Y
where G is the matrix (gii). We could write G
=
M � �(f). We call G the matrix
associated with the form f relative to the bases CB, 3 will be proved in the next section. The standard Borel subgroup B of GL 2 is the group of all matrices
wit h a, b, d E F' and ad # 0. For the Borel subgroup of SL 2 , we require in addition that ad = 1 . By a Borel subgrou p we mean a subgroup which is conjugate to the standard Borel subgroup (w hether in GL 2 or SL 2 ). We let U be the group of matrices
u(b)
=
(� �) .
wit h b E F.
We let A be the group of diagonal matrices
(� �).
with a, d E F*.
Let with a E F'* and w
=
( � �) .
For the rest of this section, we let
Lemma
8. 1 .
The matrices X (b)
=
(� n
and
Y(c)
=
(� �)
generate SL 2 (F). Proof. Multiplying an arbitrary element of SL 2 (F') by matrices of the above type on the right and on the left corresponds to elementary row and column operations, that is adding a scalar multiple of a row to the other, etc. Thus a given matrix can always be brought into a form
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X I I I , §8
by such multiplications. We want to express this matrix with a
=F
1 in the form
Matrix multiplication will show that we can solve this equation , by selecting x arbitrarily =F 0 , then solving for b , e, and d successively so that
1 + bx =
a'
x b d= e= . , 1 + bx 1 + be
Then one finds 1 + be = ( 1 + xb)  1 and the two symmetric conditions b + bed + d = 0 e + bex + x = 0 , so we get what we want, and thereby prove the lemma . Let U be the group of lower matrices
Then we see that
wuw 1 = U.
Also note the commutation relation
w (aO d0) w _ 1 = (od O)a ' so w normalizes A. Similarly, wBw1 = B is the group of lower triangular matrices. We note that
B = A U = VA
'
and also that A normalizes U. There is a decomposition of G into disjoint subsets G
= B BwB. u
Indeed, view G as operating on the left of column vectors. The isotropy group of
is obviously U. The orbit
Be 1 consists of all column vectors whose second
TH E SIM PLIC ITY O F SL2(F)/ + 1
X I I I , §8
539
component is 0. On the other hand,
and therefore the orbit Bwe 1 consists of all vectors whose second component is # 0, and whose first component is arbitrary. Since these two orbits of B and BwB cover the orbit Ge 1 , it follows that the union of B and BwB is equal to G (because the isotropy group U is contained in B), and they are obviously disjoint. This decomposition is called the Bruhat decomposition. Proposition
8.2.
The Borel subgroup B is a maximal proper subgroup.
Proof. By the Bruhat decomposition, any element not in B lies in BwB, s o the assertion follows since B, BwB cover G. Theorem 8.3 . IfF has at least four elements, then SL 2 (F) is equal to its own
commutator group. Proof. We have the commutator relation (by matrix multiplication) s(a)u(b)s(a)  1 u(b)  1 u(ba 2  b) u(b(a 2  1)). =
=
Let G = SL 2 (F) for this proof. We let G ' be the commutator subgroup, and similarly let B' be the commutator subgroup of B. We prove the first assertion that G = G ' . From the hypothesis that F has at least four elements, we can find an element a =F 0 in F such that a 2 =F l , whence the commutator relation shows that B' = U. It follows that G ' ::J U, and since G ' is normal, we get
G' ::J w uw  1 • From Lemma 8. 1, we conclude that G ' G. =
Let Z denote the center of G. It consists of + I, that is + the identity 2 x 2 matrix if G = SL 2 (F) ; and Z is the subgroup of scalar matrices if G = GL 2 (F). Theorem 8.4. If F has at least four elements, then SL 2 (F)/Z is simple.
The proof will result from two lemmas. Lemma 8.5.
The intersection of all conjugates of B in G is equal to Z. Proof. We leave this to the reader, as a simple fact using conjugation with w. Lemma 8.6. Let G
H ::J G ' .
=
SL 2 (F). If H is normal in G, then either H
Proof. By the maximality of B we must have HB
=
B or HB
=
G.
c
Z or
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X I I I , §9
If H B = B then H c B. Since H is normal, we conclude that H is contained in every conjugate of B, whence in the center by Lemma 8.5. On the other hand , suppose that HB = G. Write w =
hb
with h E H and b E B. Then because H is normal. Since U HU = G. Hence
G/H
c
=
H U and U, U generate SL 2 (F), it follows that HU/H
�
U/( U n H)
is abelian, whence H ::J G', as was to be shown. The simplicity of Theorem 8.4 is an immediate consequence of Lemma 8.6.
§9.
TH E G ROU P
SLn (F) ,
n �
3.
In this section we look at the case with n > 3, and follow parts of Art in's Geometric Algebra, Chapter IV. (Artin even treats the case of a noncommuta tive division algebra as the group ring, but we omit this for simplicity.) For i, j = 1 , . . . , n and i =F j and c E F, we let 1
E ii(c)
=
0
1
0
·· C l)
0
1
be the matrix which differs from the unit matrix by having c in the ijcomponent instead of 0. We call such E ii(c) an elementary matrix. Note that det E ii(c)
=
1.
If A is any n x n matrix, then multiplication E ii(c)A on the left adds c times the jth row to the ith row of A. Multiplication A E ii(c) on the right adds c times the ith column to the jth column. We shall mostly multiply on the left. For fixed i =F j the map
TH E G ROUP SLn (F),
XI I I , §9
is a homomorphism of F into the multiplicative group of n G L " (F).
x
n
>
3
541
n matrices
Proposition 9. 1 . The group SL " (F) is generated by the elementary matrices. If A E GL n(F), then A can be written in the form
A = SD
'
"''here S E SL n( F ) and D is a diagonal matrix of the form 1
D=
0 0 1
so D has 1 on the diagonal except on the lower right corner, where the com ponent is d = det(A). Proof. Let A E GL n(F'). Since A is nonsingular, the first component of some row is not zero, and by an elementary row operation, we can make a 1 1 # 0. Adding a suitable multiple of the first row to the second row, we make a 2 1 # 0, and then adding a suitable multiple of the second row to the first we make a 1 1 = 1 . Then we subtract multiples of the first row from the others to make a i l = 0 for i # 1 . We now repeat the procedure with the second row and column, to make a 2 2 = 1 and a i 2 = 0 if i > 2. But then we can also make a 1 2 = 0 by sub tracting a suitable multiple of the second row from the first, so we can get a; 2 = 0 for i # 2. We repeat this procedure until we are stopped at a " " = d =F 0, and ani = 0 for j # n. Subtracting a suitable multiple of the last row from the preceding ones yields a matrix D of the form indicated in the statement of the theorem, and concludes the proof. Theorem 9.2.
For n >
SL "( F ) is equal to its own commutator group. Proof. It suffices to prove that E ii(c) is a commutator. Using n > 3, let k # i, j. Then by direct computation, 3,
expresses Eii(c) as a commutator. This proves the theorem. We note that if a matrix M commutes with every element of SL "(F'), then it must be a scalar matrix. Indeed, just the commutation with the elementary matrices £l). ·( 1 ) = I + 1 l}· ·
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X I I I , §9
MATR ICES AND L I N EA R MAPS
shows that M commutes with all matrices 1 ii (having 1 in the ijcomponent, 0 otherwise), so M commutes with all matrices, and is a scalar matrix. Taking the determinant shows that the center consists of � n (F)l, where �n (F) is the group of nth roots of unity in F. We let Z be the center of SLn (F) , so we have just seen that Z is the group of scalar matrices such that the scalar is an nth root of unity . Then we define
Theorem 9.3. For n > 3, PSL n(F) is simple.
The rest of this section is devoted to the proof. We view GL " (F) as operating on the vector space E = F " . If A. is a nonzero functional on E, we let H;.
=
Ker A.,
and call H;. (or simply H) the hyperplane associated with A.. Then dim H = n 1 , and conversely, if H is a subspace of codimension 1 , then E/H has dimension 1 and is the kernel of a functional. An element T E GL"(F) is called a transvection if it keeps every element of some hyperplane H fixed, and for all x E E, we have 
..
Tx
=
x
+
for some h E H.
h
Given any element u E H;. we define a transvection Yu by
Tux
=
x
+
A.(x)u.
Every transvection is of this type. If u , v E H;. , it is immediate that If T is a transvection and A E GLn(F), then the conjugate A T A  1 is ob viously a transvection. The elementary matrices E ii(c) are transvections, and it will be useful to use them with this geometric interpretations, rather than formally as we did before . Indeed, let e 1 , , e n be the standard unit vectors which form a basis of F 3. Let G be SLninvariant , and suppose that G contains a transvection T # I. Then SL n(F') c G. Lemma
9.5.
Proof. By Lemma 9.4, all transvections are conjugate, and the set of transvections contains the elementary matrices which generate SLn (F) by Proposition 9. 1 , so the lemma follows. Let n > 3. JfG is a subgroup ofGLn(F) which is SLninvariant and which is not contained in the center of GL n (F), then SLn(F) c G. Theorem 9.6.
Proof. By the preceding lemma, it suffices to prove that G contains a transvection, and this is the key step in the proof of Theorem 9.3. We start with an element A e G which moves some line. This is possible since G is not contained in the center. So there exists a vector u # 0 such that Au is not a scalar multiple of u , say Au = v . Then u, v are contained in some hyperplane H = Ker A . Let T = Tu and let B = ATA  1 T 1• Then
A T A  1 # T and B = A T A  1 T  1 # I.
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X I I I , §9
MATR ICES AND L I N EAR MAPS
This is easily seen by applying say B to an arbitrary vector x, and using the definition of Yu . In each case, for some x the lefthand side cannot equal the righthand side. For any vector x E F we have
Bx  x E (u, v), where (u, v ) is the plane generated by u, v. It follows that BH
BH
=
c
H, so
H and Bx  x E H.
We now distinguish two cases to conclude the proof. First assume that B commutes with all transvections with respect to H. Let w E H. Then from the definitions, we find for any vector x :
A.(x)Bw
BTw x
=
Bx
Tw Bx
=
Bx + A.(Bx)w
+
=
Bx
+
A.(x)w.
Since we are in the case BTw = Tw B, it follows that Bw = w. Therefore B leaves every vector of H fixed. Since we have seen that Bx  x E H for all x, it follows that B is a transvection and is in G, thus provi ng the theorem in this case. Second, suppose there is a transvection Tw with w E H such that B does not commute with Tw . Let Then C # I and C E G. Furthermore C is a product of T� 1 and BTw B  1 whose hyperplanes are H and BH, whicn is also H by what we have already proved. Therefore C is a transvection, since it is a product of transvections with the same hyperplane. And C E G. This concludes the proof in the second case, and also concludes the proof of Theorem 9.6. We now return to the main theorem, that PSL " (F) is simple. Let G be a normal subgroup of PSL " (F), and let G be its inverse image in SL " (F). Then G is SL" invariant, and if G =F l , then G is not equal to the center of SL " (F). Therefore G contains SL n (F) by Theorem 9.6, and therefore G = PSL,lF), thus proving that PSLn(F) is simple. Example.
By Exercise 4 1 of Chapter I, or whatever other means , one sees that PSL2(F5) = A5 (where F 5 is the finite field with 5 elements) . While you are in the mood, show also that
XI I I , Ex
EXERCISES
545
EX E R C I S ES 1 . Interpret the rank of a matrix A in terms of the dimensions of the image and kernel of the linear map LA . 2 . (a) Let A be an invertible matrix in a commutative ring R . Show that ('A)  1 = 1(A 1 ) . (b) Let f be a nonsingular bilinear form on the module E over R . Let A be an Rautomorphism of E. Show that ('A) 1 = 1(A  1 ) . Prove the same thing in the hermitian case , i . e . (A * )  1 = (A  1 ) * . 3 . Let V, W be finite dimensional vector spaces over a field k. Suppose given nondegenerate bilinear forms on V and W respectively , denoted both by ( , ) . Let L : V � W be a surjective linear map and let 1L be its transpose; that is , (Lv, w ) = (v, 1Lw) for v E V and w E W. (a) Show that 1L is injective . (b) Assume in addition that if v E V, v =/= 0 then (v, v) =/= 0. Show that
V = Ker L EB Im 1L , and that the two summands are orthogonal . (Cf. Exercise 33 for an example . ) 4. Let A 1 , A , be row vectors of dimension n, over a field k. Let X = (x 1 , , xn). Let b1, b, E k. By a system of linear equations in k one means a system of type •
.
•
.
•
•
•
•
•
,
If b 1 = = b, = 0, one says the system is homogeneous. We call n t he number of variables, and r the number of equations. A solution X of the homogeneous system i s called trivial if X; = 0 , i = 1 , . . . , n . (a) Show that a homogeneous system of r li near equations 1n· n unknowns with n > r always has a nontrivial solution. (b) Let L be a system of homogeneous linear equations over a field k. Let k be a subfield of k'. If L has a nontrivial solution in k', show that it has a nontrivial solution in k. ·
·
·
5. Let M be an n x n matrix over a field k. Assume that tr( M X) = 0 for all n X I n k. Show that M = 0.
x
n matrices
6. Let S be a set of n x n matrices over a field k. Show that there exists a column vector X :1 0 of dimension n in k, such that M X = X for all M E S if and only if there exists such a vector in some extension field k' of k.
, 7 . Let H be the division ring over the reals generated by elements i, j, k such that i2 = j 2 = k2 =  I , and ij =  ji = k,
jk =  kj = i,
ki =  ik = j.
Then H has an automorphism of order 2 , given by a0 + a 1 i + a j + a 3 k t+ a0  a 1 i  a2j  a3 k. 2 Denote this automorphism by ex t+ a. What is cxa ? Show that the theory of hermitian
546
MATR ICES AND LINEAR MAPS
XI I I , Ex
forms can be earned out over H, whtch is called t he dtvtsion nng of quaternions (or by abuse of language, the noncommutative field of quaternions). 8 . Let N be a stnctly upper tnangular n Show that N n = 0.
x
n matnx, t hat ts N = ( a ;) and aii = 0 tf i > j.
9. Let E be a vector space over k, of dtmension n. Let T: E + E be a It near map such that T is nilpotent, that ts y m = 0 for some postttve integer m . Show t hat there extsts
a basis of E over k such that the matrix of T with respect to this basis is strictly upper triangular .
1 0.
If N ts a nilpotent n
x
n matnx, show that I + N is I nvertible.
1 1 . Let R be the set of all upper tnangular n x n mat nces (a ii) wtth aii in some field k, so a ii = 0 tf i > j. Let J be the set of all stnctly upper triangular matnces. Show that J is a twosided tdeal in R. How would you descnbe the factor ring R/J ?
1 2. Let G be the group of upper triangular matrices with nonzero d tagonal elements. Let H be the subgroup consisting of those matrices whose diagonal element ts 1 . ( Actually prove that H ts a subgroup). How would you descnbe the factor group G/H ?
1 3 . Let R be the ring of n x n matrices over a field k . Let L be the subset of matrices which are 0 except on the first column . (a) Show that L is a left ideal . (b) Show that L is a minimal left ideal ; that is , if L' C L is a left ideal and L ' =I= 0 , then L ' = L . (For more on this situation , see Chapter VII , §5 . )
1 4. Let F be any field . Let D be the subgrou p of diagonal matnces tn GLn(F). Let N be the normalizer of D tn G Ln(F). Show that N /D is isomorphic to the symmetnc group on n elements.
1 5. Let F be a fin tte field with q elements. Show that the order of G Ln(F) ts
(q n  1 )(q n  q ) . . . (q n  q"  1 ) = qn 0 for P =I= 0 . Show also that the monomials of degree d form an orthogonal basis . What is (M ' M)? (c) The map P � P(D) is an isomorphism of Pol(n , d) onto its dual .
X I I I , Ex
EXE RC ISES
551
(d) Let � = Dt + · · · + D� . Note that � : Pol(n , d) � Pol(n , d  2) is a linear map . Prove that � is surjective . (e) Define Har(n , d) = Ker� = vector space of harmonic homogeneous poly nomials of degree d. Prove that dim Har(n , d) = (n + d  3) ! (n + 2d  2)/(n  2) ! d ! . In particular, if n = 3 , then dim Har(3 , d) = 2 d + 1 . (f) Let r 2 = Xt + · · · + X�. Let S denote multiplicat ion by r 2 • Show that (� P , Q) = (P , SQ) for P E Pol(n , d) and Q E Pol(n, d  2) , so '� = S . More generally , for R E Pol(n , m) and Q E Pol(n , d  m) we have (R(D)P, Q) = (P , RQ) . (g) Show that [� , S] = 4d + 2n on Pol(n , d) . Here [� , S] = � o S  S o � Actually , [� , S] = 4E + 2 n , where E is the Euler operator E = 2: X;D;, which is , however, the degree operator on homogeneous polynomials . (h) Prove that Pol(n, d) = Har(n , d) EB r 2Pol( n , d  2) and that the two summands are orthogonal . This is a classical theorem used in the theory of the Laplace operator. " 2 = 0 . Let (i) Let (c 1 , , en ) E k " be such that L.J C; •
•
•
H�(X) = (c 1X1 +
· · ·
+ cnXn ) d .
Show that H� is harmonic , i . e . lies in Har(n, d ) . (j) For any Q E Pol(n , d) , and a positive integer m , show that
Q(D)H';' (X ) = m(m  1 )
· · ·
(m  d + l )Q(c)H';'  d(X ) .
34 . (Continuation of Exercise 33). Prove:
Let k be algebraically closed of characteristic 0. Let n > 3 . Then Har(n , d) as a vector space over k is generated by all polynomials H� with (c) E k" such that 2: cf = 0. Theorem .
[Hint : Let Q E Har( n , d) be orthogonal to all polynomials H� with (c) E k". By Exercise 33(h) , it suffices to prove that r 2 1 Q. But if 2: ct = 0 , then by Exercise 33 (j) we conclude that Q(c) = 0. By the Hilbert Nul lstellensatz , it follows that there exists a polynomial F(X) such that Q(X)s = r 2 (X)F(X) for some positive integer s . But n > 3 implies that r 2 (X ) i s irreducible , so r 2 (X) divides Q(X ) . ] 3 5 . (Continuation of Exercise 34). Prove that the representation of O(n) = Un (R) on Har( n, d) is irreducible. Readers will find a proof in the following: S. HELGASON , Topics in Harmonic Analysis on Homogeneous Spaces , B irkhauser, 1 98 1 (see especially § 3 , Theorem 3 . 1 (ii)) N. VILENKIN , Special Functions and the Theory of Group Representations , AMS Trans lations of mathematical monographs Vol . 22 , 1 968 (Russian original , 1 965) , Chapter IX , § 2 .
552
X I I I , Ex
MATR ICES AND LINEAR MAPS
R . HowE and E. C . TAN , NonAbelian Harmonic Analysis, Universitext , Springer Verlag, New York , 1 992 . The HoweTan proof runs as follows . We now use the hermitian product
(P, Q)
J
=
P(x) Q (x) da(x) ,
S"1
where a is the rotation invariant measure on the (n  1 )sphere s n  t . Let e1 , , en be the unit vectors in R n . We can identify O(n  I ) as the subgroup of O(n) leaving en fixed. Observe that O(n) operates on Har(n , d) , say on the right by composition P � P o A , A E O(n) , and this operation commutes with �. Let •
•
.
,1: Har( n, d)
t C
be the functional such that ,1(P) == P(en ) · Then A is O(n  I )Invariant, and since the hermitian product is nondegenerate, there exists a harmonic polynomial Qn such that
A(P)
==
(P, Qn ) for all P E Har(n , d) .
Let M c Har(n , d) be an O(n)submodule. Then the restriction AM of A to M is nontrivial because O(n) acts transitively on s n  t . Let Q,';f be the orthogonal pro jection of Qn on M. Then Q;t is O(n  1 )invariant, and so is a linear combination
Q�(x)
=
2:
j + 2k = d
ci x� �� � .
Furthermore Q/f i s harmonic . From this you can show that Q� i s uniquely determined, by showing the existence of recursive relations among the coefficients ci . Thus the submodule M is uniquely determined, and must be all of Har(n , d).
Irreducibility of sln(F). 36. Let F be a field of characteristic 0. Let g == sin (F) be the vector space of matrices with trace 0, with its Lie algebra structure [X, Y] == X Y  YX. Let Eu be the matrix having ( i, j)component 1 and all other components 0. Let G == SLn (F) . Let A be the multiplicative group of diagonal matrices over F. (a) Let Hi == Eu  Ei+ l , i+ l for i == I , . . . , n  I . Show that the elements Eu ( i i= j) , Ht , . . . , Hn l form a basis of g over F. (b) For g E G let c(g) be the conjugation action on g, that is c(g) X == gXg Show that each Eu is an eigenvector for this action restricted to the group A . (c) Show that the conjugation representation of G on g is irreducible, that is, if V i= 0 is a subspace of g which is c ( G)stable, then V == g. Hin : Look up the sketch of the proof in [JoL 0 1 ], Chapter VII, Theorem 1 . 5, and put in all the details. Note that for i i= j the matrix Eu is nilpotent, so for variable the exponential series exp( tEu ) is actually a polynomial. The derivative with respect to can be taken in the formal power series F[[t] ] , not using limits. If == ex p( X , show that X is a matrix, and 
t
t
x( t) t ) �x(t) Yx( t) 1 = XY  YX = [X , Y] . t=O
1
•
t,
C H APT E R
XIV
Represe ntatio n of O n e E n d o m orp his m
We deal here with one endomorphism of a module , actually a free module , and especially a finite dimensional vector space over a field k. We obtain the Jordan canonical form for a representing matrix , which has a particularly simple shape when k is algebraically closed . This leads to a discussion of eigenvalues and the characteristic polynomial . The main theorem can be viewed as giving an example for the general structure theorem of modules over a principal ring . In the present case , the principal ring is the polynomial ring k[X] in one variable .
§1 .
R E P R ES E NTATI O N S
Let k be a commutative ring and E a module over k. As usual, we denote by End k (E) the ring of kendomorphisms of E, i.e. the ring of klinear maps of E into itself. Let R be a kalgebra (given by a ringhomomorphism k � R which allows us to consider R as a kmodule). By a representation of R in E one means a k algebra homomorphism R � End k (E), that is a r inghomomorphism
which makes the following diagram comm u tative :
� k/ 553
554
R EPRESENTATIO N OF ONE E N DOMORPHISM
XIV, §1
[As usual, we view End k (E) as a kalgebra ; if I denotes the identity map of E, we have the homomorphism of k into Endk(E) given by a � a l. We shall also use I to denote the unit matrix if bases have been chosen. The context will always make our meaning clear.] We shall meet several examples of representations in the sequel, with various types of rings (both commutative and noncommutative). In this chapter, the rings will be commutative. We observe that E may be viewed as an Endk(E) module. Hence E may be viewed as an Rmodule, defining the operation of R on E by letting (x, v) � p(x)v
for x E R and v E E. We usually write xv instead of p(x)v . A subgroup F of E such that RF c F will be said to be an invariant sub module of E . (It is both Rinvariant and kinvariant . ) We also say that it is invariant under the representation. We say that the representation is irreducible, or simple, if E # 0, and if the only invariant submodules are 0 and E itself. The purpose of representation theories is to determine the structure of all representations of various interesting rings, and to classify their irreducible representations. In most cases, we take k to be a field, which may or may not be algebraically closed. The difficulties in proving theorems about representa tions may therefore lie in the complication of the ring R, or the complication of the field k, or the complication of the module E, or all three. A representation p as above is said to be completely reducible or semisimple if E is an Rdirect sum of Rsubmodules E; ,
such that each E; is irreducible. We also say that E is completely reducible. It is not true that all representations are completely reducible, and in fact those considered in this chapter will not be in general. Certain types of completely reducible representations will be studied later. There is a special type of representation which will occur very frequently. Let v E E and assume that E = R v. We shall also write E = (v). We then say that E is principal (over R), and that the representation is principal. If that is the case, the set of elements x E R such that xv = 0 is a left ideal a of R (obvious). The map of R onto E given by x � xv
induces an isomorphism of Rmodules, R/a � E (viewing R as a left module over itself, and Rja as the factor module). In this map, the unit element 1 of R corresponds to the generator v of E.
REPRESENTATIONS
XIV, § 1
555
As a matter of notation, if v 1, , vn E E, we let ( v 1 , , v") denote the sub module of E generated by vb . . . , v" . Assume that E has a decomposition into a direct sum of Rsubmodules •
•
•
•
•
•
As sume that each E; is free and of dimension > 1 over k. Let CB 1 , , CBs be , E5 respectively over k. Then { CB 1 , base s for E 1 , , , and let A be an n x n matrix, which we view as a linear map of E into itself. The invariants (q 1 , • • • , q ,) will be called the invariants of A (over k). Corollary 2.2. Le t k' be an extension field of k a nd let A be an n
n matrix in k. The invariants of A over k are the same as its in var ian ts over k'. x
558
R E PRES ENTATION OF ONE ENDOMORPHISM
XIV, §2
Proof. Let { v 1 , , vn } be a basis of k over k. Then we may view it also as a basis of k' over k ' . (The unit vectors are in the kspace generated by v 1 , • . . , vn ; hence v 1 , . . . , vn generate the ndimensional space k' < n> over k'.) Let E = k . Corollary 2.3. L et A, B be n
n matrices over a field k and let k' be an extension field of k. Assume that there is an invertible matrix C' in k ' such that B = C ' AC '  1 • Then there is an invertible matrix C in k such that B = CAC  1 • x
Proof. Exercise. The structure theorem for modules over principal rings gives us two kinds of decompositions. One is according to the invariants of the preceding theorem. The other is according to prime powers. Let E =F 0 be a finite dimensional space over the field k, and let A : E � E be in End k(E). Let q = q A be its minimal polynomial. Then q has a factorization,
into prime powers (distinct). Hence E is a direct sum of submodules
such that each E(p;) is annihilated by pfi. Furthermore, each such submodule can be expressed as a direct sum of submodules isomorphic to k[t]j(pe) for some irreducible polynomial p and some integer e > 1 .
(t  cx)e for some cx E k, e > 1 . Assume that E is isomorphic to k[t]j(q). Then E has a basis over k such that the matrix of A relative to this basis is of type Theorem
2.4. Let qA ( t )
=
(X
1 0 0
0
0 0
(1.
0 1
(1.
X IV, §2
DECOM POS ITION OVE R ONE ENDOMORPH ISM
559
Proof Since E is isomorphic to k [t] /( q ), there exists an element v E E such that k[t] v = E. This element corresponds to the unit element of k[t] in the isomorphism k [t]/( q ) � E. We contend that the elements
v, (t  cx)v, . . . , (t  cx)e  1 v, or equivalently,
v, (A  cx)v, . . . , (A  cx)e  1 v
,
torm a basis for E over k. They are linearly independent over k because any relation of linear dependence would yield a relation of linear dependence between v, A v, . . . ' Ae  l v, and hence would yield a polynomial g(t) of degree less than deg q such that g(A) = 0. Since dim E = e, it follows that our elements form a basis for E over k. But (A  cx)e = 0. It is then clear from the definitions that the matrix of A with respect to this basis has the shape stated in our theorem. Corollary 2.5. Let k be algebraically closed, and let E be a finitedimensional
nonzero vector space over k. Let A E End k (E) . Then there exists a basis of E over k such that the matrix of A with respect to this basis consists of blocks, and each block is of the type described in the theorem. A matrix having the form described in the preceding corollary is said to be in
Jordan canonical form.
A matrix (or an endomorphism) N is said to be nilpotent if there exists an integer d > 0 such that Nd = 0. We see that in the decomposition of Theorem 2.4 or Corollary 2 . 5 , the matrix M is written in the form Remark
1.
M=B+N where N is nilpotent. In fact, N is a triangular matrix (i.e. it has zero coefficients on and above the diagonal), and B is a diagonal matrix, whose diagonal elements are the roots of the minimal polynomial. Such a decomposition can always be achieved whenever the field k is such that all the roots of the minimal polynomial lie in k. We observe also that the only case when the matrix N is 0 is when all the roots of the minimal polynomial have multiplicity 1 . In this case, if n = dim E, then the matrix M is a diagonal matrix, with n distinct elements on the diagonal.
560
REPRESENTATION OF ON E ENDOMORPH ISM
XIV, §2
Remark 2. The main theorem of this section can also be viewed as falling under the general pattern of decomposing a module into a direct sum as far as possible , and also giving normalized bases for vector spaces with respect to various structures , so that one can tell in a simple way the effect of an endo morphism . More formally , consider the category of pairs (£, A ) , consisting of a finite dimensional vector space E over a field k, and an endomorphism A : E � E. By a morphism of such pairs
j
f: (E , A) � (E', A' ) we mean a khomomorphism f: E � E' such that the following diagram is commutative:
E _f_+ E' A
E
.
f
E'
It is then immediate that such pairs form a category , so we have the notion of isomorphism . One can reformulate Theorem 2 . 1 by stating :
Two pairs (E, A) and (F, B) are isomorphic if and only if they have the same invariants. Theorem 2 . 6 .
You can prove this as Exercise 1 9 . The Jordan basis gives a normalized form for the matrix associated with such a pair and an appropriate basis . In the next chapter, we shall find conditions under which a normalized matrix is actually diagonal , for hermitian , symmetric , and unitary operators over the complex numbers . As an example and application of Theorem 2 . 6 , we prove : Corollary 2. 7 . Let k be a field and let K be a finite separable extension of degree n. Let V be a finite dimensional vector space of dimension n over k, and let p, p' : K � Endk( V) be two representations of K on V; that is, embeddings of K in Endk( V) . Then p , p ' are conjugate; that is, there exists B E Autk( V)
such that
p '( � ) = Bp ( � )B  1 for all � E K.
Proof. By the primitive element theorem of field theory , there exists an element a E K such that K = k[ a] . Let p ( t ) be the irreducible polynomial of a over k. Then (V, p ( a)) and ( V, p'( a) ) have the same invariant , namely p ( t) . Hence these pairs are isomorphic by Theorem 2 . 6 , which means that there exists
B E Autk (V) such that
p'( a) = Bp ( a)B  1 •
But all elements of K are linear combinations of powers of a with coefficients in k, so it follows immediately that p'( � ) = B p ( � )B  1 for all � E K, as desired .
XIV , §3
TH E CHARACTER ISTIC POLYNOM IAL
561
To get a representation of K as in corollary 2 . 7 , one may of course select a b asis of K, and represent multiplication of elements of K on K by matrices with respect to this basis . In some sense , Corollary 2 . 7 tells us that this is the only way to get such representations . We shall return to this point of view when considering Cartan subgroups of GL n in Chapter XVIII , § 1 2 .
§3.
T H E C H A RACTE R I STI C PO LYN O M IAL
Let k be a commutative ring and E a free module of dimension n over k. We consider the polynomial ring k[t], and a linear map A : E � E. We have a homomorphism k[t] � k[A] as before, mapping a polynomial f(t) on f(A), and E becomes a module over the ring R = k[t] . Let M be any n x n matri x in k (fo r instance the matrix of A relative to a basis of E). We define the characteristic polynomial PM(t) to be the determinant det(t l M) 
n
where I " is the unit n x n matrix . It is an element of k[t]. Furthermore, if N is an invertible matrix in R, then M)N) = det(t /"  M). det(t/ "  N  1 MN) = det( N  1 {t/ Hence the characteristic polynomial of N  1 M N is the same as that of M. We may therefore define the characteristic polynomial of A, and denote by PA , the characteristic polynomial of any matrix M associated with A with respect to some basis. (If E = 0, we define the characteristic polynomial to be 1 .) If cp : k � k' is a homomorphism of commutative rings, and M is an n x n matrix in k, then it is clear that "

where cpPM is obtained from PM by applying cp to the coefficients of PM . Theorem 3. 1 . (CayleyHamilton).
Proof Let { v 1 , • . •
We have PA(A) = 0. , v" } be a basis of E over k . Then .
t vJ
n
=
'"' i...J
i=
1
a l.J. v.l
where (a ii) = M is the matrix of A with respect to the basis. Let B(t) be the matrix with coefficients in k[t], defined in Chapter XIII, such that
562
XIV, §3
R E PRESENTATION OF ONE ENDOMORPHISM
Then
B(t)B(t) .
0
because

Hence PA (t)E = 0, and therefore PA(A)E as was to be shown.
0. This means that PA(A)
=
=
0,
Assume now that k is a field. Let E be a finitedimensional vector space over k, and let A E End k (E). By an eigenvector w of A in E one means an element w E E, such that there exists an element A. E k for which A w = A.w. If w # 0, then A. is determined uniquely, and is called an eigenvalue of A. Of course, distinct eigenvectors may have the same eigenvalue. Theorem 3.2. The eigenvalues of A are precisely the roots of the character
istic polynomial of A.
Proof Let A. be an eigenvalue. Then A  A.I is not invertible in Endk (E), and hence det(A  A.I) = 0. Hence A. is a root of PA . The arguments are re versible, so we also get the converse. For simplicity of notation, we often write A Theorem 3.3. Let w 1 ,

A. instead of A  A.I.
, wm be nonzero eigenvectors of A, having distinct eigenvalues. Then they are linearly independent. •
•
•
Proof Suppose that we have with a i E k, and let this be a shortest relation with not all a i = 0 (assuming such exists). Then a i # 0 for all i. Let A. 1 , , A.m be the eigenvalues of our vectors. Apply A  A. 1 to the above relation. We get •
a 2 (A. 2  A. 1 )w 2
•
.
+ am (A.m  A. 1 )wm which shortens our relation, contradiction. + · ··
=
0,
Corollary 3.4. If A has n distinct eigenvalues A.b . . . , An belonging to eigen
vectors v 1 ,
•
•
•
, vn , and dim E
=
n, then { v 1 ,
•
•
•
, vn } is a basisfor E. 1he matrix
XIV, §3
TH E CHARACTER ISTIC POLYNOM IAL
563
of A with respect to this basis is the diagonal matrix : 0
It is not always true that there exists a basis of E co nsisting of
Warning.
eigenvectors ! Remark.
Let k be a subfield of k'. If M is a matrix in k, we can define its characteristic polynomial with respect to k, and also with respect to k'. It is clear that the characteristic polynomials thus obtained are equal. If E is a vector space over k, we shall see later how to extend it to a vector space over k'. A linear map A extends to a linear map of the extended space, and the character istic polynomial of the linear map does not change either. Actually, if we select ' a basis for E over k, then E � k
E
E'
�
E
f
566
XIV, §3
REPRESENTATION OF ONE ENDOMORPHISM
Then we can define the kernel of such a morphism to be again a pair. Indeed, let E� be the kernel off : E ' � E. Then A ' maps E0 into itself because
fA ' E�
=
AfE�
=
0.
We let A� be the restriction of A ' on E0 . The pair (E� , A� ) is defined to be the kernel of our morphism. We shall denote by f again the morphism of the pair (E ' , A') � (E, A). We can speak of an exact sequence
(E', A') � (E, A) � (E", A"), meaning that the induced sequence
E' � E � E" is exact. We also write 0 instead of (0, 0), according to our universal convention to use the symbol 0 for all things which behave like a zero element. We observe that our pairs now behave formally like modules, and they in fact form an abelian category. Assume that k is a field. Let ct consist of all pairs (E, A) where E is finite dimensional over k.
Then Theorem 3. 7 asserts that the characteristic polynomial is an Euler Poincare map defined for each object in our category ct , with values into the multiplicative monoid of polynomials with leading coefficient l . Since the values of the map are in a monoid, this generalizes slightly the notion of Chapter III , §8 , when we took the values in a group . Of course when k is a field, which is the most frequent application , we can view the values of our map to be in the multiplicative group of nonzero rational functions , so our previous situation applies . A similar remark holds now for the trace and the determinant. If k is a field, the trace is an Euler map into the additive group of the field, and the deter minant is an Euler map into the multiplicative group of the field. We note also that all these maps (like all Euler maps) are defined on the isomorphism classes of pairs, and are defined on the EulerGrothendieck group. Theorem 3. 10. Let k be a commutative ring, M an n
n matrix in k, and f a polynomial in k[t]. Assume that PM(t) has a factorization, PM(t)
x
n
=
n (t  cxi)
i= 1
into linear factors over k. Then the characteristic polynomial of f(M) is given by p
f ( M ) (t)
n
=
n (t  f(cx i )),
i= 1
XIV, Ex
EXERC ISES
and tr(f(M))
=
n L f(rx i), i= 1
det(f(M))
567
n
=
0 f(rx i).
i= 1
Proof. Assume first that k is a field. Then using the canonical decomposi tion in terms of matrices given in Theorem 2 . 4 , we find that our assertion is immediately obvious. When k is a ring, we use a substitution argument. It is however necessary to know that if X = (x; j) is a matrix with algebraically indepe ndent coefficients over Z, then Px(t) has n distinct roots y 1 , , Yn [in an algebraic closure of Q(X)] and that we have a homomorphism •
•
•
mapping X on M and y 1 , , Y n on rx 1 , , rxn . This is obvious to the reader who read the chapter on integral ring extensions, and the reader who has not can forget about this part of the theorem. •
•
•
•
•
•
EX E R C I S ES 1 . Let T be an upper triangular square matrix over a commutative ring (i.e. all the ele ments below and on the diagonal are 0). Show that T is nilpotent. 2. Carry out explicitly the proof that the determinant of a matrix *
*
. . 0
0
· · ·
*
0
Ms
where each Mi is a square matrix, is equal to the product of the determinants of the matrices M 1 , , Ms . •
•
•
3. Let k be a commutative ring, and let M, M' be square n x n matrices in k. Show that the characteristic polynomials of MM' and M' M are equal.
4. Show that the eigenvalues of the matrix
0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 in the complex numbers are + 1 , + i.
568
XIV, Ex
REP RESENTATION OF ONE ENDOMOR PHISM
5. Let M, M' be square matrices over a field k. Let q, q' be their respective minimal polynomials. Show that the minimal polynomial of
is the least common multiple of q, q' .
6. Let A be a nilpotent endomorphism of a finite dimensional vector space E over the field k. Show that tr(A ) = 0. 7 . Let R be a principal entire ring . Let E be a free module over R , and let E v = HomR(E, R) be its dual module . Then E v is free of dimension n . Let F be a sub module of E. Show that Ev /F .1 can be v iewed as a submodule of F v , and that its invariants are the same as the invariants of F in E. 8. Let E be a finitedimensional vector space over a field k. Let A E Autk(£). Show that the following conditions are equivalent : (a) A = I + N, with N nilpotent. (b) There exists a basis of E such that the matrix of A with respect to t his basis has all its diagonal elements equal to 1 and all elements above the diagonal equal to 0. (c) All roots of t he characteristic polynomial of A (in t he algebraic closure of k) are equal to 1 . 9. Let k be a field of characteristic 0, and let M be an n
nilpotent if and only if tr(Mv)
=
0 for 1
m
=
1 . Let r' be a subgroup of dimension n , w11} be a basis of r'. Write also. Let { v . , . . . , vn } be a basis of r, and let { w 1 , •
wi =
•
.
L a ii v i .
Show that the index ( r : r') is equal to the absolute value of the determinant of the matrix (aii). 1 7. Prove the normal basis theorem for finite extensions of a finite field. 1 8. Let A = (aii) be a square n x n matrix over a commutative ring k. Let A ii be the matrix obtained by deleting the ith row and jth column from A . Let b ii = (  1 ) i + i det(Aii), and let B be the matrix (b i)· Show that det(B) = det(A)"  1 , by reducing the problem to the case when A is a matrix with variable coefficients over the integers. Use this same method to give an alternative proof of the CayleyHamilton theorem, that PA(A ) = 0.
570
XIV, Ex
R E PRESENTATIO N OF ONE ENDOMOR PH ISM
1 9. Let (£, A) and (£', A ') be pairs consisting of a finitedimensional vector space over a field k, and a kendomorphism. Show that these pairs are isomorphic if and only if their invariants are equal. 20 . (a) How many nonconjugate elements of GL 2 (C ) are there with characteristic poly nomial t 3 ( t + 1 ) 2 ( t  1 )? (b) How many with characteristic polynomial t 3  1 00 1 t? 2 1 . Let V be a finite dimensional vector space over Q and let A : V � V be a Qlinear map such that A5 = Id. Assume that if v E V is such that Av = v, then v = 0 . Prove that dim V is divisible by 4 . 22 . Let V be a finite dimensional vector space over R , and let A : V � V be an R linear map such that A 2 =  Id . Show that dim V is even , and that V is a direct sum of 2dimensional Ainvariant subspaces . 23 . Let E be a finitedimensional vector space over an algebraically closed field k. Let A, B be kendomorphisms of E which commute, i.e. AB = BA. Show that A and B have a common eigenvector. [Hint : Consider a subspace consisting of all vectors having a fixed element of k as eigenvalue.] 24. Let V be a finite dimensional vector space over a field k. Let A be an endomorphism of V. Let Tr(A m ) be the trace of A m as an endomorphism of V. Show that the following power series in the variable t are equal: tm IX) m exp 2: Tr(A ) m) (= m
I
=
det(J  tA)
or
d   log det(J  tA) dt
=
IX)= Tr(Am) tm .
2:
m
I
Compare with Exercise 23 of Chapter XVIII. 25 . Let V, W be finite dimensional vector spaces over k , of dimension n . Let (v, w) � (v, w) be a nonsingular bilinear form on V x W. Let c E k, and let A : V � V and V : W � W be endomorphisms such that (Av , Bw) = c(v, w) for all v E V and w E W.
Show that
det(A)det( tl  B) = (  t ) n det(c/  tA)
and
det(A)det(B) = e n . For an application of Exercises 24 and 25 to a context of topology or algebraic geometry , see Hartshorne ' s Algebraic Geometry , Appendix C , §4. 26. Let G == SLn (C) and let K be the complex unitary group. Let A be the group of di agonal matrices with positive real components on the diagonal. (a) Show that if g E NorG (A ) (normalizer of A in G), then c(g ) (conjugation by g) permutes the diagonal components of A , thus giving rise to a homo morphism NorG ( A ) + W to the group W of permutations of the diagonal coordina ies. By definition, the kernel of the above homomorphism is the centralizer Cen G (A ) . (b) Show that actually all permutations of the coordinates can be achieved by elements of K, so we get an isomorphism W
�