An Introduction to Homological Algebra 9780125992503, 0125992505

Table of contents :
Content:
Edited by
Page iii

Copyright Page
Page iv

Dedication
Page v

Preface
Pages ix-xi

1 Introduction
Pages 1-22

2 Horn and ⊗
Pages 23-56

3 Projectives, Injectives, and Flats
Pages 57-107

4 Specific Rings
Pages 108-149

5 Extensions of Groups
Pages 150-165

6 Homology
Pages 166-193

7 Ext
Pages 194-219

8 Tor
Pages 220-231

9 Son of Specific Rings
Pages 232-264

10 The Return of Cohomology of Groups
Pages 265-298

11 Spectral Sequences
Pages 299-366

References
Pages 367-369

Index
Pages 371-376

Pure and Applied Mathematics: A Series of Monographs and Textbooks
Page 377
Samuel Eilenberg, Hyman Bass

Citation preview

AN INTRODUCTION TO HOMOLOGICAL ALGEBRA JOSEPH J. ROTMAN Department of Mathematics University of Illinois Urbana, Illinois

@

1979

ACADEMIC PRESS

New York San Francisco London

A Subsidiary of Harcourt Brace Jovanovich, Publishers

COPYRIGKT @ 1979, BY ACADEMIC PRESS, b C . ALL RIGHTS RESERVED. NO PART OF THIS PUBLICATION MAY BE REPRODUCED OR TRANSMITTED IN ANY FORM OR BY ANY MEANS, ELECTRONIC OR MECHANICAL. INCLUDMG PHOTOCOPY, RECORDING, OR ANY INFORMATION STORAGE AND RETRIEVAL SYSTEM, WITHOUT PERMISSION IN WRITING FROM THE PUBLISmR.

ACADEMIC PRESS, INC.

111 Fifth Avcnue. New York, New York 10003

United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD.

24/28 Oval Road, London NW1 7DX

Library of Congress Cataloging in Publication Data Rotman, Joseph J Date An introduction to homological algebra. Bibliography: p. 1 . Algebra, Homological. I. Title. QA169.R66 512'.55 78-20001 ISBN 0-12-599250-5

AMS (MOS) 1970 Subject Classifications: 18-01, 13-01,16-01,20-01

PRINTED M THE UNITED STATES OF AMERICA

79808182

9 8 1 6 5 4 3 2 1

To my mother, Rose, and to Phil

57sS1 , 7 7 1 ,7niin--’nuS

Preface

In 1968, I taught a course in homological algebra at Urbana; those lectures were published in 1970 as my “Notes on Homological Algebra”. Since much of the basic material there has now drifted into earlier courses, I have rewritten the notes, incorporating new items to take advantage of the present times. As it is convenient that a book be essentially self-contained, the old material remains, though redone. Therefore, students at different levels should find something of interest here. Learning homological algebra is a two-stage affair. First, one must learn the language of Ext and Tor and what it describes. Second, one must be able to compute these things, and. often, this involves yet another language: spectral sequences. The following exercise appears on page 105 of Lang’s book, “Algebra’‘: “Take any book on homological algebra and prove all the theorems without looking at the proofs given in that book.” My guess is that Lang was trying to foster the healthy attitude that, in spite of its cumbersome, perhaps forbidding, appearance, homological algebra is actually accessible to those who wish to learn it. If nothing else, this book is my attempt to make the subject lovable. The origins of our subject are the origins of algebraic topology. At first glance, our opening sentence, “We begin with a brief discussion of line integrals in the plane”, must appear strange; upon reflection, how could we begin any other way? A simple definition may mystify a reader if it seems to have been offered up by some oracle. There are no oracles here. Indeed, one role of the introductory first chapter is to give the etymology, both linguistic and mathematical, of the word homology.

ix

X

Preface

With Chapter 1, the next three chapters form a short course in module theory. The main subject of homological algebra is the pair of bifunctors Horn and tensor. Thus, Chapter 2 studies their behavior on all modules, whereas Chapter 3 studies how they act on certain special modules: projectives, injectives, and flats. These special modules are used at once to characterize Horn and tensor (Watts' theorems, which gave rise to adjoint functor theorems i n category theory). We also introduce localization for later applications to polynomial rings and to regular local rings. To make all more concrete, to see why certain earlier hypotheses are needed, and to make contact with "honest" mathematics, Chapter 4 examines the interplay between the ring of scalars and the special modules. In particular, we prove the Quillen-Suslin theorem that projective R [ x , , . . . , xnlmodules are free when R is a field. Chapter 5 is important etymology: group extensions are considered so that one can see the familiar formulas being born, crying for a projective resolution. At last, Chapter 6 defines homology and derived functors. The next two chapters treat Ext", the derived functors of Horn, and Tornrthe derived functors of tensor. We show that Ext was named for its relation to extensions, and that Tor was named for its relation to torsion. Chapters 9 and 10 apply Ext and Tor to rings and to groups. An introduction must show why its subject is valuable, and we have chosen these two areas as the most familiar ones in algebra. Dimension theory of rings organizes much of the material in Chapter 4 and also yields new results (we describe Serre's characterization of regular local rings, and the theorem of Auslander-Buchsbaum-Nagata that these rings have unique factorization). Applications to groups include a discussion of the Schur multiplier, Maschke's theorem, the Schur-Zassenhaus theorem, and a description of the theorem of Stallings-Swan characterizing free groups as those groups of cohomological dimension one. We have tried to keep the needs of algebraic topologists in mind, partly because one ought, and partly because the algebra is better understood when one sees its topological sources. A short discussion of Eilenberg-MacLane spaces does point out the intimate connection between topology and homological algebra; indeed, we even prove the universal coefficient theorem for group cohomology by looking at its counterpart for singular cohomology . The final chapter, spectral sequences, is almost a text by itself. This is not unexpected, for we have already mentioned that it is a second language. (Here I have not given etymology, although Lyndon [1948] is recommended for an algebraic viewpoint.) I think this exposition differs from others in its emphasis on illustrating how spectral sequences are used. Also, exact couples seem to eliminate most of the horrors of many indices. If I have succeeded, the reader will know when a spectral sequence is likely to arise, and he will know what to do with it if it does.

Preface

xi

And now the pleasant paragraph in which I can thank people from whom I have learned. I. Kaplansky, S. MacLane, J . Gray, and S. Gersten gave lectures that taught me much. L. McCulloh has made many valuable suggestions. E. Davis allowed me to use his account of Quillen’s solution of Serre‘s problem. R. Treger told me of Vaserstein’s proof of Suslin’s solution of Serre’s problem, and was kind enough to translate it for me from the Russian. An incomplete list of others who helped: P. M . Cohn, A. Fauntleroy, E. Green, P. Griffith, the late A. Learner, A. Mann, D. Robinson, M. Schacher, and A. Zaks. Of course, I am indebted to the fine books on homological algebra, especially those of Cartan-Eilenberg 119561 and MacLane 119631. I am grateful to the Lady Davis Foundation; most of this book was written while I was a Lady Davis Fellow and Visiting Professor at the Technion, Israel Institute of Technology, Haifa, and at the Hebrew University of Jerusalem. My warm thanks to both of these institutions for their hospitality. Finally, I thank Mrs. Esther Tuval for a superb job of typing my manuscript.

1

Introduction

LINE INTEGRALS AND INDEPENDENCE OF PATH

We begin with a brief discussion of line integrals in the plane. Assume X is an open disc in the plane with a finite number of points z,, . . . ,z, deleted. Fix points a and b in X . Given a curve fl in X from a to b, and given a pair of real-valued functions P and Q on X , one wants to evaluate JB P dx Q dy. It is wise to regard p as a finite union of curves, for p may only be piecewise smooth, e.g., it may be a polygonal path. For the rest of this discussion, we ignore (needed) differentiability hypotheses. An important question is when the line integral above is independent of the path p; if p’ is a second path in Xfromatob,isJBrPdx+Qdy=JBPd,x+Qdy?

+

P‘

Plainly, if y is the closed curve p - p’ (i.e., go from a to b via p and then go backward along p’ to u), then the integral is independent of path if and only if J, P d x + Q d y = 0. Thus, we are led to consideration of closed curves in X . Now this last line integral is affected by whether any of the “bad” points zl,. . . , z, lie inside y (one of the functions P or Q may have a singularity

1

2

1 Introduction

at some zi).Consider

Each yi is a (simple) closed curve in X with the point ziinside (and the other z’s outside); y is a (simple) closed curve in X having all the yi inside. If y is oriented counterclockwise and each of the yi is oriented clockwise, then Green’s theorem states that

where R is the shaded 2-dimensional region in the picture. Of course, one is tempted to, and does, write the sum of line integrals more concisely as s,+Y1+...+V”

Pdx

+Qdy;

indeed, instead of mentioning the orientations explicitly, one often writes a formal linear combination of curves (negative coefficients being a convenient way to indicate how the orientations are aligned). Observe that for simple curves the notion of “inside” always makes sense. However, we may allow curves to wind about some zi several times, assuming these curves are not so pathological as to fail to have an inside, and this introduces integer coefficients other than f 1 in the formal linear combinations. Let us now restrict attention to those pairs of functions P and Q satisfying aQ/ax = dP/ay. For these function pairs, the double integral in Green’s theorem vanishes, and we have J y + v l +. . .+?. P dx Q dy = 0. An equivalence relation on formal linear combinations (or, “oriented unions”) of closed curves in X suggests itself: call two such combinations a and a‘ equivalent if the values of

+

I

Pdx

+ Qdy

and

sa,

Pdx

+ Qdy

agree for all pairs P and Q as restricted above. In particular, note that combinations c1 whose curves comprise the boundary of a 2-dimensional region in X are trivial (the integrals have value 0). The equivalence classes of formal linear combinations of closed curves in X are called “homology classes”,

Categories and Functors

3

from the Greek “homologos” = “homo” + “logos” meaning “agreeing”. In short, integration is always independent of paths lying in the same homology class. Visibly, these homology classes reflect the complexity of the ambient space X . Our discussion mentioned three sorts of finite unions of oriented curves in X : arbitrary such (usually called “chains”); closed unions (usually called “cycles”);boundaries of 2-dimensional regions in X . We have just witnessed the origin of homology as an abelian group of classes of formal linear combinations of closed curves in X in which boundaries of 2-dimensional regions are trivial. The step from considering closed curves in a planar region X to generalized chains, cycles, and boundaries in any euclidean space, indeed, in any topological space, does not seem an awesome leap today; nevertheless, this leap was the birth of algebraic topology. We shall return to the picture of Green’s theorem, for homological algebra was born when analogies between topological homology theory and algebra were recognized.

CATEGORIES AND FUNCTORS

Let us introduce some terms to facilitate our discussion. Recall a set-theoretical distinction : our primitive vocabulary contains the words “element” and “class”; the word “set” is reserved for those classes that are small enough to have a cardinal number. Thus, we may correctly say “the class of all rationals” or “the set of all rationals”, but we may only say “the class of all groups” (this latter class cannot be a set, for there exist groups of any nonzero cardinal). Definition: A category (5 consists of a class of objects, obj 6,pairwise disjoint sets of morphisms, Hom,(A, B), for every ordered pair of objects, and compositions Hom,(A, B) x Horn,(B, C ) + Hom,(A, C), denoted (f,g) H gf, satisfying the following axioms: (i) for each object A, there exists an identity morphism 1, E Hom,(A, A ) such that fl, = f for allf E Hom,(A, B ) and 1g, = g for all g E Hom,(C, A ) ; (ii) associativity of composition holds whenever possible: if f E Hom,(A, B), g E Hom,(B, C ) and h E Horn,( C , D),then Several remarks are in order. First, the only requirement on Hom(A, B) is that it be a set; it is allowed to be empty. Second, we usually writef: A -+ B instead off E Hom(A, B);this explains our notation for composition. Finally, for each object A, the identity morphism 1, is unique: if 2, E Hom(A, A ) is

4

1 Introduction

a second morphism satisfying (i), then 2A= 2,1, = l A . There is thus a one-one correspondence A H 1, between obj 6 and the class of identity morphisms in 6,so that one could describe 6 solely in terms of morphisms and compositions. This is sometimes convenient for category-theorists, but we shall never forget objects. Examples: 1. 6 = Sets. Here objects are sets, morphisms are functions, and composition is the usual composition of functions. 2. 6 = Top. Here objects are topological spaces, morphisms are continuous functions, and composition is the usual composition. 3. 6 = .YJI. Here R is a ring (associative with l), objects are left Rmodules, morphisms are R-maps (i-e., R-homomorphisms), and usual composition. Recall that a left R-module is an additive abelian group M equipped with an action of R making it a “vector space over R : there is a function R x M + M , denoted (r,m ) w rm, satisfying: (i) (ii) (iii) (iv)

r(m

+ m’) = rm + rm‘;)

( r + r’)m = rm + r’m; (rr’)m = r(r’m); lm = m ;

where m,m‘E M

and

1, r,r’ER.

A function f:M.+ N between left R-modules M and N is an R-map if f(m

+ m’) = f(m) + f(rn’)

and

f ( r m ) = rf(m).

We will often write to mean M E obj .YJI. 4. If R = Z, the ring of integers, then R-module = Z-module

= abelian group, while Z-map = homomorphism. We usually denote the category of abelian groups by Ab rather than zYJl. 5. 6 = YJIR. Here R is a ring, objects are right R-modules, morphisms are R-maps, and composition is the usual composition.

Recall that an abelian group M is a right R-module if there is a function R x M + M , denoted (r,rn) H rm, satisfying all the axioms of a left R-module save (iii). The third axiom is replaced by (iii)’ (rr‘)m= r’(rm). If the value of R x M appealing form

-+

M on

( I , m)

is denoted mr, then (iii)’has the more

(iii)” m(rr’) = (mr)r’. We shall consistently use the latter notation (having r on the right) when discussing right R-modules. A function f :M + N between right R-modules M and N is an R-map if f(m + m’) = f(m) + f(rn‘) and f(rnr) = f ( m ) r .

5

Categories and Functors

If R is commutative, we may regard every right R-module M as a left R-module by defining rm = mr for all r E R and m E M . 6. Let S be a semigroup with (two-sided) unit, i.e., a monoid. We may construe S as a category 6 by defining obj O = { *} (i.e., 6 has only one object, *), Horn(*, *) = S, and composition as the given multiplication in S .

There are two features of this last example worth noting. First, obj 6 is a set, indeed, a rather small set. Second, morphisms are not functions between objects, for the object * is not comprised of elements. 7. Let X be a quasi-ordered set (this means X has a reflexive and transiTo say it another way, X would be partially ordered tive binary relation I). if, in addition, Iwere antisymmetric. We may construe X as a category 6 by defining obj 6 = X ,

Hom(x, y)

=

{%’

and composition i:i;

if x Iy (we have invented a new symbol) otherwise,

= iz

whenever x I

J’

I 2.

This last example has the feature that some Hom sets may be empty. The list of examples can be continued, and other categories will appear when necessary. There is a rough analogy between one’s first meeting with point-set topology and one’s first meeting with category theory. The definitions of topological space and of category are simple, abstract, and admit many examples, both natural and artificial. The raison d’etre of topological spaces is that continuous functions live on them, and it is continuity that is really interesting. Similarly, the raison tl’etru of categories is that “functors” live on them; let us define these at once. If one views a category as a generalized monoid (forget the objects, so one is dealing with morphisms that can sometimes be multiplied), then a functor is just a homomorphism. Definition: Let 6 and satisfying:

a be

categories. A functor F : 6 -, a is a function

(i) If A E obj 6,then F A E obj 3; (ii) If f : A -, B is a morphism in (5. then F f : F A + F B is a morphism in 3; (iii) if A B 5 C are morphisms in 6,then

F(d7f)= Fy Ff; (iv) for every A

E obj

6,we have F( l A ) = lFA.

Examples: 8. The identity functor F : O -,6 defined by F A = A and Ff = f . 9. The Hom functors 6 -+ Sets. Fix an object A in 6 and define

6

1 Introduction

FC = Hom(A, C); iff: C + C‘ is a morphism in 6,define Ff:Hom(A, C) + Hom(A, C ) by g H fg. One usually denotes Ff by f,. 10. If 6 = ,‘93, then the Hom functor F = Hom,(A, ) actually takes values in Ab: Hom,(A, C ) = {all R-maps A 3 C } is an abelian group if we define g h by a ++ g(a) h(a); moreover, one easily checks that every f, is a homomorphism of abelian groups. 11. A statement similar to that in Example 10 is true for 6 = %TIR. 12. If D is a fixed object in 0, define the constant functor I I :a -+ D by IC( = D for every C E obj 6 and = 1, for every morphism f in 6.

+

+

If1

Our description of the Hom functors is incomplete-what second variable?

if we fix the

Definition: Let Q: and D be categories. A contravariant functor F:C -+ D is a function satisfying: (i) If A E obj 6,then F A E obj D; (ii)’ if f:A + B is a morphism in 6,then Ff: FB -+ F A is a morphism in 3 (thus F reverses arrows); (iii)’ if A 4 B 25 C are morphisms in 6, then

F ( d ) = Ff Fg; (iv)

for every A E obj 6,we have F(1,) = lFA.

Examples: 13. Fix an object B in 6 and define F : 6 +Sets by F A = Hom(A,B); if f:A + A’ is a morphism in 6, define Ff:Hom(A’,B) + Hom(A,B) by g H gf. One usually denotes Ffby f*. 14. If 6 = R‘Bl,then F = Horn,( ,B) actually takes its values in Ab. A similar statement is true for 6 = ‘93,.

A functor is often called a covariant functor to emphasize it preserves the direction of arrows. Exercises: 1.1. A morphism f:A + B in a category 6 is called an equivalence if there is a morphism g:B A in 6 such that gf = 1, and fg = lg. What are the equivalences in the categories described above? 1.2. If F : 6 -+ D is a functor (either variance) and f is an equivalence in 6, then Ff is an equivalence in 3. -+

Definition : A category 6 is pre-additive if each HomdA, B)is an (additive) abelian group and the distributivity laws hold, when defined. Distributivity relates the addition of morphisms to the given composition in 6. As every nonempty set admits some abelian group structure, it would be foolish not to demand distributivity in the definition of pre-additive.

7

Tensor Products

Of course, R!Ul and !UIR are pre-additive. If 6 and D are pre-additive categories, then we say a functor (contra or co) F : 6 + D is additive if F(f + g) = Ff Fg for every pair of morphisms f and g lying in the same group. This says the function Hom,(C, C') + Hom,(FC, FC') given by f ~ F isf a homomorphism. Note that the Hom functors on module categories are additive.

+

Exercises: 1.3. In a pre-additive category, zero morphisms 0 : A + B are defined (namely, the zero element of Hom(A,B)).If F : 6 + D is an additive functor (either variance) between pre-additive categories, then F preserves zero morphisms. If 6 and 9 are module categories, then F(0) = 0, where 0 is the zero module. 1.4. If R is a commutative ring, then Hom,(M,N) is an R-module if we define r f by m H r . fm. Moreover, either Hom functor takes values in R!Ul. (A special case of Exercise 1.4 is R a field (so that R-modules are vector spaces and R-maps are linear transformations). The contravariant functor HornR(,R) is the dual space functor that assigns to each space its space of linear functionals and to each linear transformation its transpose.)

1.5. If R is a ring, its center is the subring Z(R) = {x E R : x r = rx for all r E R}. For every ring R, prove that Hom,(M,N) is a Z(R)-module and that the Hom functors take values in zcR,!Ul. 1.6. Let A be a left R-module and let r E Z ( R ) .Prove that the function p , : A -+ A defined by a ~ r isa an R-map. Give an example to show this may be false if r $ Z(R). (Hint: Take A = R R . ) 1.7. Let R and S be rings and cp: R + S a ring map. Every left S-module M may be construed as a left R-module if one defines r . rn = cp(r)m(similarly for right modules).

TENSOR PRODUCTS

There is a second family of functors, tensor products, that are as fundamental in homological algebra as the Hom functors. Before describing them, we give a construction for abelian groups; it will be generalized to modules in Chapter 3. Definition: Let G be an (additive) abelian group with subset X ; we say G is a free (abelian) group with basis X if each g E G has a unique expression

8

1 Introduction

of the form

where mx E Z and “almost all” m,

= 0 (i.e.,

all but a finite number of m, = 0).

The analogy between free groups and vector spaces is a good one; the next result shows that the notions of basis in each are kindred. Theorem 1.1: Let G be free with basis X ; let H be any abelian group and and let f : X -+ H be any function. Then there exists a unique homomorphism f”: G + H with f(x) = f ( x )for all x E X .

Ern,

Proof: If g E G, then g = 1 m , x . Define f’: G + H by f:g H f ( x ) .Note that f is well defined because the expression for g is unique and at most finitely many m, are nonzero. That f’ is unique follows from the fact that X generates G . I

The construction off” from f is called extending by linearity. Theorem 1.2: Given a set X , there exists a free abelian group G having X as a basis. Proof: Let ( Z , : x E X } be a family of copies of the integers Z . Consider the cartesian product Z,; its elements are “vectors” (m,)-this notation indicates that the xth coordinate is m, E Z. The cartesian product is an abelian group under coordinatewise addition :

nxoX

(m,)+ (n,) = ( m , + nx).

Define X’ = {all vectors having 1 as xth coordinate (fixed x E X ) and all other coordinates 0). Clearly X ‘ is in one-one correspondence with X . Define G’ as the subgroup of n Z , generated by X‘. It is immediate that each g E G‘ is a linear combination of elements of X ’ ; that such an expression for g is unique is the definition of equality in n Z , . Thus, G’ is a free abelian group with basis X’. If one actually wants X itself as a basis, define G = (G’ - X ’ ) u X (i.e., replace X ’ by X ) and define the obvious addition on the new set G ; the new group G is free abelian with X as basis.

9

Tensor Products

Here is a very formal but useful notion. Definition:

A diagram

A’

commutes if j3a

= a’F; a

a’

* B‘

diagram

commutes if c( = yb; a larger diagram comprised of squares and triangles commutes if each of its squares and triangles commutes. (Ofcourse,commutativity of a triangle is the special case ofa commutative square one of whose sides is an identity map.) Definition: Let R be an associative ring with 1. If A is a right R-module, B a left R-module, and G is an additive abelian group, then an R-biadditive function’ is a function f:Ax B-G

such that for all a, a‘ E A , b, b‘ E B, and r E R,

6) f(a + a’, b) = f(4b) + .f(u’, b); (ii) f(a,b b’) = f(a,b) f(a,b); (iii) f(ar,b) = f(a,rb).

+

+

We define a tensor product of A and B as an abelian group A ORB that converts R-biadditive functions A x B + G to additive maps A ORB + G. Definition: A tensor product of A E mRand B E ,$lis an abelian group A ORBand an R-biadditive function h which solves the following “universal

’

There is a related but distinct notion, R-bilinearity, that we will consider later in this chapter.

10

1 Introduction

mapping problem": A x B - A B R Bh

for every abelian group G and every R-biadditive f, there exists a unique homomorphism f ' making the diagram commute.

Theorem 1.3: Any two tensor products of A and B are isomorphic. Proof: Suppose there is a second group X and an R-biadditive k : A x B + X that also solves the universal mapping problem. There is a diagram h

A x B P A B R B

where k' and h' are homomorphisms such that k'h

=k

and

h'k

=

h.

There is a second diagram A x B

h

A @R B ;

for the dashed arrow, we may choose either the identity or h'k' (for h'k'h = h'k = h);uniqueness yields 1 = h k ' . A similar argument shows that k'h' = l,, so that k is an isomorphism A ORB7 X . I The proof of Theorem 1.3 should serve as a model for a more general fact: any two solutions of a universal mapping problem, if, indeed, there are any, are isomorphic. Having proved uniqueness of tensor product, we now prove its existence.

11

Tensor Products

Theorem 1.4: B exists.

The tensor product of a right R-module A and a left R-module

Proof: Let F be the free abelian group with basis A x B, i.e., F is the group of Z-linear combinations of all ordered pairs (a, b). Define S as the subgroup of F generated by all elements of the following three forms: (a

+ a’, b) - (a,b) - (a‘,b),

+

(u,b b’) - (a,b) - (a,b’), (ar, b) - (a,rb).

Define A ORB = F / S . If we denote the coset (a,b) + S by a @ b, then it is easy to check that h:A x B + A ORB defined by (a,b) H a 0 b is R-biadditive. Assume G is an abelian group, and we have the diagram

where f is R-biadditive. Since F is free on A x B, there is a unique homomorphism cp: F + G with cp( (a, b)) = f(a, b). Moreover, f being R-biadditive implies S c ker cp; it follows that cp induces a homomorphism f ’: F / S + G by f ’ : ( a , b ) + S w c p ( ( a , b ) ) = f(a,b). This says that f ’ : a @ b w f(a,b), and this says f ’h = f.We leave the proof of uniqueness o f f ’ to the reader (use the fact that A OR B is generated by all 11 0b). I It is important to realize that a typical element of A ORB has the form x u i 0 bi and may not have an expression of the form a 0 b; furthermore, the expression & O b i is not unique. As a practical matter, one must be suspicious of a “homomorphism” g with domain A ORB given by specifying its value on each a 0 b; since these elements merely generate A QRB, the “map” g is not well defined unless it preserves all the relations. The safest scheme is to rely on the universal mapping problem, for it is easier to define an R-biadditive function on the Cartesian product A x B. Theorem 1.5: Let f:A + A‘ be an R-map of right R-modules and let g :B + B be an R-map of left R-modules. There is a unique homomorphism A ORB + A’ QR B with a @ b H f(a) 0 g(b).

Proof: The function A x B + A’ ORB’ defined by (a,b) H f(a) 0 g(b) is obviously R-biadditive; now use universality. I

12

1 Introduction

Definition: The map A QRB denoted

+

A‘ QRB’ sending a Q b H fa @I gb is

f 09. Theorem 1.6: Assume A 5 A‘ 5 A” are R-maps of right R-modules and B 5 B’ 5 B” are R-maps of lefi R-modules. Then 9 ’ Q f ’ o s Q f = s’sQf’f.

Proof: Use Theorem 1.5.

Corollary 1.7: Zf A defined by

E

I

fDzR, then there is an additive functor

F : R m -+ Ab

F(B) = A @R B ; i f f :B + B is an R-map between left R-modules B and B‘,

Ff

=

1,

of.

Similarly, for $xed B E RYJl, there is an additive functor G:’%RR+ Ab with G(A)= A QRB and Gg = g Q Is (where g : A -+ A’ is an R-map). Proof: It is clear that F(1,) = 1” @I l B is the identity on A O R B , for it fixes the set of generators of the form a Q b. That F preserves composition follows immediately from Theorem 1.6, setting g = g’ = 1,. Finally, if f,f’: B 4 B‘, then 1, Q (f+f ‘) sends a Q b H a 0 (f f ‘ ) b = a Q (fb+f’b)=aQfb+aQf’b=(l,Qf 1,Qf’)aQb. I

+

+

We shall usually write A QR instead of F and QR B instead o f G. When no confusion can arise, we suppress the subscript R. In general, A QR B is only an abelian group; however, there is a common circumstance in which it is a module.

Definition: Let R and S be rings. An abelian group B is an (R-S) bimodule, denoted RBS, if B is a left R-module, a right S-module, and the two actions are related by an “associative law” r(bs)= (rb)s

for all r E R, b E B, and s E S.

Remark: The last condition says that, for each r E R, the map B + B given by b H rb is an S-map, and, for each s E S, the map B + B given by b H bs is an R-map. Examples: 15. R itself is an (R-R) bimodule (which is merely a fancy restatement of the ordinary associative law).

13

Tensor Products

16. If R is commutative, every left R-module B may be construed as a right R-module by defining br = rb, in which case B is an (R-R) bimodule. 17. Every left R-module B is an (R-Z) bimodule: B = RBZ;similarly, every right R-module C is a (Z-R) bimodule: C = z C R . 18. If R is commutative, then a ring S is an R-algebra if there is a ring map cp: R + Z(S), where Z ( S ) is the center of S . Note that S becomes a left R-module (Exercise 1.7) if we set

r . s = cp(r)s. Since cp(r) E Z ( S ) ,this definition also makes S into a right R-module. Finally, S is a ring that is both an (R-S) bimodule and an (S-R) bimodule. 19. Every ring is a Z-algebra. Example 16 is the special case of Example 18 in which R = S and cp = 1,. In older textbooks, one often assumes that S is an R-algebra only when R is a subring of Z ( S ) , i.e., cp:R -, Z ( S ) is an inclusion. This tends to be unnecessarily restrictive; for example Z/nZ is not a Z-algebra with the older usage. Theorem 1.8: If A is a right R-module und B is an (R-S) bimodule, (briefly, A R and RBS),then A ORB is a right S-module, where ( a 0 b)s = CI @ (bs). Similarly, in the situation , A , and S(U

RB,

then A ORB is a left S-module, where

Q b) = ( S U ) Q b.

Proof: For fixed s E S , the function i t s : B -+ B defined by b H bs is an R-map, for B is an (R-S) bimodule. If F is the functor A Q R ,then Q S ) : B + B is a homomorphism (of groups). But F ( p , ) = 1,@ p s : aQ b H a Q (bs). Thus, the formula in the statement is well defined; it is mechanical that the module axioms are satisfied. 1 Corollary 1.9: Given A R and R B S , the ,functor QRB takes values in similarly, given s A , and RB, the functor A OR takes values in s1132.

mS;

Proof: All that remains to be shown is that i f f : A 4A’ is an R-map, then f @ 1, is an S-map. This is a simple calculation on any generator a Q b, using the formulas we have established in Theorem 1.8. 1 Corollary 1.10: If R is commutative and S is an R-algebra, then S ORB is a left S-module for every left R-module B.

Proof: We know that S is an (S-R) bimodule.

I

14

1 Introduction

Tensoring an R-module by an R-algebra S is called base change, for it changes the ring of scalars from R to S. This remark should alert the reader that tensor product may behave in unexpected ways. For example, every abelian group G is a Z-module, and the rationals, Q, is a Z-algebra (as is any ring). Thus, Q Qz G is a Q-module, i.e., a vector space over Q. Since every Q-space is torsion-free as an abelian group, the Q-space Q Qz G cannot contain a copy of G if G has elements of finite order. It is thus possible that the map G --* Q Qz G defined by x H 1 Q x not be one-one! This curious phenomenon will be treated in some detail later. The following special case of Theorem 1.8 is of sufficient importance to merit its own statement. Corollary 1.11: If R is commutative and A and B are R-modules, then A ORB is an R-module with r(a Q b) = ra Q b = a Q rb. Moreover, for fixed r E R, if p,:S + B is given by b H rb, then 1 Q p , : A ORB+ A QR B is also multiplication by r (the same is true in the other variable). Theorem 1.12: r f R is a ring and B a left R-module, then there is an Risomorphism R QRB Z B with r Q b H rb. Proof: First of all, R BRB is a left R-module since R is an (R-R) bimodule. The definition of left R-module says that the function R x B + B with (r,b) H rb is R-biadditive; the definition of tensor product provides a (group) homomorphism 0:R QRZ? + B with r @ b H rb. The reader may show that 8 is an R-isomorphism by observing that it has an inverse, namely, b H 1 63 b, which is an R-map. 1

Of course, there is a similar result on the other side: A ORR R-modules.

A as right

Theorem 1.13: I f R is commutative and A and Z? are R-modules, then there is an R-isomorphism 7 :A QR B + B ORA with a @ b H b Q a. Proof: The “twist” t : A x B + B ORA defined by (a,@- b @ a is Rbiadditive, hence induces a Z-homomorphism z: A Q RB -+ B O RA that is easily seen to be an isomorphism (construct its inverse in a similar way). Theorem 1.8 shows that z is, in fact, an R-map. I

There is another consequence of Theorem 1.I 1that should be mentioned. Definition: If R is commutative and A, B, and G are R-modules, then a function f:A x B + G is R-bilinear iff is R-biadditive and rf(a,b) = f(ra, b) = f ( a , rb)

all r e R, a E A, b E I?.

15

Tensor Products

Exercise 1.8 gives an example of a biadditive function that is not bilinear. This new definition poses a new universal mapping problem: I,

A x B

if A and B are R-modules (R commutative), is there an R-module A * B and an R-bilinear function k : A x B -+ A * B such that, for every R-module G and every R-bilinearf :A x B -+ G, there is a unique R-map f "with f "k = f ? Corollary 1.14: If R is commutative and A, B are R-modules, then A ORB solves the universal mapping problem posed by R-bilinear functions as well as that posed by R-biadditivefunctions. Proof: Theorem 1.11 shows that A ORB is an R-module and that h: A x B -+ A ORB sending (a,b) I-+ a h is R-bilinear.Assume f ':A ORB -+ G is the unique Z-map provided by R-biadditivity, so that f'(a 6b) = f(a, b). It only remains to check that f ' is an R-map. This is easy:

f ' ( r ( a@ b))

= f '(ra @ b) =

m, b)

(by Theorem 1.1 1) ( f is R-bilinear)

= rf(a,b) = r f ' ( a @ b).

Exercise: 1.8. Let R = { m + no:m, n E Z, o2 = 1). (A concrete realization of R is all 2 x 2 matrices over Z of the form ():. Let A = B = Z be made into an R-module by (rn + no)a = ( m - n)a, where a E Z. Let G = Z be made into an R-module by (m + na)g = (m + n)g, where g E G. Finally, define f :A x B -+ G to be ordinary multiplication of integers: (a,b) H ab. Show that f is R-biadditive but not R-bilinear. The next exercises give associativity laws as well as a useful construction. Definition: In the situation ( A R ,RBS, &), a triadditive function f : A x B x C-+G

(where G is an abelian group) is a function that is additive in each variable and f(ar,b, c) = f(a, rb,c)

and

f(a, bs,c) = f(a, b,sc).

16

1 Introduction

Exercises: 1.9. Prove that (A ORB)OsC and (a, b, c) H ( a 0 b) 0 c is a solution to the universal mapping problem for triadditive functions from A x B x C. Prove that A OR(B C) and (a,b, c) t,a 0 (b 0 c) is also a solution. 1.10. Using the technique of the proof of Theorem 1.3, prove the associative law for tensor product: there is an isomorphism (A ORB) OsC 7 A OR(BOsC) taking ( a @ b) 0 c H a 0 (b 0 c). If either A or C is a bimodule, this isomorphism is a module isomorphism. (See Exercise 2.36.) 1.11. Prove the generalized associative law for tensor product by defining n-additive functions and showing that any association yields a solution to the corresponding universal mapping problem. 1.12. Let R be a commutative ring, I an ideal in R, and M and N Rmodules. Prove that M @ N / I ( M 0 N ) E ( & f / I M )ORII ( N I I N ) . (Hint: Show that both modules solve the universal mapping problem for R/Ibiadditive functions from M f IM x NIZN.) 1.13. Let R and S be k-algebras. Prove that R @k S is a k-algebra if one defines multiplication by ( r 0 s)(r' 0 s') = rr' 0 ss'. (Hint: This multiplication is well defined, being the composite

as

(R 6 s)6 (R 0 S ) s R 0 ( S 0 R) 6 S

@1: @

1

bR6(ROS)OS

s (R 0 R) 0 ( S 6 S ) 5R 0 s, where z is the twist map of Theorem 1.13 and p, v are the multiplications of R, S, respectively.) 1.14. Prove that k [ x ] Okk[ y] E k [ x , y ] as k-algebras. Definition: If R is a k-algebra, define its opposite R O P as follows: as a k-module, there is a k-isomorphism R -,RoP denoted r H ro; multiplication R O P x ROP -,RoP is defined by (r; ,r;) t--, (r2r1)0(where r2rl is multiplication in R). 1.15. Every (R-S) bimodule A is a left R @k Sop-moduleif one defines ( r 0 so)a = (ra)s = r(as) (we are assuming R and S are k-algebras). 1.16. Every right R-module M is a left RoP-module if one defines rom = mr. As anticipated in Exercise 1.4, the abelian group HomR(A,B) acquires a module structure if one of the variables is a bimodule. The proof of the next theorem is a dull exercise.

Theorem 1.15: (i) Given RAs and RB, then HomR(A,B) is a k ! S-module if one dejnes ($)(a) = f(as);

17

Singular Homology

(ii) given RAs and B,, then Hom,(A,B) is a right R-module if one defines (fr)(a) = f(ra); (iii) given A, and sBR, then HomR(A,B ) is a lefi S-module if one defines (sf)(a)= s(f(a)); (iv) given , A and ,BR, then Homs(A, B ) is a right R-module if one defines (fr)(a) = f ( 4 r . There is a mnemonic device for this theorem. Suppose one were to write maps on the side opposite to that on which the scalars are written. Thus, if A is a right module, write fa, but if A is a left module, write af. With this convention, each of the four parts of Theorem 1.15 becomes an associativity law. There is an analogue of Theorem 1.12 holding in one variable of Hom. Theorem 1.16: I f B is a left R-module, then there is an R-isomorphism HomR(R,B)+B with f ~ f ( 1 ) Similarly, . the same formula gives an Risomorphism of right R-modules when B is u right R-module. Proof: Theorem 1.15 shows that the formula does define an R-map. If b E B, define fb: R -+ B by r H rb; then b w .fb is the inverse. I

SINGULAR HOMOLOGY Let us return to topology to construct the (singular) homology functors H , : Top- Ab, one functor for each n 2 0: we provide more details for a generalization of our earlier discussion of curves in planar regions. For each n 2 0, consider euclidean n-space R" imbedded in R"" as all vectors whose last coordinate is 0. Let co denote the origin, and let { u l , . . . , u,} be the standard orthonormal basis of R" (ui has 1 in the ith coordinate and 0 elsewhere). For each ti 2 0, let A,, = {(tl, . . . , t,,):ti 2 0, all i, and Iti = l} be the convex set spanned by { u o , . . . , u,,}; A,, is called the standard n-simplex with vertices { u o , . . . , v , ) and is also denoted An = [ u o , . . . , u,,]. Thus, A. = [ u o ] is a point, A I = [ u o , u l ] is the unit interval [0, 1];A2 = [ u o , u I , u 2 ]isthetriangle(with interior) h a v i n g v e r t i c e s u , , ~ , , ~ ~ ; A3 is a tetrahedron, and so forth. A curve in a topological space X is a continuous map o : A , X ; a closed curve in X is a curve o with o(0) = o(1). The boundary of A1 is (0,l j ; more generally, the boundary of A,, is U~=,[co,. . . , C i , . . . , u,], where means "delete". However, we need an "oriented boundary" if we are to generalize the picture of Green's theorem. --f

A

Definition: An orientation of A,, is an ordering of its vertices.

18

1 Introduction

It is clear that different orderings may give the “same” orientation. For example, consider A, with its vertices ordered uo < u1 < u 2 . A tour of the vertices shows that A, is oriented counterclockwise. Thus, the orderings

u1 < u, < uo and 0, < uo < u1 give the same tour, while the other three permutations give a clockwise tour.

Definition: Two orientations of A,, are the same if, as permutations of { u o , . . . , un}, they have the same parity (both are even or both are odd); otherwise, the orientations are opposite.

After examining this definition for the tetrahedron A 3 , the reader will be content with it. Given an orientation of A,,, one may orient [ u o , . . . ,gi,.. . , u,,] in the sense ( - l)”uo, . . . , Ci,. . . , u,,], where - [ u o , . . . , Gi,. . . , u,,] means its orientation is opposite to that of [ u o , . . . , Gi, . . . , u,,] (vertices in displayed order). For example, consider A, oriented counterclockwise: u2

the natural way to orient the edges is:

The edges are thus oriented [uo,ul], [u1,u2], and [ u 2 , u o ] . Since [ u 2 , u o ] = - [ u 0 , c 2 ] , the oriented boundary ofA2 is [ c , , u 2 ] u - [ u 0 , z ~ , ] u [ I ~ ~ , U=~

]

19

Singular Homology

[6,,ul,u2] u -[11,,6,,u,] u [ U ~ , I ~ ~ , P The ~ ] . oriented boundary of A,, should thus be ( - l)”u,, . . . , L3,, . . . , u.].

u:=,

Definitions: If Xis a topological space, an n-simplex in X is a continuous function o:A,, X . The n-chains in X comprise S,,(X ) , the free abelian group with basis all n-simplexes in X . For convenience, set S- ,(X) = 0. --f

Observe that S , ( X ) is precisely the group of chains suggested by line integrals: all formal linear combinations of curves in X . The group S,(X) is the n-dimensional generalization of S,(X ) we anticipated. If o:A,, -+ X , its boundary So should be ( - l)’(aI [u,, . . . ,C i , . . . ,u,,]). A technical problem arises. It would be nice if do were an (n - 1)-chain; it is not because the domain of o I [ u , , . . . D,, . . . , u,,] is not the standard (n - 1)-simplex A,,- 1. To state the problem is to solve it. For each i, define ei:A,,-l + A,, as the affine map sending the vertices {u,, . . . , u,- 1 } to the vertices {u,, . . . , Gi,.. . , u,} that preserves the displayed orderings:

I;= , .

ei(t,, . . . , t , - ,)

= (tl,

. . . , ti- ,,O. t i , . . . , t,- ,)

Definition: If o:A,, -+ X , then Sfla = Theorem 1.17:

ano =

,(

-

I:=, ( - l)’oei

E

E

A,,.

S,,-l(X).

There is a unique homomorphism L?,,:S,(X) + S,- , ( X ) with l)’oeifor every n-simplex o in X.

Proof: Extend by linearity.

I

The homomorphisms d, are called boundary operators; usually one omits the subscript n. We now have a sequence of homomorphisms

~~~-s~~x)~s~~,(X)+’’’-.s,(x)1:s0(x)+0.

,

.

Let us denote ei:A,,- -+A, by [u,, . . . P,. . . . , u,]. Lemma 1.18: The following ,fr~rmulrrshold jbr ei ’~ c ~ : A , , -+~ A,,: if i < 1 , then e, r j = [z),,. . . , Pi, . . . , P i , . . . , u , ] : if i 2.j, then e, ,: e j = [ I : ~ , .. . , D j , . . . , Ci, 1 , . . . , u,].

Proof: The maps e, and ej, hence their composite, are completely determined by their values on the vertices . . . , u , , - ~ } .The computation showing that the two displayed vertices are the deleted ones is routine. I {/lo,

Theorem 1.19: For each n 2 1, we have i n - ?,

=0

20

1 Introduction

Proof: It suffices to show d do = 0 for every n-simplex o:A,

+X .

d d a = d(C( - I)’oei)= C(- l)’d(aei)= =

C (-

( - l)’+joeiej id

l)1+j,[UO,. . . , u i , . . . , Cj, . . . , u,] A

i is a family of submodules of M , then a submodule of M.

njEJMi is also

Definition: Let X be a subset of a module M. The submodule ofM generated by X is MJ,where { M J : j EJ } is the family of all submodules of M that contain X. We denote this submodule by (X).

n,,

25

Modules

Theorem 2.1 : Let X be a subset of M . If X then ( X ) = { x r i x i : r E i R, x i E X } .

=

0, then ( X ) = 0; i f X

#

0,

Proof: If X = 0, then 0 is a submodule of M containing X , from which it = 0. If X # then the subset S = { x r i x t : r E i R, xi E X } follows that (0) is defined (it is defined when X = 0if one enjoys summing over an empty index set). Since R contains 1, we have X c S . An easy check shows S is a submodule of M , so it follows at once that ( X ) c S . For the reverse inclusion, it suffices to show that if M' is any submodule of M containing X , then S c M ' (for then S is contained in the intersection of all such M ' , which is ( X ) ) . This is clear: xi E M', all i, implies x r i x i E M ' for all ri E R. 1

a,

Definition: A module M is finitely generated (f.g.) if there is a finite subset { x , , . . . , x , } o f M with { x , , . . . , xn) = M ; a module M is cyclic if there is a single element x E M with (x) = M . Definition: Let f :M + N be an R-map. We say ,f is monk (or is a monomorphism) iff is one-one; we say f is epic (or is an epimorphism) iff is onto.

Of course, f is an isomorphism if and only i f f is both monic and epic. Definition: If M ' is a submodule of M , the quotient module M j M ' is the quotient group M / M ' made into an R-module by

r(m

+ M ' ) = rm + M'.

One must assume M ' is a submodule in order that the action of R on M / M ' be well defined. Examples: 8. If M' is a submodule o f M , the inclusion i: M' + M is monic. 9. If M' is a submodule of M , the natural map n:M + M / M ' defined by m H m + M' is epic, and ker 71 = M'. 10. I f f : M + N , then f is monic if and only if ker f = 0. 11. I f f : M + N , then f is epic if and only if coker f = 0 (cokernel f is defined as the quotient module Njim f). 12. (First Isomorphism Theorem) I f f : M + N , then the map rn ker f t,f ( m ) is an isomorphism Mfier f 7 im f . 13, (Second Isomorphism Theorem) If M , and M , are submodules of M , then m , M , n M , H m , + M , is an isomorphism

+

+

M I I M I n M2 % ( M i

+ Mz)/Mz.

The Second Isomorphism Theorem follows easily from the First: let n : M + M j M , be the natural map, and let f = n1 M I . It is easy to see that kerf = M , n M , and im f = ( M , M 2 ) / M 2 .

+

26

2 HornandB

14. (Third Isomorphism Theorem) If M , c M , are submodules of M , then ( M / M 2 ) / ( M l / M 22) M / M , . The Third Isomorphism Theorem also follows easily from the First: the map f : M / M , + M / M , given by m + M , H m + M , is epic with kernel MlIM,. 15. (Correspondence Theorem) If M ’ is a submodule of M , there is a one-one correspondence between the submodules S of MfM’ and the “intermediate” submodules of M containing M‘ given by S H n-’(S) (where n : M -+ M / M ’ is the natural map). Theorem 2.2: A module M is cyclic if and only if M z RfI for some left ideal 1. Moreover, if M = ( x ) , then I = { r E R : r x = 0 ) . Proof: First of all, R/I is cyclic with generator 1 + I ; i f f :R/I -+ M is an isomorphism, then M = (x), where x = f(1 + I ) . Conversely, assume M = ( x ) . Define f rR -+ M by f ( r ) = rx. Since f is epic, M z R/ker f . But ker f is a submodule of R, which is a left ideal; indeed, ker f = { r E R : r x = 0). 1 Definition: Two maps

are exact at M if im f = ker g. A sequence of maps (perhaps infinitely long)

. . . + M n +/ . + Il d M nf A M , - l

-+“’

is exact if each adjacent pair of maps is exact. Exercises: 2.1. If 0 -+ M’ s M is exact, then f is monic (there is no need to label the only possible map 0 M ’ ) ;if M 4 M ” -+ 0 is exact, then g is epic; if 0 -+ M M’ -+ 0 is exact, then f is an isomorphism. 2.2. If M ’ M 3 M ” is exact with f epic and g monic, then M = 0. Conclude that exactness of 0 -+ M -+ 0 gives M = 0. 2.3. Prove that a map cp is monic if and only if cpf = cpg implies f = g

s

-+

s

f

(the diagram is A 3 B C ) ; prove that cp is epic if and only if h q = kq 8 implies h = k. 2.4. If M , L M 2 -+ M , 5 M , is exact, then f is epic if and only if g is monic. 2.5. If M , L M 2 -+ M , -+ M , 5 M , is exact, then f epic and g monic imply M , = 0 (use Exercises 2.2 and 2.4).

27

Modules

2.6. If 0 + M ' A M + M" 0 is exact, then M ' E iM and MIiM z M". Such sequences are called short exact sequences. 2.7. Consider the commutative diagram with exact rows -+

0-A-B-C

Prove that there exists a unique map A + A' making the diagram commute. Similarly, one can uniquely complete the commutative diagram with exact rows

Remark: There is a categorical translation of Exercise 2.7. Let 2I denote the category whose objects are all R-maps; define a morphism cp: f + g as a pair of maps cp = (q1,cp2) making the following diagram commute:

.t

A

One may now see that ker and coker are functors 'u + .1132. Exercises: 2.8. Iff: M

-+

N is a map, there is an exact sequence

O + ker f

+M

N +coker f +O.

2.9. (Restatement of Third Isomorphism Theorem) If M , c M , are submodules of M , there is a short exact sequence 0 + M , / M 2 --* M I M , + M / M , 4 0.

28

2 Homand

63

2.10. (Another Version of Third Isomorphism Theorem) Consider the commutative diagram

where a is monic and p is epic. Then kerp # 0 if and only if coker a # 0, i.e., aK‘ is a proper submodule of K . (Hint: Write C = M / K and C’ = M / K ’ , so that ker B = K ’ / K . )

SUMS AND PRODUCTS Definition : Let { A j : jE J } be a family of modules. Their product, denoted the module whose underlying set is the Cartesian product of the “vectors” a = (aj)where aj E A,, and with module operations defined by (0,) + (bj) = ( a j + bj), r(aj)= (raj).

Definition: The sum of the denoted ujeJ is the submodule of n,,, consisting of all “almost all” of whose coordinates 0 (i.e., all but a finite number of are 0). Aj,

A,

A,,

(a,) aj

a, =

Other names and notations are common. Product is often called “direct product”, “complete direct sum”, or “strong direct sum”; sum is often called Aj. In “direct sum” or “weak direct sum” and is usually denoted category theory, sums are called “coproducts”. A, = A,; in If the index set J is finite with n elements, then this case we also write A , 0 .. * 0 A,. If the index set J is infinite (and infinitely many A , # 0),then u A j is a proper submodule of n A j .

n,,,

xi,,

UjEJ

Exercises: 2.11. If R is a field (or a division ring), every module is a sum of copies of R (this is just the statement that every vector space has a basis).

29

Sums and Products

2.12. If MI and M , are submodules of M with M , n M , = 0 and M I + M , = M , then M = M , 0 M , . (In this case, we write M = MI 0 M,: “equal” instead of “isomorphic”.)

Functors only recognize objects and morphisms, so our next task is to characterize product and sum in these terms. If A = n A j , define projections pi: A + A j and injections , I j : A j + A by pj:(aj) F-+ a j

and Aj(aj) is that element having a j in the jth coordinate and 0 elsewhere.

Note that p j l j = l A jand p j & = 0 if j # k . These maps are also defined when A is a sum, in which case c l j p j = 1,; this means that for each a E A, almost all pja = 0 and xAjpja = a.

ujEJ

Theorem 2.3: Let modules A and { A j j: E J} be given. Then A A j if and only if there are maps l j :A j + A such that, given any module X and any maps A j + X , there is a unique map cp: A X with q,Ij = fj, all j E J.

fi:

--f

4

A;

Proof: Suppose A is the sum of the A j , and let { p j :A + A j } be the projections. Define cp: A + X by rpa = c f i p j a .Note that since A is a sum, almost all pja are 0, so the formula for rpa makes sense. The diagram above does commute: cpAjaj = 1 f k p k i j a j= h a j . k

We claim cp is the unique such map. If I): A + X satisfies +Aj = fi, all j, then $a = C$,Ijpja = C f j p j a = rpa. Suppose, conversely, that a unique p:A -+ X always exists. Consider the diagrams

30

2 Homand @

where c l j : A j + D A j is the jth injection. Now cp exists by hypothesis, while $ exists by the first part of the proof. Assemble these diagrams into

and note that we have commutativity: $cpAj = +aj = i j .The hypothesis applies t o this last diagram if we set X = A and fj = A j , and gives uniqueness of $cp. But, visibly, 1,:A + A completes the same diagram, so that $cp = 1,. If we assemble the original pair of diagrams a second way, then we obtain q$ = 1L1 A , . Hence cp is an isomorphism. I Remarks: 1. The maps i j must be monic. 2. The proof above should remind the reader of the proof of Theorem 1.3. Indeed, we have shown that sum is the solution to a universal mapping problem. Theorem 2.4: If A j is the j t h injection A j + U A j and if B is a module, then the mup

8:Hom(UAj,B)+nHom(Aj,B) given by H(cpij)

is an isomorphism.

Proof: It is clear that 8 is a homomorphism (of abelian groups). 8 is epic: if (4)E Hom(Aj,B), then h:Aj-+ B for each j ; by Theorem 2.3, there is a map q : U A j + B with c p l , = fj, all j , i.e., 8(q)= (fj). 8 is m o n k suppose 8(cp) = 0 = (cpl,), i.e., each c p i j = 0. Thus, cp completes the diagrams

fl

31

Sums and Products

Since any completion is unique and since the zero map also completes, we havecp = 0. I In terms of functors, we have just seen that Hom( ,B ) converts sums to products; in particular, it preserves jnirc) sums.

Exercises: 2.13. If all the Aj are bimodules or if B is a bimodule, then the isomorphism 8 of Theorem 2.4 is a map of modules. 2.14 Give an example in which H o m ( n A j , B ) Hom(Aj, B). 2.15 Give an example in which H o m ( n A j , B ) 8 Hom(Aj, B). 2.16 If k is a field and V is a vector space over k, let V* = Hom(V,k), its dual space.Assume dim V = 111. I f m is finite. then dim V* = dim V(whence V* 2 V ) ;if m is infinite, then dim V * = 2" (whence V* $ V ) .

n

We are also able to describe product as the solution to a universal mapping problem.

nJEJ

Theorem 2.5: Let modules A tinti [ A , : I E J I he given. Then A 2 A , if and only i f there are maps pJ:A + A, such thrit. giuen any module X und any maps f,: X + A , there is a unique mup cp :X -+ A with pip = f,, all j E J .

\ ///

A, 4

PJ

A

?

X

Proof: If A is a product, then one easily shows that the projections p J have the property described. The converse is proved exactly as Theorem 2.3 if one merely reverses direction of all arrows. 1

One may prove that the maps p, must be epic.

Theorem 2.6: Zf p j is the jth projection n A , --* A, and i f B is a module, then the map Q:Hom(B,HAj)-+ n H o m ( B , A j ) given by

CP H (PjV) i s an isomorphism.

Proof: Repeat the proof of Theorem 2.4 rniirtrtis mufundis.

1

32

2

Homand 8

We have just seen that Hom(B, ) preserves products; in particular, it preserves jinite sums.

Exercises: 2.17. If all the A j are bimodules or if B is a bimodule, then the isomorphism 8 of Theorem 2.6 is a module isomorphism. Hom(B, Aj). 2.18. Give an example in which Horn(& u A j ) 2.19. Give an example in which Horn(& H A j ) Hom(B, Aj).

$u n

We have seen that the universal mapping problems describing sum and product differ only in that all arrows have reversed direction. This phenomenon is common enough to deserve a definition. Recall that a statement S (in the first order propositional calculus) is a well formed formula involving the usual logical connectives (“and”, “or”, “not”, “implies”, “equality”) and the quantifiers 3 and V. If % is a category, then we say S is a statement about % if the variables occurring in S are interpreted as objects and morphisms of %. Definition: If S is a statement about a category %, then its dual S* is the statement about % obtained by reversing the direction of each morphism (i.e., interchange “domain” and “target”) and replacing each composite as of morphisms by PLY. The notion of “dual” extends in an obvious way to diagrams; reverse every arrow and retain commutativity of any square or triangle.

Exercises: 2.20. Show that the zero module is a solution to the universal mapping problem: for every module X , there is a unique map X -,0. 2.21. Show that the zero module is also a solution to the dual universal mapping problem :for each module X,there is a unique map 0 -+ X . Conclude that 0 is “self-dual”. It is easy to see “identity morphism” and “isomorphism” are self-dual. In Exercises 2.41 and 2.49, it is shown that “cokernel” and “kernel” are dual. From this, it follows that “monic” and “epic” are dual. Finally, “exact” is self-dual, for we may describe exactness of A B 4 C as ker(B + cokerf) = ker g or as coker(ker f -+ B) = coker 9. It is not true that the dual of a theorem must be a theorem. However, should both be true, it is often (but not always) the case that the proofs of each are dual. Two such examples are the dual pairs: Theorems 2.3 and 2.5; Theorems 2.4 and 2.6. In future instances, we will supply only one proof (as we did for Theorems 2.3 and 2.4) and merely say the other proof is dual. The special case of the sum of two modules merits more discussion.

33

Sums and Products

Theorem 2.7: Given A and B with i: A 4 B monic, then A is a summand of B (i.e., B = iA 0 C for some submodufe C o f B) if and only if there is a map p : B -+ A with pi = I,, .

Proof: If A is asummand of B, define p : B -+ A as the projection on A having C = kerp. Conversely, given p : B + A with pi = l,,, define C = kerp. We claim B = iA 0 C. If b E B, then b = ipb

+ ( b - iph).

The first term is in iA, and the second is in C, for p(b - ipb) = 0. Finally, it is clear that iA n C = 0, so Exercise 2.12 completes the argument. I Definition: A short exact sequence 0 4 A a mapj:C -+ B with p j = 1,.

A B 5 C -+ 0 is split if there is

Exercises: 2.22. The short exact sequence above is split if and only if there is a map q : B A with qi = l A . Conclude that in a split short exact sequence, A is a summand of B and C is a summand of B. 2.23. Assume A = B 0 C and B c M c A . Prove that M = B 0 ( M n C). 2.24. If F:,m + Ab is an additive functor (either variance) and if 0 A -, B C -+ 0 is a split exact sequence, then FA is a summand of FB. Conclude that F preserves finite sums. (Hint: 1, = jp + iq, where the notation is as in Exercise 2.22.) -+

-+

Having seen how the Hom functors treat sums, let us now see how tensor product behaves. Theorem 2.8: The map

Let A be a right R-module und let (€3,: j E 3 ) be lefi R-modules. 0:A

UBj

+

u ( A 8, B j )

given by a 0(bj)H

0b j )

is an isomorphism. There is a similar isomorphism i f the sum is in thefirst variable. Proof: Observe first that 8 is well defined, for the function A x LIB,-+ U ( A0 B,) given by (a, (b,))w ( a Q b,) is R-biadditive. Let E.,:B, LIBj be the jth injection. In order to define a map 9:u ( A Q Bj) + A Q U B , , it suffices to give maps & : A 0 B j -+ A @ u B j . Define fi so that a Q bj H a 0 Ajbj (i.e., fi = 1,@ A,). If p, is the jth injection A Q B, + U ( A Q Bj), -+

34

2

Homand @

Theorem 2.3 gives a map cp with cppj = f j , all j . It is a straightforward check that Bcp and cp6 are identities. 1 We have just seen that the functor A OR preserves sums; the functor O R Balso preserves sums. Exercises: 2.25. Ifeither A or each Bj is a bimodule, then the isomorphism 8 in Theorem 2.8 is a map of modules. 2.26. Give an example of modules A and {Bj:j E J } for which A @ , ( n B j ) $ n ( A ORBj). (Hint: Take R = Z and p E Z a prime; define A = Q and B, = Z/p“Z, n = 1,2,. . . .)

EXACTNESS The next question is how functors affect exactness. All functors are additive functors between categories of modules. Definition: A functor F is left exact if exactness of O+A:B~C

implies exactness of O-+FAZFBZFC; a functor F is right exact if exactness of A~BLC-tO implies exactness of

FA

5 FB 2 F

C ~ O .

If F is left exact, then F a :F A + FB is monic and im FCY= ker Ffl; thus F preserves monomorphisms and F “preserves kernels” in the sense that Fa:F(ker p) S ker FP. Similarly, if F is right exact, then F preserves epimorphisms and F “preserves cokernels”. It follows that FC z FB/im FA. There are analogous definitions for contravariant functors. Definition: A contravariant functor F is left exact if exactness of A:B5C-+O

implies exactness of o-+FcFB-FBF”.FA;

35

Exactness

a contravariant functor F is right exact if exactness of O+A:B>:C

implies exactness of FC 3F B

FA

4

0.

Thus, a contravariant functor F is left exact if and only if F(coker x ) z ker Fa via FP, and is right exact if and only if F(ker P) z coker FB via Fa. Definition: A functor is exact if it is both left exact and right exact.

Clearly a left exact functor that preserves epimorphisms is exact, as is a right exact functor that preserves monomorphisms. Now that we have the words, we may state the theorems. Theorem 2.9: Hom(M, ) is a left exactfunctor and Hom( ,M ) is a leji exact contravariant functor, for every module M . Proof: We only show F = Hom(M, ) is left exact: the proof for Horn(, M ) is similar. If 0 + A 5 B -% C is exact, we must show exactness of

0 -+ Hom(M, A ) 2Hom(M, B ) 3 Hom(M, C), where F a :f H af and Fp :g H By. (i) Fa is monic. If ( F a ) f = 0, then xf = 0 and af(a) = 0 for all a E A. Since a is monic, f ( a ) = 0 for all a E A, whence f = 0. (ii) imFa c ker FP. Suppose g E im F x , so that g = af for some f E Hom(M, A). Then (FP)g = Pg = Pgf = 0, since Pa = 0. (iii) ker F p c im Fa. Suppose g E Hom(M,B) and pg = 0; we must show g = af for somef: M + A . If rn E M , then Bgtn = 0 and grn E ker P = im a ; hence there is a unique a E A with aci = g m (since r is monic). Definef:M -+ A by f ( m ) = a = a- 'g(rn);clearly af = g. I Examples: Let R = Z, so that we deal with abelian groups, and consider the exact sequence 0 + Z 5 Q -% Q/Z -+ 0. 16. F = Hom(M, ) need not be right exact. Choose M = Z/2Z. Note that Hom(A4,Q) = 0 and Hom(M, Q/Z) # 0, so that

FP:Hom(M,Q) + Hom(M,Q/Z) cannot be epic. 17. F = Horn(, N ) need not be right exact. Choose N = Z. Note that Hom(Q, Z) = 0 and Hom(Z, Z) # 0, so that Fa:Hom(Q, Z) + Hom(Z, Z) cannot be epic. Theorem 2.10: The functors M

OR unrl

R@

N trre right exact ,functors.

36

2

Homand @

Proof: We only give the proof for M @ ; the other proof is similar. If % C + 0 is exact, then we must show M Q A - M1 @@a B % M @ C + O

A5B

is exact.

(i) im 1 @ a c ker 1 @ B. It suffices to prove (1 @ B)(1 Q a) = 0. But (1 @ p)(l @ a) = 1@ j3a = 1 @ 0 = 0. induces a map (ii) ker 1 @ /3 c im 1 @ a. If E = im(1 8 a), then 1 0 /I

B : ( M @ B ) / E + M @ C given by r n @ b + E H r n @ / ? b (for E c k e r l @ B , by (i)). It is easily seen that 1 @ B = Bn, where n:M @ B + ( M @ B)/E is the natural map. Suppose we show is an isomorphism. Then

B

kerl@p=kerBn=kern=E=iml@a and we are done. We construct a map M @ C -+ ( M @ B)/E inverse to The function f:M x C -,( M @ B)/E given by f ( m ,c) = rn Q b

+ E,

B.

where fib = c

is well defined: such an element b exists because B is epic; if pb’ = c = Bb, then B(b’ - b) = 0 and there is an element a E A with aa = b’ - b; it follows that m @ b’ - m @ b = in @ ( b - b) = 1 0 a(m@ a) E im 1 @ a = E . Since f is biadditive, the universal property of tensor product gives a map f:M @ C --t ( M 0 B)/E with f(m Q c) = in Q b + E . Visibly, f and B are inverse functions. (iii) 1 Q B is epic. Let Crni@ ci E M @ C. Since fl is epic, there are bi E B with Bbi = c i , all i. Hence 1 Q P(xmi @ bi) = Emi @ c i . 1 Example:

18. The functor M

OR need not be left exact.

Let R = Z and let M = (x) = 2/22. Exactness of 0 -+ Z + Q -+ Q/Z + 0 does not give exactness of 0 + M @ Z + M @ Q, for M @ Z % M # 0, by Theorem 1.12, while M @ Q = 0 (x @ q = x @ 2q/2 = 2x Q 4/2 = 0 @ 4/2 = 0). Exercises: 2.27. If M is a torsion group (i.e., every element has finite order), prove that M @ Q = 0. 2.28. Prove that QJZ0 Q/Z = 0. ADJOINTS

There. is a remarkable relationship between Hom and @, which has a very simple explanation: every function of two variables f : A x B + C can be regarded as a one-parameter family of functions of one variable fa: B + C,

37

Adjoints

where a E A, merely by fixing the first variable. For technical reasons, see below, fixing the second variable leads to a slightly different description. Theorem 2.11 (Adjoint Isomorphism): For rings R and S , consider the situation ( R A , SBR, J). Then there is an isomorphism T :Horns@ ORA,

c)5 HornR(A, Homs(B, c)).

In the situation (A,, RBS, Cs),there is an isomorphism

t’: Homs(A @R B, C) 7 HOmR(A, Homs(B, C)). Proof: We only discuss the first isomorphism; the second is similar. Since B is a bimodule, Theorem 1.8 shows B @ , A is a left S-module [s(b0 a) = (sb) 0 a ] and Theorem 1.15 shows Homs(B, C) is a left R-module [(rf)(b)= f ( b r ) ] .Thus, both sides make sense. If f : B @ , A + C is an S-map, for each a E A define f , : B + C by f,(b) = b 0 a; one checks easily that a H f, is an R-map A + Horn#, C), which we denote f.One further checks that 7 :f H f is a homomorphism. To see that T is an isomorphism, we exhibit its inverse. Assume g :A -+ Hom,(B, C) is an R-map. Define g ‘ : BORA -+ C by b @ a H g,(b). Again computations are left to the reader: g‘ is a well defined S-map; g H g’ is the inverse of T. I

Theorem 2.11 is the proper context in which to view the relation between R-modules and representations of R. Every abelian group M is a (Z - Z) bimodule, so the second isomorphism above gives t’:Hom,(R OZM , M ) 3 Hom,(R, Hom,(M,M)). If a:R x M -+ M is the scalar multiplication defining the module structure on M , then this Z-biadditive function determines a map a‘: R 0, M + M , and t’:C H ~ ,where p:R + Hom,(M,M) = End,(M) is a representation. The reason the isomorphism of Theorem 2.1 1 is called “adjoint” is quite formal. Let F = B OR and G = Homs(B, ). With thisnotation, Theorem 2.1 1 becomes Hom,(FA, C) 2 HomR(A,GC). If one pretends that Hom( , ) is an inner product, then, in the parlance of linear algebra, F and G are adjoint. We remind the reader of a convenient notation introduced in Chapter 1. If u : A + B is a map and F = Hom(X, ), then let us write Fu = a*; if G = Hom( , X ) , let us write Gu = a*. Thus, lower star is the map induced by the covariant Horn functor, upper star is the map induced by the contravariant Hom functor, u J f ) = af,

and

ol*(g) = gu.

Definition: Let F : 2l+ 6 and G :6 + 2l be functors. The ordered pair ( F , G) is an adjoint pair if, for each A E obj 2l and C E obj 6 there is a bijection 7 = T ~ , Hom,(FA, ~ :

C) + Hom,,(A, C C )

38

2

Hornand8

which is “natural”’ in each variable, i.e., the following diagrams commute: Hom,(FA, C )

IT

Hom,(A, GC) HomJFA, C )

Theorem 2.12: I f B adjoinf pair.

(Ff)*

.f *

Y*

= SBR

+ HomJFA‘, C)

1.

Horn,(”

all f : A ’ + A in%; GC)

HomJFA, C ’ )

is a bimodule, then ( B OR , Horn,(& ) ) is an

Proof: For each A and C, we have constructed the isomorphism 7 = zA,c in Theorem 2.11. It only remains to check naturality, and this is straightforward because there is an explicit formula for every map.

Before we give an application of adjointness, we give a converse of Theorem 2.9.

>

Lemma 2.13: Let B’ B B ‘ + 0 be a sequence (not necessarily exact); suppose that for every module M , we have exactness of

0 + Horn,(#’, M ) 2 HOmR(B, M ) 5 HOmR(B’,M ) . Then the original sequence is exact. Remarks: 1. A dual statement is true if we apply covariant Hom functors to 0 - B + B + B‘. 2. We shall have a better version of this in Lemma 3.51. Proof: (i) b is epic. Let M = coker b = B”/im 3 and let f:B” + M be the natural map. Then b*f = fb = 0, so that f = 0 (since b* is monic). It follows that coker b = 0, whence b is epic. (ii) ima c kerb. We know a*b* = 0. In particular, if M = B‘ and f = lBt,,then a*b*(f) = 0. But a*P*f = f b a = pa, since f = 1,-, and so Pa = 0. (iii) kerb c ima. Choose M = coker a = B/ima, and let g : B + M be the natural map. As before, a*g = ga = 0, so there is f E Horn(#’, M ) with

’ A precise definition of naturality is given later in this chapter.

39

Direct Limits

p*f

= g, i.e.,fB = g. If im OL ker fl, there is an element b E B with pb = 0 and b $ im a;this second fact gives gb # 0. But yh = f p h = 0, a contradiction. I

Theorem 2.14: Let F : + ,‘Dl and G :S\IR +J !.R be functors. Zf ( F , C ) is an adjoint pair, then F is right exact and G is left exact.

- -

Proof: Assume A’+ A +A”+O is exact; we must show FA‘+ F A +FA”+O is exact. Consider the commutative diagram (the naturality condition)

0 + Hom,(FA”, B )

1

0 + HOmR(A”, GB)

Hom,(l-‘A, B )

Hom,(FA’, B )

HOmR(.A, CB)-

Hom,(A‘, GB)

I

I

where B is any S-module. Since HornR(, G B ) is contravariant left exact, the bottom row is exact; since all the downward arrows are isomorphisms, commutativity gives exactness of the top row. As B is arbitrary, Lemma 2.13 applies to give exactness of FA’ + FA + FA“ + 0. The proof of the left exactness of C is similar, using the other naturality condition in the definition of adjoint pair. I Corollary 2.15:

B OR is right exact for any right R-module B.

Proof: We may consider the right R-module B as a bimodule ZBR, so that ( B OR,Homz(B, ) ) is an adjoint pair. I

There are two remarks necessary here. First, there are many other examples of adjoint pairs. Second, if ( F , G ) is an adjoint pair, much more is true than Theorem 2.14. Actually, F preserves all direct limits and G preserves all inverse limits. In order to prove this, and because the constructions are useful, we now investigate limits.

DIRECT LIMITS Definition: Let I be a quasi-ordered set and (5 a category. A direct system in (5 with index set Z is a functor F : I -+ B (of course, we construe I as a category as in Example 7 of Chapter 1). Let us elaborate this definition. For each i E I , there is an object F , and, whenever i, j E I satisfy i 5 j , there is a morphism q$: Fi + Fj such that: (i) cpi:Fi + F iis the identity for every i E I ;

40

2

Homand 8

(ii) if i 5 j Ik, there is a commutative diagram

Remarks: 1. There is no reason to restrict oneself to quasi-ordered sets; one could speak of direct systems with an index category. 2. After our discussion of direct systems and direct limits, we shall consider the dual notions of inverse systems and inverse limits. In the following examples, the category Q; will be a category of modules (and so will not be mentioned). Examples: 19. For any I, fix a module A and set A i= A, all i E I, and ‘p: = 1, for all i lj.This is the constant direct system with index set I, denoted IAl. 20. Let I have the trivial quasi-order : i Ij if and only if i = j . A direct system with index set I is an indexed family of modules {Fi:iE I}. 21. Let I have exactly three elements, I = { 1,2,3}, with quasi-order 1 < 2 and 1 c 3. A direct system with index set I is a diagram

F,

F3

F2

22. Let I be the positive integers with the usual (quasi-) order. A direct system is a sequence A, -,A 2 --f A , -, . . . . 23. If A is a module, then the family of all f.g. submodules of A is quasi-ordered by inclusion. This family together with all possible inclusion maps is a direct system (over itself). 24. If A = UieJA j , then the family of all finite “partial sums” A j , 0 .. . @ Ajn is quasi-ordered by inclusion; this family together with all inclusion maps is a direct system (over itself). 25. Let R be a domain with quotient field Q. The family of all cyclic R-submodules of Q of the form (l/r) is quasi-ordered by (l/r) 5 (l/s) if and only if r 1 s, i.e., rr‘ = s for some r’ E R. This family with inclusions is a direct system (over itself).

41

Direct Limits

Definition: Let F = ( F i ,cp)} be a direct system in 6.The direct limit of this system, denoted @ P i , is an object and a family of morphisms a i : F i + 1 5Fi with ai = a j q f whenever i Ij satisfying the following universal mapping problem:

for every object X and every family of morphisms J : F i + X with J = fjcpj whenever i I j, there is a unique morphism p: 1 5Fi -+ X making the above diagram commute. As usual, 1% F iis unique to isomorphism if it exists. Although 1% F iis the standard notation, we admit it is deficient in that it omits the morphisms cpj. Theorem 2.16:

The direct limit of a direct system of modules (Fi, cpj>exists.

Proof: For each i E I , let Ai:Fi -P u F i be the ith injection into the sum. Define I$ Fi = ( U F J / S , where S is the submodule generated by all elements Ai'p;ai - Aiai, where ai E Fi and i j . If one further defines a, :Fi -P lim Fi by aiF+ Aiai + S, then + he may routinely verify that we have solved the universal problem. 1 Let us now determine the direct limits of the particular direct systems above. 19'. The direct limit of the constant direct system IAl is A . Fi = u F i . There are two 20. If I has the trivial quasi-order, then ways to see this. Since there are no q: with i # j , the universal mapping problem is exactly that of Theorem 2.3. Alternatively, the submodule S in the construction in Theorem 2.16 is just 0, for 'pi is the identity. 21'. If I is the three point quasi-ordered set above, then the direct limit is usually called the pushout. Let us restate the universal mapping

1 5

42

2

Homand 8

problem. Given a pair of maps

C the pushout is a module L and maps a, B such that

.I’

A

C

U

’B

’L

commutes; moreover, given any other pair of maps a’, B’ making the diagram commute, there exists a unique 0 as below making all commute:

I’

A-B

The construction of L as in Theorem 2.16 yields a certain quotient of AOBOC. Exercises: 2.29. Given

A

C

Y

+B

43

Direct Limits

prove that the following diagram is a pushout : A f

B

.

C

61’

+B

+D

+

where D = (C 0B)/W, W = ( ( f a , - g u ) : u E A } , f ’ : b H (0, b) W , and g ’ : c c1 (c, 0) + W . (This is not the construction of Theorem 2.16.) 2.30. In the pushout diagram in Exercise 2.29, if g is monic (or epic), then g‘ is monic (or epic). Moreover, parallel arrows have isomorphic cokernels. 2.31. Assume I is the positive integers and that we have an ascending sequence A , c A , c A , c . . . . Prove that lim Ai. + Ai 2 2.32. Prove every module A 2 I@ ,4’, where A’ ranges over all f.g. submodules of A. 2.33. Prove u A i 2 l@(Ai, 0 . . . @ Ai,), the direct limit of all finite “partial sums.” 2.34. If R is a domain with quotient field Q, then Q 2 l$(l,/r), r # 0, so that Q is a direct limit of cyclic submodules each isomorphic to R.

Ur:

Definition: Let E and F be functors: E : % 23 and F : % + 23.A natural transformation t : E + F is a class of morphisms t,: EA + FA, one for each A E obj 2I,giving commutativity of

EA-EA’

FA

E/

Fl

b

FA’

for every f : A -+ A‘ in %. There is a similar definition if both E and F are contravariant. Exercises: 2.35. (i) The maps t , : A + R ORA defined by a ++ 1 @ a constitute a natural transformation t : 1 -+ R OR , where 1 is the identity functor. (ii) The maps s,:A + HomR(R,A ) defined by a H f, (where f,(r) = ra) constitute a natural transformation s: 1 + HomR(R, ).

44

2 Homand

8

2.36. Fixing two variables in the expression A @ B @ C gives a functor of the remaining variable. Prove that the associativity isomorphism of Exercise 1.10 is natural in each variable. 2.37. If 2I is the category of all vector spaces over a field,then t,: V -+ Y** (second dual) defined by x c-, ( ,x) ( = evaluation at x) constitute a natural transformation 1 -+ **. 2.38. Let (F,G) be an adjoint pair, where F : % + 6 and G:E + a,and let r :Hom(FA, C) + Hom(A, GC) be the natural bijection. Now set C = FA, so that z:Hom(FA,FA)+Hom(A,GFA). Ifonedefines tA = z ( l F A ) : A + G F A , prove that the tA constitute a natural transformation 1+ GF. A similar construction yields a natural transformation FG + 1. 2.39. Let 1 denote the identity functor on R%U. Show that the class of all natural transformations 1 -+ 1 is a ring isomorphic to Z(R), the center of R. 2.40. Show that all functors a+% form a category if we define Hom(F, G) = all natural transformations F + G. (This is not quite correct, for Hom(F, G) may be a class and not a set. To eliminate this difficulty, one usually assumes that is a small category, i.e., the class of all morphisms in 2I is a set.) We continue our discussion of direct systems with index set I. Since I is a small category, Exercise 2.40 shows that all such direct systems form a category, Dir(I). We repeat the definition of a morphism (natural transformation) r:{F,, (pi} + {Gi,IC/:}; t is a family of maps ti:Fi+ Gi making all the following diagrams commute (when i Ij ) : Fi f i Gi

Fi

‘1

’Gi

We claim that l$: Dir(I) -, RYJl is a functor. We already know l@({Fi, (pi}) = ( U F i ) / S . If r :{Fi,cpj} + { Gi, tji} and if lim( + { G i ,@i})= ( uGi)/S’,then define ?:l$ Fi + 1% Gi by C;liai+ S I+ CA;riai+ S’ (where ai E Fi, and li,2: are injections into u F i , U G i respectively). Definition: A quasi-ordered set I is directed if, for each i, j E I, there exists k E I with i I k and j I k. Our first two examples of quasi-ordered sets (trivial and three point) are not directed; the other examples are directed. In particular, a sum u A i is

45

Direct Limits

a direct limit of the Ai ; if one wants the index set directed, however, then u A i must be taken as the direct limit of its finite partial sums. The reason we have introduced directed index sets is twofold: we shall have a simpler description of lim F , ; lim: Dir(1) R!Bl is an exact functor. -+

-+

-+

Theorem 2.17: Let { A i , q j } be a direct system with directed index set I , let A, be the ith injection l i : A i+ H A i , and let lim A , = (UA,)/S (as in + Theorem 2.16). (i) lim A , consists of all liai+ S ; -+ (ii) Aiai S is 0 if and only ij‘ qfai = 0 for some t 2 i.

+

Proof: Let us observe at once that Theorem 2.16 tells us that 1% Ai consists of all elements of the form x = I l i a i + s. Since I is directed, there is an index j 2 i, all i occurring in the sum for x. For each such i, define b’ = qiai E A j , so that b = C b ’ E A j . It follows that

C;liai - l j b = C(liai- Ajbi)= C(Aia,- ljq;ai)E S,

so that x = ‘&$a, + S = Ajb + S ; this proves (i). If Ajq;ai = 0, then liai = Aiai - Ajqfu, E S, so that liui + S = 0. For the converse, assume Aiai + S = 0 (by (i), &ai + S is a typical element of lim + A,), i.e., Aiai E S. Since any scalar multiple of a relator &g’,aj - Ajaj (where j 5 k ) has the same form, there is an expression = x ( A k q ’ , a j - Ajaj) E

s.

j

Choose an index t E I larger than any of the indices in this expression. C1ear1y , A,qfa, = (&;ai - AjU,) = (A,qfa,

+ E.,a,

- Aiai) + x(i.,q’,aj

- Ajaj).

i

We may rewrite each of the terms on the right as a sum of relators in which the lower index is t : A k q i a j - l j a j = (,&q/aj - l j a j )

+ [j.,q:(- q i a j )

- Ak(-qf$j)],

for q$qi = q/, by definition of direct system. Therefore, we may write ;lrq:ai= I ( l , C p j U j - i j U j )

after a harmless change of notation; moreover, we may assume that we have combined all terms having the same top index j (for the sum of such terms has the same form). This last equation relates elements in U A i , where each element has a unique expression of the form x l , a , . We conclude that if

46

2

Homand 8

j # t, then Ajaj = 0, hence a j = 0, for the element il,cpfaihas jth coordinate 0; if j = t, then i,cp:a, - Atat ,= 0 because cp: is the identity. Thus, each term on

the right is 0, so that Atcp:ai = 0, whence cppfai= 0.

I

It is now a simple matter to verify that the following gives an alternative description of direct limit when the index set is directed. First of all, recall the standard set-theoretic construction of disjoint union u A i : given a family of sets {Ai: i E I } , one may pretend no two of them overlap, Ai n A j = 0for i # j , and then take their union. Definition: Let I be a directed index set and let {Ai, cp;} be a direct system over I . If X is the disjoint union jl)Ai, define an equivalence relation on X by

ai

- a.

Ui€

if there exists an index k 2 i, j with is denoted [ai].

Ai,

Uj€

Aj,

viai = v i a j . The equivalence class of ai

If { A i ,cpi} is a direct system over a directed set I , define an R-module L as follows. The elements of L are the equivalence class [ai] defined above, r [ a i ] = [rail

and [ail

if r E R,

+ [ail = C a k +

4

7

where k 2 i, j , ak = viai, and a; = cp&. Using Theorem 2.17, the reader may check that when the index set I is Ai -, L defined by &ai + S H [ai] is an isomorphism. directed, the map

1 5

Theorem 2.18: Let I be a directed quasi-ordered set. Suppose there are morphisms of direct systems over I {Ai,cpj} 1,{Bi& such that is exact for each i

5 {Ci,dj}

0 + A i5 Bi 9 ci-+ 0 E

I . Then there is an exact sequence of modules

o -+

lim -+

5 lim B~ 5 lim ci-+ 0. --* -+

Remark: The hypothesis is the appropriate notion of exactness in Dir(1). Proof: The only possible difficulty (and the only place where one must assume I is directed) is showing i is monk; therefore, we prove this and leave the remainder to the reader (but see the first paragraph following this proof).

47

Direct Limits

Assume x E 1 5Ai and fx = 0 in 1 5Bi. Let us set l$ Ai = (UAi)/S and set l i : A i u A i the ith injection; let us set l$Bi = ( u B i ) / T and set pi:& -+ LIBi the ith injection. Thus, x = iiai S, by Theorem 2.17(i), and f x = pitiai T. Since 2x = 0, Theorem 2. I7(ii) says there is some j 2 i with 3/ftiai = 0. Since t is a morphism between direct systems, we have tjq;ai = 0. But t j is monic, whence qfai = 0, and this gives x = Aiai S = 0. I --f

+

+

+

Here is a sketch of an argument proving that 1 5is right exact (the tedious part of Theorem 2.18 that we left to the reader). “Constant” is a functor I I :$JI -,Dir(Z): to each module A, it assigns the constant direct system with index set I having Ai = A , all i E I ; i f f : A + A’, then lfl:lAl-+(A’( is the natural transformation lfli = f for all i E I. Another way of stating that I$ is a solution to a universal mapping problem is to say there is a natural bijection HomR(l$Ai,C)-+ Hom.ir,,,((A,.qj),ICI)

(it is instructive to give all necessary details). But this just says that ( 1 5 , I I) is an adjoint pair. Therefore lim is right exact, by a simple generalization of Theorem 2.14 replacing module categories by more general categories (e.g., Dir(Z)). Indeed, the next theorem will actually do this.

Theorem 2.19: Let ‘% and 6 be categories, and let F : ’% -+ 6 and G :6 -+ Ql be functors. 1s ( F , G) is an adjoint pair, then F preserves all direct limits (with any, not necessarily directed, index set). Remark: Once we introduce inverse limits, we shall see that G preserves them.

Proof: If I is a quasi-ordered set and { A i , qf)is a direct system in B with index set I, then it is easy to see { F A , ,Fqf) is a direct system in 6 with index set I . Consider the commutative diagram in 6

FA,

48

2

Homand @

where ai:Ai-+ 1% Ai are provided by the definition of 1%. We require a unique morphism y:F(l@ Ai) .+ X making all commute. Since (F,G) is an adjoint pair, there is a natural bijection z:Hom,(Fl@ A i , X )-+ Hom,(l$ A i , CX).Consider now the commutative diagram in %: B lim A:_-___-__---+ GX

Aj By definition of lim, there is a unique B E Hom,(l@Ai, CX) making the -+ diagram commute. Define y E Hom,(F 1 5A i ,X) by y = T - ‘(p). S’ince T (hence T - ’ ) is natural, y makes the original diagram commute. Finally, y must be unique: if y‘ were a second such morphism, then z(y’) would be a Ai -+ CX, contradicting the uniqueness of p. Thus second morphism lim -+ F(1$ A i ) and l@ FA, both solve the same universal mapping problem, and so they are isomorphic. I

Corollary 2.20: For any right R-module B, the functor B BR preserves direct limits. Proof:

As in Corollary 2.15, (BBR ,Hom,(B, )) is an adjoint pair.

I

We now have a second proof of Theorem 2.8 that B gR preserves sums, for sum is a direct limit. We can also reprove Theorem 2.15 that B BR is right exact, i.e., B BR preserves cokernels. It only remains to show that cokernel is a pushout, for pushout is a direct limit. Exercises: 2.41. Prove that iff:A diagram

A 0

0

-+

1

B, then cokerf is the pushout of the ,B

49

Inverse Limits

2.42. If F is a right exact functor that preserves sums and if F is additive (so that F(0) = 0), prove that F preserves direct limits. (Hint: Regard a direct limit as a cokernel of a suitable map between sums.) 2.43. A subset K of a directed set I is cofinal if, for each i E I , there i E I , (pi} is a direct system with index set I , exists k E K with i 5 k. If {Ai, then (Ai, i E K , (pi} is a direct system with index set K ; moreover, the direct limits of these two systems are isomorphic. 2.44. If ( A , , q ; } is a direct system with index set I , and if I has a top element, 00 (i.e., co E I , i 5 co for all i E 1. and cc is the unique such element), then 1% Ai r A , . (Hint: 1 is automatically directed.) 2.45. Show that Exercise 2.43 may be false if the index set is not directed (Hint: Pushout.) The reader may wonder why we were not content to prove Theorem 2.19 for module categories. Aside from the fact that the proof of the more general version given is the same as its particular case, there is a nice application to modules. Theorem 2.21 : Any two direct limits (perhaps with distinct index sets) commute.

(IF,I)

1 is an adjoint pair of functors Proof: We have already noted that (of course, the domains and ranges of the functors are not module categories). Theorem 2.19 applies to show I& preserves all direct limits. I We have refrained from giving precise notation for Theorem 2.21, for it would make a simple result appear complicated. We only make two remarks to aid the reader. First, if I and K are quasi-ordered sets, then the modules in the direct system should be doubly indexed Fi,. Second, here is a modest example of Theorem 2.21. If A , c A , c A, c . . . is a sequence of submodules of A, then the cokernel of U A , (namely, A/&&) is isomorphic to the “union” of the cokernels l@ A / A i .

INVERSE LIMITS Let us now consider inverse limits, which will appear as the dual of direct limits. Since all the proofs of the next theorems are dual to the ones just given, we leave the details to the reader’s mirror.

50

2 HornandB

Definition: Let I be a quasi-ordered set and 6 a category. An inverse system in 6 with index set I is a contravariant functor F : I -+6 (as usual, regard I as a category). In more detail, for each i E I there is an object F iand whenever i 5 j there is a morphism $!:Fj F , such that -+

(i) $j:Fi Fi is the identity for every i E I ; (ii) if i I j I k, there is a commutative diagram -+

In each of the following examples, the category 6 is a category of modules (and so will not be mentioned). 26. For any I , the constant direct system / A \with index set I (where A is a module) is also an inverse system with index set I. 27. If I has the trivial quasi-order, then an inverse system with index set I is a family of modules {Ai:i E I} (so again this is the same as the corresponding direct system over I ) . 28. If I is the three point quasi-ordered set we have been considering, an inverse system over I is a diagram

29. If I is the positive integers with the usual (quasi-) order, then an inverse system over I is a sequence A,

+ A , + A , +-. . *

.

Definition: Let F = {Fi, $!} be an inverse system in 6. The inverse limit of this system, denoted lim Fi,is an object and a family of morphisms +. cli: I@ F i.+ Fiwith ai= $+xi whenever i Ij satisfying the following universal

51

Inverse Limits

mapping problem :

for every object X and morphisms A: X + F , making the diagram commute whenever i I j , there is a unique morphism 8 : X + Fi making the diagram commute.

12

As usual 1% F iis unique to isomorphism if it exists. Although lim Fi is c: the standard notation, it is deficient in that it omits the morphisms $1.

Theorem 2.22: The inverse limit of an irwerse system of modules {Fi,${} exists. Proof: Since direct limit is a quotient of a sum, the dual notion of inverse limit should be a submodule of a product. For each i E I , let pi be the ith projection p i : n F i + F i . Define limF, = {(ai)E n F , : a i = ${iij, whenever i I j } . c If one further defines cci:l$ F i+ Fi as the restriction pillipFi, then he may routinely verify that he has solved the universal problem.

1

Let us now determine the inverse limits of some particular inverse systems. 26’. The inverse limit of the constant system IAl is A . 9Fi = n F i . There are two 27‘. If 1 has the trivial quasi-order. then 1 ways to see this. Since there are no ${ with ,j f i, the universal mapping problem is exactly that of Theorem 2.5. Alternatively, the submodule of n F i is everything, for there are no constraints on the coordinates. 28’. If I is the three point quasi-ordered set above, then the inverse limit is usually called the pullback (or fiber product). Let us restate the

52

2

HornandB

universal mapping problem. Given a pair of maps C

B-A

f

the pullback is a module L and maps a, such that L-C

commutes; moreover, given any other pair of maps a’, making the diagram commute, there exists a unique 6 as below making all commute:

The construction of L as in Theorem 2.22 yields a certain submodule of A@B@C. Exercises: 2.46. Given

,Ig C

B

s

A

53

Inverse Limits

prove that the following diagram is a pullback:

D

B

Ig

a

'C

'A

.f

where D = {(b,c) E B 0 C:fb= gc}, a : (b,c) I-+ c and /?:(b,c) H b. (This is not the construction of Theorem 2.22.) 2.47. In the pullback diagram in Exercise 2.46, if g is monk (or epic), then /? is monic (or epic). Moreover, parallel arrows have isomorphic kernels. 2.48. In the pullback diagram of Exercise 2.46, if both f and g are monic, then we may identify D with B n C . Moreover, in this case, there is a pushout diagram .A €3

A

A/C

A/(B

2.49. I f f : B -,A, then

B

J

+C)

,A

is a pullback diagram. (Thus, cokernel is a pushout, hence a direct limit, while kernel is a pullback, hence an inverse limit.) 2.50. Let { A i : iE I } be a family of submodules of a module M . If one quasi-orders { A i:i E I } by reverse inclusion, then this family and the various inclusion maps form an inverse system, and l p A i2 E, Ai. 2.51. If K is a cofinal subset of a directed quasi-ordered set, and if {Ai, is an inverse system with index set I , then l@ A iis isomorphic to the inverse limit of the corresponding system with index set K (see Exercise 2.43). 2.52. If {Ai,+{}is an inverse system with index set I , and if I has a top element 00, then l@ Aiz A , .

ni

54

2

HornandB

2.53. Show that Exercise 2.5 1 may be false if the index set is not directed. (Hint: Pullback.)

It would be a sin not to mention completions. If I is an ideal in a commutative ring R, then I" = { C U l U 2 . . . a,:ai E Z } ;

-

it is easy to see I" is an ideal and that I = I' 3 I' 3 * . . If M is an R-module, then M 3 I M =I I'M =I . . . . The family of quotient modules M I I ' M , i = 1,2,3, . . . ,and maps $!: M / I i M + M / I ' M , for j 2 i, given by x ijM H x + I ' M , is an inverse system indexed by the positive integers.

+

Definition: The I-adic completion of M , denoted A,is I@ MII'M.

Consider the diagram

where x i is the natural map, and p is provided by the definition of l i p It is a fact that p is monic if ( ) I ' M = 0; it is another fact that if M = R, then i? is a commutative ring and every I-adic completion is an i?-module. There is a more down-to-earth description of fi (we leave details to the interested reader). Assume n Z ' M = 0. If x E M , and x # 0, there exists i with x E I'M but x $ I i f l M ;define llxll = 2-'; if x = 0, define JJxII = 0. It is easy to see that IIx - yII is a metric on M . It turns out that M is the completion of M in the sense of metric spaces ( M is a dense subspace of fi and every Cauchy sequence in M converges in fi). Given two contravariant functors % one -may + define (I, a natural transformation between them (no surprises). Let us only give the special instance of a morphism between inverse systems. Definition: If I is a quasi-ordered set, a morphism t : { F i , $ ! ] + {Gi,cp!} between inverse systems over I is a family of maps ti:Fi+ G imaking the

55

Inverse Limits

following diagrams commute when i I j : Fj

I.

Gj

It is easy to see that all inverse systems with index set I and their morphisms form a category, Inv(Z). We claim that 15:Inv(Z) + R(m is a functor. I)!} + It only remains to define t:l@ F , -, I@ G, for a morphism t : (Fi,

{C,,cpi), and this is done by f : ( a , )H ( C l d l ) .

We have already noted that we may regard I I:,IIJZ -+ Inv(Z), for the constant system is, at our pleasure, either a direct system or an inverse system. Theorem 2.23:

(I 1, lim) is an adjoint pair of functors. t

Proof: Dual to the argument that (I$,

I I) is an adjoint pair. I

Theorem 2.24: Let 2I and CS. be curegories. trnd 1t.r F : 2I + 0.and G :E + ‘.Lf be functors. If ( F , G) is an adjoint pair, then G prrserws all inverse limits. Proof: Dual to the proof of Theorem 2.19. Corollary 2.25: limits.

Zf B is a left R-module,

tken

1 Hom,(B, ) preserves inverse

Proof: We may regard B as a bimodule *BZ,so Theorem 2.12 gives ( B OZ ,Horn,(& ) ) an adjoint pair. 1

There is another way to prove Corollary 2.25, dual to Exercise 2.42. Exercises: 2.54. If F is a left exact functor that preserves products, then F preserves inverse limits. (Hint: Regard an inverse limit as the kernel of a suitable map between products.) 2.55. If K and L are submodules of M , then Hom(B, K ) and Hom(B, L ) may be regarded as subgroups of Hom(B. M ) and

Hom(B, K n L ) z Hom(B. K ) n Hom(B, L).

(Hint: Use pullback.)

56

2

Hornandm

2.56. I@ is left exact (in the sense of Theorem 2.18). One need not assume the index set is directed. Theorem 2.26: Any two inverse limits (perhaps with distinct index sets) commute. Proof: Dual to the proof of Theorem 2.21.

I

9Hom(Aj,B). Theorem 2.27: For any module B,H o m ( l 5 A j , B)z 1 Proof: Use the analogue of Exercise 2.54: if F is a left exact contravariant functor converting sums to products, then F converts direct limits to inverse limits. I

One last remark about limits: usually, direct limit and inverse limit do not commute. We give an easy example which the reader may formalize. Construct an example of three subgroups A, B, C of an abelian group for which A n ( B e C ) # ( A n B ) O ( A n C),

and note that intersection is an inverse limit (even a pullback in this case) and sum is a direct limit. A more interesting example is that the completion of a product is the product of the completions, but this is not so if one replaces “product” by “sum”.

3

Projectives, Injectives, and Flats

FREE MODULES We begin by generalizing the notion of free abelian group to modules. Definition: A left R-module A is free if it is a sum of copies of R. If Rai z R Rai, then the set {ai:i E Z> is called a basis of A . and A =

uipl

From the definition of sum as “vectors” almost all of whose coordinates are 0, it follows at once that given a basis {ai:iE Z}, each a E A has a unique expression a = Criai, where ri E R and almost all ri = 0.

Theorem 3.1: Let X = {ai:iE I} be a busis of a free module A. Gioen any module B and any function f :X -+ B, there is a unique mapE A -,B extendingf.

Proof: For fixed i E I, define f;.:Rai Theorem 2.3 gives a unique map f :A

-+

B by rai H rf(ai). Since A = URai, with Tai = Lai = fai, all i. I

-,B

Note that it is easy to describe J/ explicitly: f:Criai H

xii

f(ai).

57

58

3 Projectives, Injectives, and Flats

Theorem 3.2: I f X is a set, there exists a free module A having X as a basis. Proof: Exactly as Theorem 1.2 (with the advantage that all necessary constructions are already available). I

Theorem 3.3: Every module M is a quotient of a free module. Proof: Let UM be the underlying set of M (forget addition and scalar multiplication), and let A be the free module with basis UM. Define f : UM + M by m H m. By Theorem 3.1, there is a map f’:A + M extending f , and T is epic because f is onto. I

Of course, the free module with basis UM may be extravagantly large; there may be many, smaller, free modules mapping onto M. In down-toearth language, Theorem 3.3 says that A4 may be described by generators and relations: if A is free with basis X and f : A -+ M is epic, then X is called a set of generators of M and kerf is called its submodule of relations. This definition conflicts with our previous notion of generators of M ; the earlier usage (which we shall maintain) says M is generated by f ( X ) . Exercises: 3.1. Let U : , m + Sets be the “forgetful” functor: regard every module as a mere set and every map as a mere function. Let F : Sets -,m , be the “free functor”: F X is the free module with basis X ; if f : X + Y is a function, Ff = 7 defined as in Theorem 3.1. Prove that U and F are functors. 3.2. Prove that ( F , U ) is an adjoint pair. (This really says that modulevalued functions on a set X may be identified with homomorphisms from the free module with basis X . )

Before pursuing generators and relations further, let us consider an obvious question. Do every two bases of a free module have the same cardinal? We do know this is so for left R-modules when R is a field or R is a division ring. Definition: A ring R has IBN (invariant basis number) if, for every free left R-module A, every two bases of A have the same cardinal. In this case, rank A is defined as the cardinal of some basis of A. Theorem 3.4: Every commutative ring R has IBN. Proof: By Zorn’s lemma, R has a maximal ideal M, whence RIM is a field. Let A be free with basis {ai:i E I } . By Exercise 2.4, the quotient module A/MA is an RIM-module (being annihilated by M), i.e., A/MA is a vector space over RIM. Since MA = UMa,. we see that AIMA z U(Rai/Mai). Since Rai/Mai 2 R/M, it follows that A/MA has dimension card(Z).If { b j : jE J } is a

59

Free Modules

second basis of A, the same argument gives dim A / M A = card(J). Since fields have IBN, we conclude that card(Z) = card(J). I Exercises: 3.3. A ring R has IBN if and only if, for every f.g. free left R-module A, every two bases of A have the same cardinal. 3.4. If R has IBN and cp:S -+ R is a ring map (by definition, cp(1) = l), then S has IBN. (Hint: If A is a free left S-module, consider R OSA.) 3.5. Let k be a field and V an infinite-dimensional vector space over k. Prove that R = End,(V) does not have IBN. (Hint: Show that, as left Rmodules, R 2 R 0 R by applying Horn,( V , ) to an isomorphism V 2 V 0 V.) 3.6. Assume R has IBN, and there is an exact sequence

0 +F ,

+ F,-

1

+. . . + F ,

-+

0,

where each Fi is a f.g. free left R-module. Prove that

c:=,

( - 1)' rank Fi = 0.

In the next chapter, we shall introduce a large class of (not necessarily commutative) rings having IBN: left noetherian rings. It is fair to say that most rings one encounters do have IBN. Here is a standard use of determinants. Lemma 3.5: Let S be commututii~eand let M he LZ f g . S-module. I f M f o r some ideal J of S, then (1 - a ) M = 0 Jor some u E J.

=

JM

Proof: If {xl, . . . , x,,} generates M , then each xi = I a i j x j ,where aij E J. In terms of matrices, these equations become B X = 0, where X is the column vector of the xi's and B is the n x n matrix B = 1 - ( u , ~(ofcourse, ) I is the n x n identity matrix). Let B* be the classical adjoint of B, i.e., the matrix of cofactors, so that B*B = (det B)I. Then B*BX = (det B)X and B*BX = 0 (since B X = 0). We conclude that (det B ) s , = 0 for all i, hence (det B)M = 0. But obviously det B = 1 - a for (I E J, because each aij E J . I

Theorem 3.6: Let R be commufutire and /er M be a $g. R-module. If f i : M + M is epic, then B is an isomorphism. Proof: Let S = R[x], the polynomial ring in one variable x, and make M into an S-module by defining (r,xn

+ . . . + r I x + r,)m

= r,ljOnt

+ . . . + r , fim + ram,

where fl is the composite of p with itself i times. If J = Sx, then j? epic implies J M = M . By Lemma 3.5, (1 - u)M = 0 for some u E J. Since a = rip, where ri E R, it follows that y = r , / Y is fi-' (for fi and y commute, and 1 - y f i = 0). I

c:=

'

60

3 Projectives, Injectives, and Flats

Corollary 3.7: Let R be commutative and let A be a free R-module of jinite rank n. If { a l , . . . ,a,} generates A, then it is a basis of A. Proof: Let Fbefreewith basis { x l , . . . ,x,} anddefinecp: F + A by xi I+ ai; it suffices to prove cp is an isomorphism. Now cp is epic because A is generated by { a l , . . . , a , } . There is thus an exact sequence

0 +K

+F

3A

+ 0,

where K = kercp. Since A is free, hence projective, this sequence splits. Therefore, F = K @ A’, where A’ z A z F (both A and F are free of rank n). Define P : F .+ F as the composite F 1: A’ + F , where p is the projection having kernel K . Clearly P is epic with kernel K . By Theorem 3.6, K = 0 and cp: F + A is an isomorphism. 1 In the next chapter, we shall see that Corollary 3.7 holds when R is left noetherian. Indeed, it is not difficult to see that a ring R satisfying Corollary 3.7 must have IBN (where one replaces “ A of rank n” by “ A a sum of n copies of R”). The converse is false [P. M. Cohn, 19661. Let us return to generators and relations. In the next chapter we shall prove that every subgroup S of a free abelian group F is itself free; moreover, rank S 5 rank F . If a f.g. abelian group G is given as F/S, where F is f.g. free (and Theorem 3.3 says G can always be so given), then one may use the freeness of S to obtain information about G. The “Simultaneous Basis Theorem” [Fuchs, 1970, p. 781 states that if {xl,. . . , x,} is a basis of F, then there is a basis { y l , . . . , y,,,} of S , m 5 n, and integers { k l , . . . ,k,} such that kixi = yi, all i, and each ki divides ki + 1. Of course, the Simultaneous Basis Theorem implies that G is a sum of cyclic groups (in which the order of any summand divides the order of the next summand). If one considers R-modules instead of Z-modules (abelian groups), then submodules of free modules need not be free; also, f.g. R-modules need not be sums of cyclic modules. Rather than surrender, we iterate the process of generators and relations. Definition: A free resolution of a module M is an exact sequence . . . + F , + dn F,-l

4“’ -+

Fl

3 F , 5 M -+O

in which each F , is a free module. Theorem 3.8: Every module M has a free resolution. Proof: By Theorem 3.3, there is a free module F , and an exact sequence O+ So -+F , 5 M

+ 0.

61

Free Modules

Similarly, there is a free module F, and an exact sequence O+S,

+

F , +So+O,

and, by induction, a free module F , and an exact sequence O+S,-tF,+S,-l+O.

\/\A/

Assemble all these sequences into the diagram -+ F

...

~

d

A

. . . 0/s2\

F,

0

d2

F,

L F~ E M + ti

0

0

where the maps d, are just the indicated composites. For every n, ker d, = S , and im d, = S,- 1. Hence im d,, = ker d, and the top row is exact. I The diagram above makes it plain how to decompose a long exact sequence into short exact sequences. We emphasize that a module has many free resolutions. Let us return to free modules. Theorem 3.9: Consider the diagram with fl epic: F

If F is free and

a : F + C is any map, then there exists y:F + B with tl = by.

Remark: We do not assert y is unique.

Proof: Let X = {xi:i E I} be a basis of F. Since fi is epic, each axi can be lifted, i.e., there is an element bi E B with fibi = axi. The axiom of choice provides a function cp:X + B with cpxi = bi,all i E I. By Theorem 3.1, there is a map y:F B with yxi = q x i , all i. To see that a = fly, it suffices to check each on X,and by(xi) = &(xi) = ahi = a x i . I Corollary 3.10:

If F is free, then thefunctor Hom(F, ) is exact.

62

3 Projectives, Injectives, and Flats

Proof: Since Hom(F, ) is left exact (Theorem 2.9), we need only show that it preserves epimorphisms. If B -+B C +. 0 is exact, then we must show Hom(F, B) 2 Hom(F, C)+0 is exact, i.e., iff€ Hom(F, C ) ,then f = P,(g) = Pg for some g E Hom(F, B). But this is precisely the situation of Theorem 3.9: F

PROJECTIVE MODULES Definition: A module P is projective if it behaves as F does in Theorem 3.9 :

P

Theorem 3.11 : A module P is projective

if and only if Hom(P, ) is exact.

Proof: The proof of Corollary 3.10 shows that Hom(P, ) is exact if P is projective. Assume, conversely, that Hom(P, ) is exact. Consider the diagram

P

B-C-0

4

P

Since p,: Hom(F,B) + Hom(F, C ) is epic, there is g E Hom(F, B) with f = P,(g) = Pg; this says that P is projective.

Is every projective module free? Let us first gather a bit more information. Theorem 3.12: If P is projective and where P' z P .

P:B +P

is epic, then B = ker P @ P ,

63

Projective Modules

Proof: Consider the diagram

P

-.

Since P is projective, there is a map )’:P B with fly y is monic, and so Theorem 2.7 gives the theorem. I

=

1,. By set theory,

It is useful to restate Theorem 3.12 in two ways Corollary 3.13: I f A is a submodule of B with BIA projective, then A is a summand of B. Every short exact sequence 0 -+ A + B + P + 0, with P projective, is split.

Theorem 3.14: A module P is projective If and only if it is a summand of a free module. Moreover, every summcmd of c i projective is projective. Proof: By Theorem 3.3, there is a free module F and an epimorphism F -+ P . By Theorem 3.12, P is (isomorphic to) a summand of F. For the converse, we show that any summand of a projective module F is itself projective. There are maps i : P -+ F and p : F + P with p i = 1,. Consider the diagram

F-P

P

I I

I

., I

{ I I I

.I

‘

i.

B-C-0 Ir Since F is projective, there is a map y : F -, B with B y = f p . Define g : P by g = iji. Since pg = pyi = fpi = f , it follows that P is projective. I

+B

We can now give an example of a ring R for which projectives may not be free. If R = 2/62, then R = 2 / 2 2 02/32. so that 2 / 2 2 is projective, being a summand of the free module R . But 2 / 2 2 is not free, for every nonzero free R-module has at least 6 elements. We shall consider the question when R-projectives are free in the next chapter. The next theorem is often useful; it characterizes projectives by means of “linear functionals” that act like coordinate functions.

64

3 Projectives, Injectives, and Flats

Theorem 3.15 (Projective Basis): A module A is projective if and only if there exist elements { a k : kE K } c A and R-maps { q k : A+ R:k E K} such that

(i) if x E A , then almost all q k X = 0; (ii) i f x E A , then x = xk&((PkX)ak. Moreover, A is then generated by { a k : kE K } . Proof: Assume A is projective, and let $ : F A be an epimorphism from some free module F. By Theorem 3.12, there is a map q :A -+ F with $q = 1, . Let { ek:k E K } be a basis of F. If x E A, then rpx has a unique expression qx = xrkek,

where rk E R and almost all r k = 0. Define q k : A -+R by (pkx = rk. Visibly ( P k X = 0 for almost all k. If we define ak = $ek, then $ epic implies {ak:k E K } generates A . Moreover, if x E A, then

Conversely, assume the existence of {ak:kE K ) and { q k : A+ R:k E K). Let F be a free module with basis ( e k : kE K }, and define a map $:F + A by ek H a k . It suffices to exhibit a map q : A + F with $q = 1,. Define q:A + F by x H x(qkx)ek; this sum is finite, by condition (i), so q is well defined. By condition (ii), $px = $ c ( q k x ) e k

=x(qkx)$ek

= x ( q k x ) a k = x.

Therefore, $q = 1,. Exercises: 3.7. A module P is projective if and only if every short exact sequence 0 + A + B + P + 0 splits. 3.8. If { P j : j E J } is a family of projectivemodules, then U P j is projective. (We shall see in the next chapter that a product of projectives need not be projective.) 3.9. If R is commutative and P and Q are projective R-modules, then P ORQ is a projective R-module. 3.10. Assume R and S are rings, with S an (S-R) bimodule. If P is a projective R-module, then S ORP is a projective S-module. 3.11. Assume P is a projective left R-module and I is a two-sided ideal in R. Prove that P / l P is a projective left R/l-module. 3.12. A complement of a projective module P is a module Q (necessarily projective) with P O Q free. Prove that every f.g. projective has a f.g. complement. 3.13 (Eilenberg) Every projective module P has a free complement. (Hint: First, prove this when R = 2 / 6 2 and P = 2/22.)

65

Injective Modules

INJECTIVE MODULES We now dualize the notion of projective. Definition : A module E is injective if, for every module B and every submodule A of B, every f : A -,E can be extended to a map g : B -,E. The diagram is E

0 There are no obvious examples of injective modules, but we shall see that they exist in abundance. Let us derive their first properties; one should expect duals of the first properties of projectives. Theorem 3.16: is exact.

if the functor Hom( ,E )

A module E is injective zf and only

.

Proof. Assume E is injective. Since Hom( E ) is contravariant left exact, it suffices to show it converts monomorphisms to epimorphisms: if a : A -+B is monic, then a*:Hom(B, E ) + Hom(A, E ) is epic. Therefore, we must show that if f E Hom(A, E ) , there exists g E Hom(B, E ) with a*(g) = f . Since cc*(g) = gcc, this is precisely the definition of injectivity of E . The converse is just as easy and is left to the reader. I

Theorem 3.17: injective.

If { E j :j

EJ

} is a famil). of injective modules, then n E j is

Proof: Let Aj and p j be the injections and projections of the product n E j . Consider the diagram

IlEj

T

+Ei PI

I

0-A-B Since E j is injective, there is a map q j : B+ E j with gjcc = p jf . Define h : B + n E j by b H (gjb).Then

haa so that ha

=f

= (gjaa) = ( p j f u ) = f a ,

and n E j is injective.

I

66

3 Projectives, Injectives, and Flats

We shall see in the next chapter that a sum of injectives need not be injective. Theorem 3.18: Every summand D of an injective module E is itself injective. Proof: Consider the diagram

0-A-B

where I and p are injection and projection, respectively. Since E is injective, there is a map g : B + E with y a = If. Define h : B + D by h = p g . Then ha

= pga = p

since p i = 1,. Thus, D is injective.

i . =f

I

Theorem 3.19: A module E is injective if and only if every short exact sequence 0 + E 5 B + C 0 splits. I n particular, E is a summand of B. --f

Proof: Assume E is injective; consider the diagram E

There exists a map g: B .+ E with g i = l,, so the sequence splits. For the converse, consider the diagram

E

0 - M ' y M

Injective Modules

Construct the pushout diagram

E-P

I’

O-M‘-M By Exercise 2.30, the map E‘:E P is monic. By hypothesis, there is a map p : P + E with per' = 1,. Define g : M + E by g = /3f‘. Then --f

gE = Pf’a = p r y = j-,

and so E is injective.

I

Now Theorem 3.19 is the dual of Theorem 3.14. If one compares the proofs, one sees that we could give a second proof of Theorem 3.19 if we knew the dual of the statement “Every module is a quotient of a projective” were true. The dual, “Every module can be imbedded in an injective,” is, indeed, true; it is an important result (being the dual of “generators and relations”) and we now begin working toward it. Theorem 3.20 (Baer Criterion): An R-niotlde E is injective if and only w e r y inup j : I + E, where I is (1 left ideal of R , ciin he extended t o R.

if

Proof: Suppose E is injective. As a left ideal is a submodule of R, the hypothesis is just a special case of the definition of injective, Suppose we have the diagram

E

0-A-B For notational convenience, let us assume I is an inclusion map. We approximate a map g : B + E by looking at all modules between A and B that do possess an extension of f . More precisely. let Q consist of all pairs ( A ’ ,g’), where A c A‘ c B and g’:A’+ E extends f . Note that (5 # 121,for ( A , f )E (5. Partially order Q by saying (A’,g’)I (A”,q’‘)if A’ c A” and g” extends g’. By Zorn’s lemma, there is a maximal pair ( Ao , go) in (5. If A, = B, we are done.

68

3 F’rojectives,Injectives, and Flats

Suppose A , # B and let x E B - A,. If I = { r E R : r x E A o ) , then I is a left ideal in R . Define h:I -+ E by h(r) = go(rx). By hypothesis, there is a map h’:R -+ E extending h. Define A, = A , + Rx, and define g l : A , -+ E by a, + IX H goao + r!z’(l), where r E R. First, g 1 is well defined: if a, + rx = ab + r’x, then (r - r’)x = ab - a, E A, and r - r‘ E I . Therefore, go( ( r - r’)x) and h(r - r’) are defined, and we have go(ab - a,)

= go( ( r - r’)x) = h(r -

r’) = h’(r - r’) = ( r - r’)h’(l).

+

Thus, go(ab) - go(ao)= rh’(1) - fh’(1) and go@;) + r‘h’f1) = gO(a,) rh’( 1). Second, g 1 does extend g o , for glao = goao for all a, E A,. The pair ( A , , g , ) lies in (3 and is larger than the maximal pair (A,,g,), a contradiction. Therefore, A, = B and E is injective. I The following diagram lemma will be useful.

Lemma 3.21 : The diagram with exact row

E can be completed to a commutatiue diagram with exact rows

O

-

+

A

a

B

BB c

+

O

in which t h e j r s t square is a pushout. Proof: Augment the given diagram by forming the pushout of the arrows emanating from A; note that a’ is monic, by Exercise 2.30. Recall (Exercise 2.29) that we may assume P = (E 0 B)/W, where W = {(ya, -aa):a E A}, y’:bH(O,b) W,anda’:ew(e,O) W.Definefi’:P+Cby(e,b) WH/3b. We let the reader check that 8’ is well defined, the diagram commutes, and the bottom row is exact. I

+

+

+

69

Injective Modules

Theorem 3.22: A module E is injective if und only if every short exact sequence 0 -+ E + B + C -+ 0, with C cyclic, splits. Proof: If E is injective, Theorem 3.19 says the sequence splits for every (not necessarily cyclic) module C. For the converse, let Z be a left ideal of R and let f:Z + E be an R-map. By Lemma 3.21, there is a commutative diagram with exact rows

1 fl

O-I-R-RII-o

j1

O-EI~,P-R/Z-O

Since R/Z is cyclic, our hypothesis is that the bottom row splits. There is thus a map s : P + E with sa' = 1,. If we define g: R -+ E by g = sf', then it is easy to see gi = f . By Theorem 3.20, E is injective. I Definition: Let M be an R-module, m E M , and r E R. We say m is divisible by r if rm' = rn for some m' E M ; we say the module M is divisible if each m E M is divisible by every non-zero-divisor' r E R (i.e., there is no nonzero s E R with sr = 0). Example: 1. The additive group of rationals, Q, is a divisible abelian group ( = Z-module). Theorem 3.23: Every injective module E is divisible. Proof: Let m E E and let ro E R be a non-zero-divisor. Define f :Rr, + E by rr, = rm; note that f is well defined because r is not a zero divisor. Since E is injective, there is a map g: R + E extending f . In particular,

rn = f ( r o )= g(ro) = rogfl), so that m is divisible by r , .

I

Exercises: 3.14. Every quotient of a divisible module is divisible. 3.15. Every summand of a divisible module is divisible. 3.16. Every product of divisible modules is divisible. 3.17. The sum of divisible modules is divisible.

'

It is clear that r must be restricted: for example. we must exclude r Theorem 3.23 shows why zero divisors are excluded

=

0. The proof of

70

3 Projectives, Injectives, and Flats

3.18. Let R be a domain with quotient field Q. Prove that Q is divisible. 3.19. A module M over a domain R is torsion-free if rm = 0 implies r = 0 or m = 0 (where r E R and m E M ) . Prove that a torsion-free module M is injective if and only if M is divisible. 3.20. Let Q be the quotient field of a domain R. Prove that a torsion-free module is divisible if and only if it is a vector space over Q. It is obvious that divisible modules share many properties of injectives; however, the converse of Theorem 3.23 only holds for a restricted class of rings. Theorem 3.24: I f R is a principal ideal domain, then an R-module D is divisible if and only if it is injective. Proof: By Theorem 3.20, it suffices to extend every map , f : I -,D to R, where I is an ideal. Since R is a PID, we know I = Rr,; clearly we may assume r , # 0, and thus ro is not a zero divisor. Since D is divisible, there is an element d E D with rod = f ( r o ) . Define g : R + D by r H rd, and note that g extends f . The converse is Theorem 3.23. I

We finally have some examples of injective modules, at least for principal ideal domains. Here is a special case of the imbedding theorem we seek. Theorem 3.25: Everjl abelian group G cun be imbedded in an injective ubelian group. Proof: Write G = F / S , where F is a free abelian group; thus, F = LIZ. If we imbed each copy of Z in a copy of the rationals, Q, then we have

G

= F / S = (uZ)/Sc

(UQ)/S.

Since Q is divisible, so are HQ and (UQ)/S, by Exercises 3.17 and 3.14. By Theorem 3.24, (HQ)/S is injective as a Z-module. I Theorem 3.26: I f D is a divisible abelian group, then Hom,(R,D) is an injective left R-module. Proof: First of all, R is a bimodule: R = zR,; by Theorem 1.15(i), Hom,(R, D) is a left R-module via uf: r' H f(r'r). We shall show that the contravariant functor Horn,( ,Hom,(R, D ) ) is exact; it is only necessary to show it converts monomorphisms to epimorphisms. The adjoint isomorphism (Theorems 2.1 1 and 2.12) gives a

71

Injective Modules

commutative diagram whenever 0 -, A Hom,(B, Hom,(R, D))-Hom,(A,

i

Hom,(B, D )

-+

B is an R-monomorphism:

Hom,(R, D ) )

*I

Horn,(A, D )

(we have identified R ORBwith B and R OR.4 with A). Since D is divisible, it is Z-injective, and so the bottom arrow is epic, by Theorem 3.16. It follows that the top arrow is also epic (by commutativity and the fact that the vertical arrows are isomorphisms). I Theorem 3.27: Every left R-module M c ~ l nhe imbedded in an injective module. Proof: lfwe regard M only as an abelian group, there is a Z-monomorphism 0 -+ M 5 D for some divisible group D (Theorem 3.25). If m E M , define f,:R -,M by r H rm. It is easy to see that cp: M --f Hom,(R,D) given by m H if, is an R-map that is monic. I

Exercises: 3.21. Fix a module E . Use Theorem 3.27 to prove that if 0 E + B -+ C + 0 always splits, then E is injective. 3.22. IfS is commutative, D an injective S-module, and R is an S-algebra, then Hom,(R, D) is an injective left R-module. 3.23 Assume R is a left principal ideal domain (i.e., R has no zero divisors and every left ideal is principal). Prove that every divisible left R-module is injective. -+

We now dualize the notion of generators and relations. Definition: An injective resolution of a module M is an exact sequence 0 -, M

-, E O -,E' -, . . . + E" -+ E " + I

-+

..

in which each E" is injective. Theorem 3.28: Every module M has an injective resolution. Proof: Dual to the proof of Theorem 3.8. using Theorem 3.27.

I

We may also define a projective resolution of a module M in the obvious way; since free modules are projective, free resolutions are a special kind of projective resolution. Therefore, every module has projective resolutions. A natural question is whether there is a "smallest" injective containing a given module. The affirmative answer was first discovered by Baer (who

72

3 Projectives, Injectives, and Flats

was the first to discover injectives), and the characterization below was first given by Eckmann and Schopf. Our exposition follows that of Lambek [1966]. Definition: An essential extension of a module M is a module E containing M such that every nonzero submodule of E meets M (i.e., if S c M and S # 0, then S n M # 0); if, in addition, M 5 E, we say E is a proper essential extension of M. The additive group of rationals, Q, is an essential extension of the integers, Z; indeed, every intermediate subgroup G with Z c G c Q is an essential extension of Z. Here are some easy exercises we shall soon use. Exercises: 3.24. Let M c E c E , ; if E is essential over M and E , essential over E, then E , is essential over M. 3.25. Let M c E . Show that E is an essential extension of M if and only if, for each e E E , either e = 0 or there is an r E R with re E M and re # 0. 3.26. Let M c E, and let ( E , :i E I > be a simply ordered (under inclusion) family of submodules of E, each E, being essential over M. Prove that U E i is essential over M. (Hint: Use Exercise 3.25.) 3.27. If M c E' c E and both E' and E are essential over M, then E is essential over E'. 3.28. If E is essential over M and cp: E -+ D is a map with cp I M monic, then cp is monic. (Hint: M n kercp = 0.) Theorem 3.29: A module M is injective i f and only if M has no proper essential extensions. Proof: Assume M is injective and M S E, where E is an essential extension. By Theorem 3.19, M is a summand of E ; there is thus a nonzero submodule N of E with E = M 0 N . As N n M = 0, this contradicts E being an essential extension. Assume M has no proper essential extensions, and let E be an injective module containing M. By Zorn's lemma, there is a submodule N of E maximal with M n N = 0. The composite M 4 E -+ E / N is monic (since N n M = 0). Indeed, E / N is essential over M: if S / N is a nonzero submodule of E / N , then S 2 N and maximality of N yields S n M # 0; it follows easily that SIN meets M. By hypothesis, the composite M -P E -+ E / N is an isomorphism, whence E = M + N . As M n N = 0, we have M a summand of E, and so M is injective.

a

Theorem 3.30: The following conditions on a module E containing a module M are equivalent:

73

Injective Modules

(i) E is a maximal essential extension of M (i.e., no proper extension of E is an essential extension of M ) ; (ii) E is an essential extension of M and E is injective; (iii) E is injective and there i s no injective E' with M c E' S E . Moreover, such a module E exists.

Proof: (i) * (ii) Since "an essential extension of" is a transitive relation (Exercise 3.24), the hypothesis is that E has no proper essential extensions. By Theorem 3.29, E is injective. (ii) (iii) If such an injective E' exists, then E' would be a summand of E, say, E = E' 0 E". Since M c E', we have E' n E" = 0 implying M n E" = 0, contradicting E being an essential extension of M . (iii) (i) Assume only that E is an injective containing A4 and consider the family of all essential extensions of M that lie in E. By Exercise 3.26, Zorn's lemma yields a maximal such, say, E'. We claim E' is a maximal essential extension of M . Assume N is an essential extension of M that contains E'. There is a diagram

-

E

0-El-N

I

\

where i is the inclusion. Since E is injective, there is a map cp:N -,E with VIE' = i; thus, cp fixes E', hence M, pointwise. By Exercise 3.28, cp is monic (for N must be an essential extension of E', by Exercise 3.27). But now cp(N) is an essential extension of M contained in E . Maximality of E' forces cp(N) = E , whence N = E . We conclude that E' is a maximal essential extension of M (as defined in statement (i) ) and that E' has no proper essential extensions. By Theorem 3.29, E' is injective. If we now invoke the full hypothesis of (iii), then E' = E, as desired. To prove the existence of E, imbed M in an injective module and take the submodule E' just constructed.

a

Definition: A module E satisfying any of the equivalent conditions of Theorem 3.30 is called an injective envelope (or injective hull) of M.

We have already proven existence of injective envelopes; we now prove uniqueness. Theorem 3.31 : Let E be an injective envelope of a module M .

74

3 Projectives, Injectives, and Flats

(i) I f D is an injective containing M , then there is a monic cp:E + D fixing M pointwise; (ii) any two injective envelopes of M are isomorphic (by an isomorphism fixing A4 pointwise).

pq

Proof: (i) Injectivity of D allows us to complete the diagram

D

\\

\

0-M-E

where i is the inclusion. Exercise 3.28 gives cp monic, since E is an essential extension of M . (ii) Assume D is an injective envelope of M . Using the notation of part (i), we claim cp is an isomorphism, i.e., cp is also epic. Were cp not epic, then q ( E ) would be a summand of D containing M , and this would contradict D being an essential extension of M . I One may thus speak of the injective envelope of M . Exercises: 3.29. If R is a domain, considered as a module over itself, then its injective envelope is its quotient field. 3.30. Theorem 3.31(ii) characterizes injective envelopes: a monomorphism i : M -+ E, where E is injective, is an injective envelope if and only if a monic dashed arrow always exists below 0

D whenever j is an imbedding of M into an injective D. Exercise 3.30 allows us to dualize the notion of injective envelope; also see Exercise 3.31 for the usual (equivalent) description.

75

Watts’ Theorem

Definition: A projective cover of M is an epimorphism E : P+ M , where P is projective, so that an epic dashed arrow always exists below 0

\ cp\

whenever

\

\

I+ Q

II/ is an epimorphism from a projective Q.

Exercises: 3.31. (Usual Definition of Projective Cover) A submodule N of M is superfluous if, whenever L is a submodule of M with L + N = M , then L = M . Prove that an epimorphism E : P + M , where P is projective, is a projective cover if and only if ker E is superfluous. (Hint: P = irn cp ker E . ) Prove that any two projective covers of a module M are isomorphic. 3.32. Prove that the 2-module 2 / 2 2 has no projective cover. Conclude that projective covers may not exist. 3.33. If c i : P i+ M i are projective covers, i = 1, . . . , n, then

+

.

is a projective cover, where U E ~ :. (. .P p,) ~ ,+-+ ( E ~ P .~. ., , E,P,). WATTS’ THEOREMS

There is a lovely application of frees and injectives in the coming proofs, due to Watts, of characterizations of the Hom and tensor functors. First, we need the appropriate notion of isomorphism of functors. Definition: If F , G:% -+ 23 are functors of the same variance, then F and G are naturally equivalent, denoted F z G, if there is a natural transformation z: F + G with each T ~F A : -,G A an equivalence.

Of course, if 23 is a category of modules, each r A is an isomorphism. Examples: 2. The natural transformation 1 + R OR of Exercise 2.35 is a natural equivalence.

76

3 Projectives, Injectives, and Flats

3. The natural transformation 1 + Hom,(R, ) of Exercise 2.36 is a natural equivalence. 4. If '2l is the category of finite-dimensional vector spaces over a field k, and if * = Horn,( ,k), then the usual evaluation 1 + ** (see Exercise 2.37) is a natural equivalence. 5. If 9I is the category of finite abelian groups and * = Horn,( ,Q/Z), then there is a natural equivalence 1 + ** (this is a special case of Pontrjagin duality of locally compact abelian groups). 6. Let R be a ring and r E Z(R) (so that multiplication by r is an R-map). Define F:,!DI+,!DIby F(M)=M/rM and,if f :M+N,then Ff :M/rM-+N/rN is given by m + rM H f m + rN. If we apply 0, M to the exact sequence

R 1,R + R/Rr + 0, we obtain an exact sequence M 5 M -+(R/Rr) 0, M + 0. There is thus an isomorphism oM:M/rM + (R/Rr) 0, M (first isomorphism theorem), and o:F + (R/Rr) 8, is a natural equivalence. This seems an appropriate opportunity to introduce an elementary and common technique, called diagram-chasing. At the moment, we will only need a special case of the coming lemma, but we give a general form for later use as well as to illustrate the technique. The important thing to observe is that each step of the proof is automatic in the sense that there is only one, obvious thing to do.

Lemma 3.32 (Five Lemma): Consider the commutative diagram with exact rows

B1

h,

*B2

h2

'8 3

h3

'B4

h,

'B5

(i) If t2 and t4 are epic and t5 is monic, then t , is epic; (ii) If t2 and t , are monic and t , is epic, then t3 is monic. In particular, if t , , t 2 , t,, t , are isomorphisms, then t , is an isomorphism.

Proof: We only prove (i); the proof of the dual (ii) is similar. Let b, E B,. Since t , is epic, 3a4 with t,a4 = h3b3.Commutativity of the last square gives t,f,a, = h4t4a4 = h4h3b3= 0 (exactness at B,). Since t ,

77

Watts’ Theorem

is monic, f4a4 = 0, so a4 E ker f4 = im f 3 . Hence 32,with a4 = fJas, and so h3t3a3= t4f3a3= t4a4= h,b3 (definition of a,).Therefore b3 - t3a, E ker h, = im h,, whence 3b, with h2b, = b, - t3a3.Since t , is epic, 3u2 with t2a2 = b2. Commutativity gives t,f2a2 = h2t2a, = h2b, = b3 - t3a3.Therefore and t , is epic.

I

Theorem 3.33 (Watts): preserves sums. Then F choose B = F(R).

t3(f2a2

+

= b3

Let F:mR -+ Ab be a right exact functor that Q R B for some left R-module B. Moreover, we may

Proof: First of all, let us make the abelian group FR into a left R-module. For M = MR and m E M, define q m : R M by r H mr. Since qm is an R-map, we have Fqm:FR+ FM. Define tM:M x F R + F M by (m,x) H (Fq,)x, where x E FR. Ifwe set M = R , then TR:R x FR + FR may easily be seen to define a left R-module structure on FR. The same calculation now shows that TM:M x F R F M is R-biadditive, and T~ induces a map I J ~M : O R FR .+ FM. Ifwe denote FR by B, then one checks that IJ: ORB -+ F is a natural transformation. Now GR: R O R B+ FR is an isomorphism (for B = FR); moreover, since both Q R B and F preserve sums, a,:A QR B + F A is an isomorphism for every free R-module A . Let M be any right R-module. By Theorem 3.8, there is an exact sequence C + A M -, 0, where both C and A are free. Since both ORBand F are right exact, there is a commutative diagram with exact rows: .+

. +

. +

FC

- lA FA

11

FM

0 -

Since oCand IJ, are isomorphisms, the Five Lemma gives oMan isomorphism. Therefore, IJ: O R B+ F is a natural equivalence. I Remark: If F:mR -+ W,,then one may modify the first paragraph of the proof to see that the right S-module FR can be construed as a bimodule; thus the theorem remains true if we replace Ab by ‘Jn,.

Example 6 illustrates this theorem: the functor F there (F = “dividing by 8’)is easily seen to be right exact and sum preserving, and F z FR OR = (R/R~)QR.

78

3 Projectives, Injectives, and Flats

Corollary 3.34: Assume R is a ring for which every submodule of a f-g. module is fig. I f 8, is the category of all fig. right R-modules and if F : SR-+ %Rs is right exact additive, then there is a bimodule B = R B S and a natural equivalence ORB z F. Moreover, we may choose B = F(R). Proof: The proof is almost the same as that of Theorem 3.33 (together with our remark at its end). As we are looking at SR,we need only consider f.g. free R-modules A ; moreover, the hypothesis on the ring R guarantees that there is an exact sequence C + A -+ M -+ 0 in which both C and A are free and f.g. (were C not f.g., then FC would not be defined). Finally, we need not assume F preserves sums, for every additive functor preserves finite sums (Exercise 2.24). \

In the next chapter, we shall identify the rings R satisfying the hypothesis of the Corollary (left noetherian again !). The following theorem ties several ideas together.

Theorem 3.35: T h e following are equivalent for a functor F : m R + Ab:

( i ) F preserves direct limits; (ii) F is right exact and preserves sums; (iii) F g ORBfor some lejl R-module B ; (iv) there is a functor G : Ab -+ %RR so that ( F ,G) is an adjoint pair. Proof: (i) 3 (ii) Cokernel and sum are direct limits. (ii) (iii) Theorem 3.33. (iii) * (iv) Theorem 2.12; set G = Hom,(B, ). (iv)*(i) Theorem 2.19. \

-

Once we give a characterization of covariant Hom functors, we shall have the dual version of Theorem 3.35. These theorems are special cases of “Adjoint Functor Theorems” [MacLane, 1971, pp. 116-1271 which give necessary and sufficient conditions that a functor between more general categories be half of an adjoint pair. Before we do this, however, let us characterize contravariant Hom functors, for the proof is so similar to that of Theorem 3.33.

Theorem 3.36 (Watts): Let F :% ,R Ab be a contravariant left exact functor converting sums to products. Then F 2 Horn,(, B ) for some lest R-module B. Moreover, we may choose B = F(R). -+

Proof: First of all, we make the abelian group F R into a left R-module. If M = R M and m E M , define ym:R M by cp,(r) = rm. Since ymis an R-map, Fy,: FM FR is defined. Define S M :M x FM + FR by (m,x) H (Fy,)x, where x E F M . In particular, r R :R x F R + FR equips F R with a left R-+

--f

79

Watts' Theorem

module structure: note that if r, s E R, then cprS = qscpr: R + R and contravariance of F gives Fq,, = Fcp,Fqp,. Define o,:FM -+ Horn,(M,FR) by o M ( x ) : m H (Fcp,)~. It is easy to check that (T:F + Horn,( ,FR) is a natural transformation, and that g R is an isomorphism. The remainder of the proof proceeds, mutatis murandis, as that of Theorem 3.33. I Remarks: 1. If one assumes F: ,$JJI + $J1, then one may modify the proof above by showing FR is a bimodule. 2. The analogue of Corollary 3.34 is also true, if one assumes the same condition on the ring R that appears there. Definition: A module C is a cogenerator forYJI.', if, for every module M and every nonzero m E M, there exists f :M + C with f m # 0. Lemma 3.37: There exists an injective cogenerator for ,Iln. Proof: Define B = U(R/I), where the sum is over all left ideals I in R, and define C to be some injective module containing B. Let M be a module with m E M, m # 0. Now ( m ) is cyclic, so that ( m ) z R / I for some left ideal I (Theorem 2.2). Since R/I is contained in C , there is a monomorphism C

g : ( m ) + C ; since C is injective, there is a map f : M -+ C with fm=gm#O. 1

The construction above of an injective cogenerator may not be the most efficient. For example, we will see later that Q/Z is an injective cogenerator for Ab (Lemma 3.50). Theorem 3.38 (Watts): Let G :RIln+ Ab be a functor preserving inverse limits. Then G Z Hom,(B, ) f o r some left R-module B. Remark: The module B has no easy description as in Theorems 3.33 and 3.36. Proof: For a module A and a set X , let A X denote the product of copies of A indexed by X . We may regard AX as all functions X + A ; in particular, we

80

3 Projectives, Injectives, and Flats

may regard e = 1, E A*. If x E A, then the xth coordinate of e is x, so that pxe = x,

where p x : A A+ A is the xth projection. Choose an injective cogenerator C for ,%, and let I7 = CGC.Since products are inverse limits, G preserves products and

e = 1Gc E G l l

= (GC)GC.

Define z : H o m ( n , C ) + G C by fr-*(Gf)e. The map r is epic, for if x E GC, then the xth projection p x :ll + C exists. Since G preserves products, Gp, is the xth projection G l l = GCGC+ GC, so that (Gp,)e = x. Therefore z(p,) = x and r is epic. We now describe ker z. If S is a submodule of ll with inclusion i: S + n, we may identify GS with the submodule im(Gi) of G l l , for G is left exact (kernel is an inverse limit). Define B as the intersection of all submodules S of n for which e E GS. By Exercise 2.50, B = Em S; it follows that GB = Km GS = nGS, using our identification and Exercise 2.50 again; we observe that e E GB. The following are equivalent for f :ll -+ C : f E ker z; 0 = z(f) = ( G f ) e ;e E ker G f ; e E G(kerf) (G is left exact); B c kerf (definition of B). If J :B + ll is the inclusion, then j * :Hom(n, C) + Hom(B, C ) , and f E ker j * if and only if J * ( f ) = fj = 0. Thus f E kerj* if and only if B = im j c kerf, so that ker j* = ker z. Exactness of 0 + B ll -+ lT/B + 0 and injectivity of C give exactness of the top row of the diagram 0 + Hom(ll/B, C )+ Hom(ll, C)

j*

Hom(B, C ) + 0

our previous calculation shows the lower sequence is exact. Since kerr = ker j * , the two cokernels are isomorphic via o,:Hom(B, C ) -+ GC given by f H ( G f ) e (this is a general fact: if a:A -,X and fl: A + Y are epimorphisms with the same kernel, then X Y via x Hfl(a-’x)). It is easily checked that o:Hom(B, ) + G is a natural transformation, where a,:Hom(B, M ) -+ GM is given by f I+ (Gf)e. For any module M, there is an imbedding 0 -+ M -+ CHom(M*C) given by m H (fm), that “vector” whose f t h coordinate is fm (here we use the fact that

81

Watts' Theorem

C is a cogenerator). There is thus an exact sequence O+ M

-+

C x -+ C"

- -

for suitable sets X and Y . Since G and Hom(B, ) are left exact, there is a commutative diagram with exact rows 0 -Horn(&

M)

Hom(B, C x )

Horn(& C")

The vertical maps CT are isomorphisms, since oC is an isomorphism and both functors preserve products; the Five Lemma shows oM is an isomorphism. We conclude that Hom(B, ) E G. I Corollary 3.39: The following are equivalent ,for a .functor G:,)IJZ+ Ab.

(i) G preserves inverse limits; (ii) G Hom,(B, ) f o r some /eft R-module B ; (iii) there is a functor F : Ab + R9Jl so that ( F , G ) is an adjoint pair.

-

Proof: (i) 3 (ii) Theorem 3.38. (ii) = (iii) Define F = B Bz,noting that B is an ( R - Z) bimodule. (iii) (i) Theorem 2.24. I

Remark: Exercise 2.54 gives a fourth condition equivalent to (i): (iv) G is left exact and preserves products.

There are several comments, whose proofs are more appropriate in a book emphasizing category theory, that should be made. First, the hypothesis in Corollary 3.34 that F be additive is not necessary in the other Watts' theorems, for any functor between module categories that preserves finite sums must be additive. Second, the more general adjoint functor theorems stress characterization of Hom functors, for these always exist, by the very definition of category. Third, one may give a precise definition of a solution to a universal mapping problem, "universal construct ion", and this is intimately related to adjoint functors. There are also two uniqueness results. If (F,G) and ( F , G ' ) are adjoint pairs, then G 1G'; the same result holds in the other variable. Second, if Horn@, ) and Hom(B', ) are naturally equivalent, then B z B'. For details, the reader may consult [MacLane, 19711.

82

3 Projectives, Injectives, and Flats

There is one more result in this circle of ideas which is within our reach and which is rather interesting. if there Definition: Two categories C and 3 are equivalent,denoted C 3, are functors F : C + and G : 3 + C such that GF z 1, and FG Z 1,. We consider whether it is possible that ,%TI z ,9X. Should this obtain, then homological algebra cannot distinguish between the rings R and S . Exercise: 3.34. If R%R Z ,m, then Z ( R ) Z Z(S). (Hint: Use Exercise 2.39.) Conclude that if R and S are commutative rings and R%TI z ,m, then R E S. Definition: A module Pis a generator for Rmif every module M is a quotient of a sum of copies of P .

Clearly R itself is a generator for ,m, for every module is a quotient of a free module. Also, the sum of n copies of R, denoted R", is a generator for Rm. Exercise: 3.35. Prove that a module Q is a cogenerator for ,llJ1 if and only if every module can be imbedded in a product of copies of Q. Conclude that generator and cogenerator are dual. Definition: A module P is small if Hom(P, ) preserves sums.

Note that R" is a small projector generator for RYJ2. Theorem 3.40: Let P be a small projective generator for R9Jl and let S = End,(P). Then Z sm. Proof: Clearly P is a right S-module (if we write maps on the right); moreover, P is an (R-S) bimodule, for if X E P , then (rx)f= r ( x f ) is just the condition that f is an R-map. By Theorem 1.15, the functor F = Hom,(P, ):Rm -+Ab actually takes values in ,%TI. Define G:s9Jl + R% by G = P 0, . Since ( F , G ) is an adjoint pair, Exercise 2.38 provides natural transformations l R+ GF and FG -+ l,, where 1, is the identity functor on RYiand 1, is the identity functor on ,W. It suffices to prove these natural transformations are equivalences. Since P is projective, F = Hom(P, ) is exact; since P is small, F preserves sums. The functor G is right exact and preserves sums. It follows that the composites GF and FG preserve sums and are right exact. Let us evaluate : GF(P) = G(Hom(P,P ) ) = G ( S ) = P 0, S

z P.

83

Watts' Theorem

If M is any R-module, there is an exact sequence of the form M -0

UP-uP-

(where the index sets of the sums may be distinct) because P is a generator for R9Jl. There is thus a commutative diagram with exact rows

We know the first two vertical maps are isomorphisms, so the Five Lemma gives M E GF(M).This gives 1, z GF. For the other composite, note that F G ( S ) = F(P Bss) Z F ( P ) = HomR(P,P) = S.

If N is any S-module, there is an exact sequence of the form

u s -+ us

-+

N

+ 0,

for every module is a quotient of a free module. The argument concludes just as for the first composite. I Corollary 3.41: If R is a ring and S in R, then RYX E s9Jl.

=

R , , ull n x n matrices with entries

Proof: We know that R" is a small projective generator for $JIJz, and End,(R") z R,. I

Corollary 3.42: Let P he a small projrctiw generator for R9Jl and let s = EndR(P). If F = H0mR(P, ):,'JJZ $JJ, [hen F is an exact functor preserving projectiues and injectives. -+

Proof: Assume Q is R-projective, and consider the diagram of S-modules

84 If G

3 Projectives, Injectives, and Flats =P

as , then we have the diagram of R-modules GFQ = Q

the dashed arrow exists because Q is projective. Now apply F to this diagram to obtain the desired arrow FQ + B. A similar argument shows F preserves injectives. I The Morita theorems go on to say, among other things, that every equivalence between module categories arises as above, that is, from a small projective generator P. We should also point out that the functor F = Hom,(P, ) in this case is (right) exact and preserves sums, so Watts’ Theorem 3.33 gives F = Hom,(P,R) OR . An exposition of the Morita theorems may be found in DeMeyer-Ingraham [1971].

FLAT MODULES

A module P is projective if and only if Hom,(P, ) is exact, and a module E is injective if and only if HornR(,E ) is exact. Let us now fix a variable of tensor. Definition: A right R-module B is flat if the functor B OR is exact. Of course, there is a similar definition for left R-modules C and ORC. Since B OR is always right exact, a module B is flat if and only iff monic implies 1,O f monic.

Theorem 3.43: R is aflat R-module. Proof: Let f:A’-+A be monic. If t : 1 -P R 0, is the natural equivalence of Theorem 1.12, then there is a commutative diagram

A’

f

85

Flat Modules

Since the vertical maps are isomorphisms, the map 1, @ f is monic and R is flat. I An example of a nonflat module is contained in Example 18 of Chapter 2: 2 / 2 2 is not a flat 2-module. The next theorem generalizes Theorem 3.26 and Exercise 3.22. Theorem 3.44: Assume B is a bimodule SBR that is R-flat and C injective. Then Homs(B, C ) is an injective IeB R-module.

= ,C

is

Proof: First of all, Hom,(B, C) is a left R-module, by Theorem 1.15. We must prove Horn,( ,Hom,(B, C ) ) is an exact functor. The adjoint isomorphism shows this functor is naturally equivalent to Hom,(B OR , C), which is the composite Horn,( ,C) (B OR ). Each of these is exact, since B is R-flat and C is S-injective, and the composite of exact functors is exact. I 0

Setting B = R, we obtain the theorem and exercise mentioned above. Theorem 3.45: Let {B,: k E K } be a family i s j a t if and only if each B, i s j u t .

of right R-modules. Then u B k

Proof: Note first that if {fk:Ak --t A k ] is a family of maps, there is a unique map f k : u A ;+ u A k with x u ; H c f k u ; : moreover, f k is monic if and only if each f k is monic. Assume f :A’ + A is monic. There is a commutative diagram

u

1

LI(BkhA’) where 1,

=

U(lL@/,

+

1

u ( B i0 4

lB, and the vertical maps are the usual isomorphisms

o a C(hko a). H

Our initial remarks give 1 @ f monic if and only if each 1, Of is monic, i.e., Bk is flat if and only if each Bk is flat. I Corollary 3.46: Every projectiue module i.s.fEut. Proof: Theorems 3.43 and 3.45 show that every free module is flat; Theorem 3.45 shows that every summand of a free module is flat. I It follows that every module A has a flat resolution, i.e., there is an exact sequence . . . --* F , + F , + A -+ 0 in which each F,, is flat.

86

3 Projectives, Injectives, and Flats

Theorem 3.47: If (Bk, q$} is a direct system of flat modules over a directed index set K , then lim Bk ispat. -+

Proof: If f :A’ -+ A is monic, there is a commutative diagram where the (1% B k )@ A’

of

lim(B, + @ A’)

rp

(limB,)@ A _--L

vertical maps are the isomorphisms of Corollary 2.20, and q is the monomorphism of Theorem 2.18. It follows that 1 @ f is monic. I This theorem may be false if the index set is not directed; for example, coker(flat -+ flat) need not be flat. Corollary 3.48: Zf R is a domain, then its quotientfield Q is a j a r R-module.

Proof: We saw in Exercise 2.34 that Q is a direct limit of copies of R, indexed by a directed set. I Corollary 3.49:

I f every f.g. submodule of B is .put, then B is .flat.

Proof: We saw in Exercise 2.32 that every module is a direct limit of its f.g. submodules, and the index set there is directed. I

Let us give another proof of Corollary 3.49 avoiding direct limits. Assume f :A’ -+ A is monic and that x E ker 1 Q f.As x E B Q A’, we have x= bi @ a;. Let F be the free abelian group with basis B x A, and let S be the subgroup of relations (as in Theorem 1.4) so that F / S = B @ A. To say that x b i @ f a : = 0 in B @ A is to say that C ( b i , f a i )E S. Let B‘ be the submodule of B generated by b l , . . . , b, together with the (finite number of) first coordinates of relators exhibiting x ( b i ,fa:)E S. If Js = C b i @ a; in B’ @ A‘, then X E ker(B’ @ A’ -, B’ 0 A). As B’ is f.g., the hypothesis gives X = 0 in B’ 6A’, from which it follows that x = 0 in B 6A (ifj:B’ + B is the inclusion, then x = ( j @ 1&). We now aim toward a characterization of flat modules that links them to injectives.

cy=

Lemma 3.50: The group Q/Z is an injective cogenerator for Ab.

Proof: Since Q/Z is divisible, being a quotient of Q, it is injective (Theorem 3.24). Let M be an abelian group, and let m E M , m # 0. If rn has infinite order, define f:( m ) --+ Q/Z by m H 5 Z; if m has finite order n, define

+

87

Flat Modules

f :(m) + Q/Z by m H l/n + Z. In either case, f(m) # 0. Now use injectivity of Q/Z t o extend f to all of M . 1 Definition: If B is a right R-module, its character module B* is the left R-module Hom,(B, Q/Z). Since B is a bimodule ,BR, Theorem 1.15(i) gives the left R-module structure on Hom,(B,Q/Z) by rf:b H f(br). We now improve Lemma 2.13. Lemma 3.51 : A sequence of right R-modules O+A~B%,f+O is exact if and only i f the sequence of charucter modules

is exact. Proof: If the original sequence is exact, then the exact contravariant functor Horn,( ,Q/Z) carries it into an exact sequence. To prove the converse, we shall show that kera* = imp* implies ker b = im a without assuming b* monic or a* epic; the result will then follow. Is im a c ker B? Suppose a E A and aa 4 ker p, i.e., baa # 0. Since Q/Z is a cogenerator, there is a map f : C + Q/Z with f b a a # 0. Thus, f E C* and f P a # 0, i.e., a * P * ( f ) # 0. This contradicts a*p* = 0. Is kerb c i m a ? If not, there is an element b E kerb with b $ im a. Thus, b im a is a nonzero element of B/im a, so there is a map g: B/im a + Q/Z withg(b +ima) #O.Ifn:B+B/imaisthenaturalmap,thenf= gx:B+Q/Z is such thatfb # 0 andf(im a) = 0. Therefore 0 = f a = a*(f), s o f E ker a* = im b*. Thus f = b*(g) = gp for some g E C*. In particular, f b = gpb. This is a contradiction, for f b # 0 while b E ker 1. 1

+

Theorem 3.52: A right R-module B isflat if and only if its character module B* is an injective left R-module.

Proof: As every right R-module, B is a bimodule B = ,BR. Since Q/Z is an injective Z-module, Theorem 3.44 shows B* = Hom,(B, Q/Z) is injective. For the converse, assume B* is R-injective and ,f :A‘ + A is monic. There is a commutative diagram with vertical isomorphisms HOmR(A, B*) I

HOmR(A’,8*) 1

0

88

3 Projectives, Injectives, and Flats

so that exactness of the top row gives exactness of the bottom. In character module notation, (B Q A)* + (BQ A‘)* + 0 is exact, and Lemma 3.51 applies to give exactness of 0 + B 0 A’ -P B @ A. Therefore, B is flat. I This theorem may be exploited by coupling it with Baer’s criterion. Theorem 3.53: I f B is a right R-module such that 0 + B Q I exact for everyxg. left ideal I of R, then B isJat.

+B

0 R is

Proof: Since every left ideal is a direct limit (with directed index set) of f.g. left ideals, Theorem 2.18 implies that 0 -+ B 0 I -+ B Q R is exact for every left ideal I of R. This gives exactness of (B Q R)* -+ (B0 I)* + 0, which, in turn, gives exactness of

HOmR(R,B*)+ HOmR(1,B*)+ 0 as in the proof of Theorem 3.52. By Baer’s criterion (Theorem 3.20), we have B* injective, and this shows B is flat. I

Of course. the converse of Theorem 3.53 is true. Theorem 3.54: If B , is Jut and I is a lefi ideal, then the map B QR I given by b Q i H bi is an isomorphism.

+ BI

Proof: Recall that BZ = { x b j i j :bj E B,ij E I} is a subgroup of B. The composite B ORI -+ B ORR + B, the second map being the usual isomorphism, has image BI and is monic, by Theorem 3.53. I

In general, neither quotients nor submodules of flats are flat (the same is true if we replace “flat” by “projective” or “injective”),but we must wait until we examine specific rings before examples can be given. The following instance, however, guarantees that a quotient of a flat is flat if the kernel is “nicely” imbedded; the situation is related to the notion of “purity” in abelian groups. Theorem 3.55: Let F be Jlat and 0 K + F 5 B -+ 0 be an exact sequence of right R-modules. The following conditions are equivalent: -+

(i) B isJat; (ii) K n F I = KI for every left ideal I ; (iii) K n F I = K I for every f i g . lefi ideal I . Proof: In conditions (ii) and (iii), note that K I c K n F I , so only the reverse inclusion is significant. We give a preliminary discussion before proving the implications.

89

Flat Modules

Tensoring the original exact sequence by a left ideal I gives exactness of

K@I+F@I-

OQ

1

B@I--+O.

By Corollary 3.54, we may identify F @ I with FI via f @ i ++ f i ; this identifies im(K 0 I -,F 0 I ) with K I . There results an isomorphism given by f i

+ KI BI

y:FIIKI% B @ l H

bf 0 i. Next,

=

{c(&)i,:f, E F , ij E I ;

=

{ P ( c f j i j ) )= f l ( F 1 ) .

since b is epic

The second isomorphism theorem provides an isomorphism

6:BI = /3(FI)-+ F I / F I n K given by bi Hf i + F I n K , where bf the composite G :

= b. We assemble these

F I / K I A B @ 15 BI 4, F I J F I

A

maps to obtain

K,

where 8: b 0 i H bi. Explicitly, a : f i + K I w f i + FI n K . Since K I c FI n K , CJ is an isomorphism if and only if K1 = F1 n K . Moreover, since the flanking maps y and 6 are isomorphisms, G is an isomorphism if and only if 8 is. (i) S. (ii) If B is flat, Corollary 3.54 says 8 is an isomorphism. Thus a is an isomorphism and KI = F I n K . (ii) 3 (iii) Trivial. (iii) * (i) If K I = F I n K for every f.g. left ideal I , then 8 is an isomorphism for every such I , and Theorem 3.53 gives B flat. I Another technical lemma and we will be able to relate flat modules to projecti ves. Lemma 3.56: Let F be free with basis { x , :j E J } , and let 0 + K + F be an exact sequence of right R-modules. I f c E F is written e

= x.J I r 1

+ B -+

0

+ .. +.~,~r,,

dejne I(v) as the left ideal of R generated by the “coordinates” r , , . . . , r,. Then B ispat if and only if v E KI(t1)for ecery c E K . Proof: If B is flat and u E K, then L’ E K n F l ( v ) = K I ( v ) , by Theorem 3.55. Conversely, let I be any left ideal and let I‘ E K n F I . Then I(v) c I ; by hypothesis, v E KI(v) c K I . Therefore, K n F I c KI. As the reverse inclusion always holds, Theorem 3.55 gives B flat. I

90

3 Projectives, Injectives, and Flats

Theorem 3.57 (Villamayor): Let F be free and let 0 -+ K + F 3 B be an exact sequence of right R-modules. The following are equivalent.

-+

0

(i) B i s p a t ; (ii) for every v E K , there is a map 8 : F -+ K with 8u = u ; (iii) for every u l , . . . , u, E K , there is a map O:F + K with 8ui = ui. i = l , ..., n. Proof: (i)* (ii) Suppose B is flat and u E K ; choose a basis { x j : jE J } of F. Let u = xj,rl . . + xjfrt and let Z(u) be the left ideal generated by r l , . . . , r,. Since B is flat, Lemma 3.56 gives u E KZ(u). Hence u = Ck,s,, where kAE K, sa E Z(u); as s, E i(u), s1 = EaLiri,where aAiE R. Thus u = Ck:ri, where kf = CkaaaiE K . Define 0 : F -+ K by x j i I-+ ki and the other basis elements go to 0. Clearly Ou = u. (ii)- (i) Choose a basis { x j :j E J } of F, and let u = xj,rl + . . . + xjtrrE K ; let 0: F + K satisfy 8u = u. Then u = 8u = O(xjl)rl+ . - . + 8(xjt)r,E KZ(u). By Lemma 3.56,B is flat. Since ( 5 ) trivially implies (ii), it only remains to show (ii)=. (iii). We do an induction on n, the case n = 1 being our hypothesis (ii). Suppose n > 1 and u l , . . . , u, E K. By (ii), there is a map 8,: F -+ K with 0,u, = u,. Define u: E K, i = 1 , . . . ,n - 1, by ui = ui - &pi. By induction, there is a map B':F-+K with t?'ut = u;, i = 1,. i . , n - 1. Finally, define 0 : F - r K by

+

e u = enU+ eyu - e,u).

It is a routine check that Oui = ui, i

=

1, . . . , n.

I

Definition: A right R-module B is finitely related (or finitely presented) if there is an exact sequence Fl+Fo-+B-+O where F1 and F , are f.g. free. It is easy to see that B is finitely related if and only if there is an exact sequence 0 K -+ F -+ B -,0 in which F is free and both F and K are tg. Note that every f.g. projective module B is finitely related: choose a fg. free module F mapping onto €3, say, with kernel K ; this sequence must split, F z B 0 K , and this shows that K is f.g. If B is flat, then the converse is true. -+

Corollary 3.58: Every finitely related p a t R-module is projective. Proof: There is an exact sequence O+ K

1 -+

F

-+

B -0,

where F is free and both F and K are Eg. If K = ( u l , . . . , u,), then Theorem

91

Flat Modules

3.57 provides a map 8 : F + K with 8vi = i = 1, . . . , n (we are assuming, without loss of generality, that iis an inclusion). Therefore 8,l = l,, so the sequence splits and B is projective. I ~

3

~

.

There is a quicker way to prove this last corollary. Lemma 3.59: Let R and S be rings and consider the situation ( R A , R B S , CJ. I f A is a kg. projective R-module, then there is LI natural isomorphism

Horn@, C ) ORA 2 Hom,(Hom,(A, B),C ) . Proof: Observe that B being a bimodule allows us to regard Hom,(B, C ) and HomR(A,B ) as modules, so the above terms do make sense. Define cA:Hom,(B, C)ORA + Hom,(HomR(A,B),C ) by

f 0a b-+

OA(f

0a),

where, for g E Hom,(A, B), OA(f

0a ) : 9- . m a ) )

E

c.

It is a routine check, with no hypothesis on A other than A = RA, that o is a homomorphism, natural in A . Moreover, if A = R, then cAis visibly an isomorphism; indeed, it is just as clear that oA is an isomorphism when A is f.g. free. Finally, we let the reader check that if c Ais an isomorphism and A’ is a summand of A , then oA,is also an isomorphism. I Let us record another isomorphism as above, even though we do not need the result for the new proof of Corollary 3.58. Lemma 3.60: Let R and S be rings and consider the situation ( R A , R B S , Cs). l f A is finitely related and C is injective, then there is a natural isomorphism

Hom,(B, C )

A

Z

HOms(HOmR(A,B), C ) .

Remark: The map oAof Lemma 3.59 is the desired isomorphism; we shall prove this in the midst of the next proof. I Theorem 3.61 : Every jinitely related j u t R-module E is projective. Proof: Let F , + F , + E -+0 be exact, where F , , F , are f.g. free. It suffices to show that Hom(E, ) is exact, i.e., if A + A ” + O is exact, then Hom(E, A ) + Hom(E, A ” ) 0 is exact. By Lemma 3.51, it suffices to show exactness of -+

0 + (Hom(E,A ” ) ) *-+ (Horn(&A ) ) * , where (Horn(& A))* is the character module Hom,(HomR(~,A), Q/Z).

92

3 Projectives, Injectives, and Flats

Consider the following diagram, for any module B and for C injective:

I

c)@ FI

+

-,

Hom,(B, C) 0 F ,

I

Hom,(B, C ) @ E

I

+0

By Lemma 3.59, the diagram commutes and the first two vertical maps are isomorphisms. The top row is exact, for tensor product is right exact; the bottom row is exact because Horn,( , B ) is left exact and Horn,( ,C) is exact (for C is injective). The Five Lemma gives the final vertical map an isomorphism for every B : Horn,(& C ) 0 E 3 Hom,(Hom(E, B), C ) (we have proved Lemma 3.60). Finally, consider the diagram

0

+

Hom,(A”, Q/Z) 0 E

-,

Hom,(A,Q/Z) @ E

0 + Hom,(Hom(E, A”), Q/Z) + Hom,(Hom(E, A),Q/Z)

Naturality of the maps o gives commutativity, and we have just seen that the maps CJ are isomorphisms. Recall that we began with an epimorphism A+A”-+O, so that we have exactness of O-+Hom,(A”, Q/Z)+Hom,(A, Q/Z). Since E is flat, the top row is exact, and this implies exactness of the bottom row, which is precisely what we had to show. I Are there any f.g. modules that are not finitely related? Let R = k [ x , , x , , . . .], polynomials in infinitely many variables over a field k ; if I is the ideal generated by the indeterminates, then R / l is cyclic (hence f.g.), and we shall show it is not finitely related. Theorem 3.62 (Schanuel’s Lemma) : Given exact sequences O+K,+P,+B-+O

and

O+K,+P,+B-+O,

where PI and P , are projective, then K , 0 P ,

%

K , @ P,.

93

Flat Modules

Proof: Consider the diagram with exact rows

0-

K 2 i P 2 - B - O

Since P , is projective, there is a map b : P , + P , making the right square commute; diagram-chasing shows there exists a map a : K , + K , making the left square commute. There is an exact sequence 0 +K

5 P I 0 K f P , + 0,

where O : k , H (ik,,ak,) and $ : ( p l , k 2 ) H /3p, - j k , (verification ofexactness is routine). Since P , is projective, this last sequence splits, giving P , 0 K , F KlOP,. I Corollary 3.63: If B is jnitely related und

O+K+MI.B-O is exact, where M is j g . , then K is .fg.

Proof: Assume first that M is free. Since B is finitely related, there is an exact sequence 0 - K , -+ F , + B + 0 with F , free and both F , and K , f.g. By Schanuel's lemma, M 0 K , z F , 0K . But M 0 K , is Eg., hence so is its summand K (summands are images). Let us drop the assumption that M is free; choose a f.g. free F mapping onto M . There is a commutative diagram with exact rows 0-

kerfbI

F-

18

I

I

Now ker fp is fg., by the first part of the proof, and the induced map a is epic, by the Five Lemma. Therefore K is f.g. \

It follows immediately from the Corollary that the cyclicmodule exhibited above is not finitely related. The real significance of Schanuel's lemma is

94

3 Projectives, Injectives, and Flats

that a strong connection exists between any two descriptions of a module by generators and relations. Exercises: 3.36. State and prove the dual of Schanuel’s lemma involving injectives. 3.37. Given two exact sequences, where the P‘s and Q’s are projective, 0 +K

+. . . + P o

+ P, + P,-

+B -0

and 0- L+

Q,,+ Q nP 1

-r.

. - + Q o + B -r 0,

then KOQ,OP, - , O . . . ~ . L O P , O Q_, , O . . . . The dualisalsotrue. 3.38. Let B be a bimodule RBS and let C be a left S-module. If B is a flat R-module and C is a flat S-module, then B OSC is a flat R-module. (Hint: The composite of exact functors is exact.)

PURITY In this brief section, we examine loss of exactness from a different viewpoint. 1fO-r A’+ A + A” + O is exact and 0- B @ A’+ B @ A + B @ A”+O is not exact (of course, this means the 0 at front cannot be there), then we have made B the culprit. Good modules preserve exactness, and we have called them flat. Perhaps the fault is not in our modules but in our sequences. Definition: An exact sequence of left R-modules i.

0 + A ‘ - +A + A “ + 0

is pure exact if, for every right R-module B, we have exactness of

O + B @ A’

3B O A + B @ A ” +

0.

We say that AA’ is a pure submodule of A in this case.

Of course, one need only worry whether 1,@ A is monic for every B. It is easy to see that every split short exact sequence is pure exact; every summand is a pure submodule. Exercises: 3.39. Prove that a right R-module A is flat if and only if whenever &ri = 0 in A (where ai E A and ri E R), there exist elements b j E A and sij E R such that ai =

bjsij

and

1 sijri = 0.

95 Purity

3.40. 0 + A’ + A + A” + 0 is pure exact if and only if it remains exact after tensoring by every finitely related module B. (Hint: To say an element lies in ker(B @ A’ + B @ A ) involves only finitely many elements of B.) 3.41. If 0 + A’ A -,A” -+ 0 remains exact after applying Hom(B, ) for all B,then the original sequence is split. --f

Lemma 3.64: Let B be a finitely related right R-module with generators b l , . . . , b, and relations Cbjrji,i = 1 , . . . , m. I f A is a left R-module with ” bj@aj= 0 in B Q R A,

1

j= 1

then there are elements hi E A with a j = x r j i h i ,all j . Proof: We have varied the definition of generators and relations a bit. We really mean there is an exact sequence

O+K3FZB+O with F free on x , , . . . , x,, K generated by c.x,rji, i = 1,. . . ,m, and pxj = b j . Tensoring by A gives exactness of

-

K 0 A 8 8 1 F 0 A= B 0A + 0, and, by hypothesis, E x j @ aj E ker p @ 1 = imp @ 1. Now every element of K @ A has an expression of the form xjrjiQ hi,

where hi E A.

i,j

In particular, C x j @ a j = (pQ 1 )

C xjrji0hi = C x,@ i.j

i

Since F is free on the xis, every element of F Q A = U ( x j R )@ A has a unique expression of the form C x j @ a j , where c i j € A. It follows that aj = x r j i h i ,as desired. I Theorem 3.65 (P. M. Cohn): Let 1:A‘ -,A be monic. Then AA’ is a pure submodule of A if and only if, given any commutative diagram with F , , F , j g . free, there is a map F , -+ A’ making the new diagram commute.

96

3 Projectives, Injectives, and Flats

Remark: We restate the condition in terms of equations and then in terms of matrices. 1. If a;, . . . , a; in A’ satisfy a system of equations la; = x r j i a i ,where a, E A, then there exist hi E A’ with a; = Crjihf for allj. 2. Assume V’ is a column vector in IA‘ and D is a rectangular matrix over R with V’ = DV for some column vector V in A ; then there is a column vector H’ in 1A’ with V’ = DH‘.

Proof: Assume AA’ is pure in A and IaJ = c r j i a i for elements ai E A. Define a finitely related module B = F / K , where F is free on x j , j = 1, . . . , n, and K is generated by c x j r j i , i = 1,. . . , m ; let bj = x j + K . In B Q A, we have

Since 1 Q I is monic, by purity, we have c b j Q a) = 0 in B Q A‘. By Lemma 3.64, there are elements hi E A’ with a; = c r j i h : ,allj. For the converse, we must show 1 @ I : B @ A’ + B Q A is monic for every B. By Exercise 3.40, we may assume B is finitely related, say, with generators b l , . . . , b, and relators c b j r j i , i = 1, . . . , m. A typical element of B Q A’ can be written c b j @ a; for a; E A’. If (1 Q I ) C b j Q a) = 0 in B O A , is x b j Q a ; = 0 in B Q A‘? By Lemma 3.64, there are elements hi E A with AaJ = c r j i h i , all j. By hypothesis, there are elements hf E A‘ with aJ = xrjihf, allj. Therefore C b j Q a> =

j

bj Q (Crjih;)=

1 i

Q hi = 0

in B Q A’.

Hence 1 Q I is monic and LA’ is a pure submodule of A . Exercises: 3.42. If R = Z, prove that a subgroup A‘ of A is pure if and only if A’ n nA = nA‘ for all n E Z. 3.43. If A and B are torsion abelian groups, prove that A Q B is a sum of cyclic groups. (Hint: Use Kulikov’s theorem [Rotman, 1973, p. 1973: if A is a torsion abelian group, there is a pure exact sequence O+C+A-+D+O,

where C is a sum of cyclic groups and D is divisible.) 3.44. Let A’ be a subgroup of a torsion-free abelian group A . Prove that A’ is pure in A if and only if A/A‘ is torsion-free.

97

Localization

3.45. If { A i:i E I} is a family of pure subgroups of a torsion-free group A, then n A i is pure in A. (Hint: There is an exact sequence

o -+ pi-+A

-+

~(A/A~).)

(If A is not torsion-free, an intersection of pure subgroups need not be pure.)

LOCALIZATION A very important construction in commutative algebra is “localizing” a ring R (it is the ring-theoretic way to focus on the behavior of an algebraic variety at a neighborhood of one of its points). All rings in this section are commutative; I saw the elegant exposition below in unpublished lecture notes of M. Artin. We assume the reader accepts the existence of R[U], the ring of all polynomials in (possibly infinitely many) variables U . Moreover, we assume the reader knows R [ V] is the free commutative R-algebra with basis U : R[UI \

T “\< m

U

b

9

R’

if R’ is an R-algebra and cp: U + R’ any function, there exists a unique R-algebra map i7, extending cp (that @ is an R-algebra map means i7, is a ring map with @(rx)= r$(x) for all r E R and x E R[V]). This merely says R-algebra maps from R[U] are completely determined by their values on the variables U .

Definition: Let S be a subset of a commutative ring R. The localization S-’R is a commutative R-algebra and an R-algebra map O:R -+ S-’R such that (i) 0(s) is invertible for every s E S; (ii) S - ’ R is universal with this property: R-SIR

\

0

/

/

98

3 Projectives, Injectives, and Flats

if R' is a commutative R-algebra and $:R + R' is an R-algebra map with $(s) invertible for all S E S, then there exists a unique R-algebra map cp:S-'R + R' with cp6 = $. As any solution to a universal mapping problem, S-'R is unique if it exists. Theorem 3.66: For every subset S of a commutative ring R, the localization S-'R exists. Proof: Let R[ U] be the polynomial ring over R in variables U = {u,:s E S], a set in one-one correspondence with S. Define S-'R = R[U]/Z, where I is the ideal generated by all elements of the form su, - 1, s E S ; define 0 : R -+ S-'R as the composite R + R[U] + S-'R, the first arrow imbedding R as the constants, the second being the natural map. It is clear that S-'R is a commutative R-algebra, 0 is an R-algebra map, and that 6(s) is invertible for every s E S . Assume R' is a commutative R-algebra and $:R + R' is an R-algebra map with +(s) invertible for all s E S. Now the definition of 6 is precisely

R'

commutativity of the top triangle in the diagram. Define cpo:RIU] + R' as the unique R-algebra map with us w +(s), all s E S. Since $(s) is invertible for all s E S, kercp, =) I ; it follows that cpo induces an R-algebra map (p:S-'R = R[U]/Z + R' with 96 = $. Uniqueness of cp follows from S-'R being generated, as an R-algebra, by the image of R [ U ] . I We now know S-'R exists, but what does it look like? Here are some simple observations, whose verifications are left as exercises. Definition: A subset S of a ring R is multiplicatively closed if it is a monoid: 1 E S; ifs,,s2 E S , then sls2 E S.

If S is any subset of R, let S be the monoid in R generated by S.

99

Localization

Exercises: 3.46. For any subset of S of R , there is an isomorphism S - ’ R z ( S ) - ’ R . (Hint: Both are solutions to the same universal mapping problem.) 3.47. If S is finite, say, S = { s l , . . . , sn), and if s = nl=lsi, then S - ’ R z s - ’ R . (Hint: If s-’ exists, so does s - ’ ( s , . . . Si.. . s,) = s; where means delete.)

’,

A

Theorem 3.67: If S is a subset of R, then each x

x

= 6(r)O(s)-’,

E

S - ’ R has a factorization

s E S, r E R .

Remark: This factorization is not unique.

Proof: The set A of all x E S - ’ R admitting such a factorization is a subalgebra containing im6; in particular, B(s) is invertible in A for every s E s. If i : A -P S - ’ R is the inclusion, there is a commutative diagram

where cp exists by universality. As the identity on S - ’ R makes the large triangle commute, uniqueness gives icp = 1. By set theory, i is onto and A=S-’R. I It follows that we may write the elements of S - ’ R as “fractions” r/s, where r E R and s E S; moreover, addition and multiplication of fractions have the usual formulas. The only difficulty is that “cross multiplication” is not the answer to the question of when r/s = r’/s’.This is plain in case 0 E S, for 0 invertible in S - ’ R forces 0 = 1 there; in this case, S - ’ R = (O}, a ring whose existence we admit, but whose abode we choose to be outer darkness. In particular, ifwe denote 6 : R + S- R by r t* r l l , then 0 need not be monic. Theorem 3.68: If 6 : R + S - ’ R , then

kerf9 = { r E R : s r = 0 for some s E S>. Proof: If sr = 0 for some s E S, then 6 ( r )= 6(rs)6(s)-’= 0. Conversely, assume O(r) = 0. As S - ’ R = R [ U ] / Z ,we have r E I , the ideal generated by all su, - 1. Writing what this means gives an equation involving

100

3 F'rojectives, Injectives, and Flats

only finitely many us, say, us,, . . . ,us,. If So = Is1,. . . , s,,}, then r E ker B o , where 8,:R -,So'R. By Exercise 3.47, we may assume the finite set So consists ofjust one element, s, whence s-'R = R [ u ] / ( s u- 1). But B&) = 0 says r = (su - l)f(u), where f(u) E R[u] = (su - l)(ao a,u . .. + a m p ) .

+

+

Expanding and equating coefficientsof like powers of u give equations: r

= -ao,

This gives s"+ r

= 0, as

sa,

= a,, . . . , s u m - , = a,,

sum = 0.

desired.

Corollary 3.69: (i) If S contains no zero divisors, then R -, S-'R is monk. (ii) If R is a domain with quotientjeld Q, then S-'R is a subring of Q. (5) If R is a domain and S = R - {0},then S-'R = Q.

Corollary 3.70: Let r/s and r'ls' there is CT E S with o(rs' - r's) = 0.

E

S I R . Then r/s

= r'Js'

if and only

if

Proof: If r/s = r'/s', then (rs' - r's)/ss' = 0 in S - ' R . Since ss' is invertible, rs' - r's = 0 in S-'R; by Theorem 3.68, a(rs' - r's) = 0 for some 0 E S. The converse is left to the reader; it is easy and does not need Theorem 3.68. 1

The usual construction of S-'R mimics the standard construction of the quotient field of a domain. Consider all ordered pairs (r, s) E R x S and define an equivalence relation by the, a priori, mysterious condition in Corollary 3.70 (the usual relation of cross multiplication does not work, for the possible presence of zero divisors in S prevents it from being an equivalence relation). The standard definitions of addition and multiplication of equivalence classes are given,and the myriad easy calculations (the operations are well defined and all the ring axioms hold) are left as a tedious exercise. Usually one assumes S is multiplicatively closed ( S = S ) when constructing S-'R; by Exercise 3.46, the same localizations arise. The two most popular examples of multiplicatively closed subsets S of R are: (i) S = R - P, where P is a prime ideal of R ; (ii) S = {r":n 2 0 } for some fixed I E R. The ring S-'R is an R-module (if we forget some multiplication). One of its most useful features, and the reason localization appears in this chapter, is that S - R is flat as an R-module. We now prove this. Definition: If S is a subset of a commutative ring R and if M is an R-module, then

101

Localization

There is an obvious map O M : M + S - M given by m H 1 Q rn. The next result generalizes Theorem 3.68, which describes kerO, (we wrote 6 for d R then). Theorem 3.71 : ker O M

=

{ m E M : am

= 0 ,for some r7 E

s}.

Proof: If om = 0 for some a E S, then O,(m) = l/ad,(am) = 0.The proof of the reverse inclusion parallels that of Theorem 3.68. As there, we may reduce to the case S = {s}, so there is an exact sequence S- 'R

0 -+ (su - 1) A R[u]

-

-+

0.

There is a commutative diagram (su - 1) @ M

it31

/I

R [u] Q M

n@I

S- ' M -+ 0

i.M

where the top row is exact because of right exactness of Q R M , and R:m H 1 Q m is monic (because, as R-modules, R is a summand of R[u]). It follows that ker OM

=

{m E M :1 @ m E ker 7c Q 1> = { m E M: 1 Q m E i m j @ l}.

Now R[u] is a free R-module with basis (1,u,u2,. . .}. Therefore R[u] 0 M 2 U ( R u i0 M ) , so that every x E R[u] Q M has a unique expression

x=

n

1 u i Q mi,

where mi E M.

i=O

Hence, ker0,={mEM:lQm=(su-

1)

n

1 uiQmi}.

i=O

We now proceed as in Theorem 3.68, expanding and equating coefficients. There are equations: l a m = -l@rno; u Q . s m o = ~ @ m , ;. . . ; un+l @ sm, = 0. un 0 s m n _ , = d' 0 nz,;

These give u"" desired. I

Q s n f l m = 0 in

QM 2

M , whence sn+'m = 0, as

102

3 Projectives, Injectives, and Flats

Corollary 3.72: Every element of S-'M has the form s-' Q m, where s E S. Moreover, s; Q m , = s; Q m2 if and only if there is some a E S with o(s2ml - slm2) = 0.

'

Proof: A typical element of S - ' M = S-'R QR M has the form E x i Q mi, where xi E S-'R and mi E M . But xi = ri/si,where si E S, so that E x i Q mi = xri/siQ mi = E l l s i Q rimi = l o i / sQ rimi

(where s = n s j and ai =

nj,i

=

sj)

11s Q Cairimi= 11s Q m.

If s; Q m l = s;' 0m 2 , then (l/sls2)0(sZml - s,mL) = 0 and hence 1 Q (s2ml - slm2) = 0. By Theorem 3.71, there is o E S with a(s2ml slm2) = 0. I As a consequence of Corollary 3.72, one writes the elements of S - ' M as m/s, where m E M and s E 3, with addition and scalar multiplication as suggested by the notation.

Theorem 3.73: For any subset S of R, the localization S-'R is Jlat as an R-module. Proof: Assume f :M' -+ M is monic; we must show 1 6 f :S-'M' -+ S - M is monic. Let (1 Q f)(s-' Q m') = s-' Q fm' = 0 in S - ' M . Then multiplying by s gives 1 Q fm' = 0. By Theorem 3.71, there is o E S with afm' = 0. But ofm' = f(am'), so that f monic implies am' = 0. Since o is a unit in S-'R, however, 0 = s-' Q am' = o(s-' 0 m') implies s - ' 0m' = 0. I

Corollary 3.74: S-'

:,m

+ s - L RYJ

is an exact functor.

Proof: S-' is, by definition, the functor S-'R QR ; to say this functor is exact is precisely to say that S-'R is a flat R-module. I

We remark that if f : M - + M ' , then S - ' f : S - ' M - + S - ' M ' is given by mls I-+

(fm)ls.

Exercises: 3.48. If S1 c S c R, then S - ' ( S ; ' A ) = S - ' A = S;'(S-'A) for every R-module A. 3.49. If A, and A 2 are submodules of B, then S - ' ( A , n A 2 )= S-'A, n S-'A2. 3.50. Every R-map f : B -+ B , where B and B are S - ' R modules, is automatically an S-'R map. (Hint: Consider sf(m/s).) 3.51. The maps 8 ,: M -+ S - ' M define a natural transformation 8: 1 -+ s-1. 3.52. Find a functor G so that ( S - G) is an adjoint pair.

',

103

Localization

3.53. Prove that S - 'R ORS - ' R 2 S - 'R as S-'R-modules. (Hint: Consider multiplication on S - ' R.)

'

Lemma 3.75: I f B is an S - 'R-module, then B 2 S - B ; moreover, ij- f :B + B is an S-'R-map between S-'R-modules, then f = S - l f , Proof: Define cp:B -+ S - ' B by b H 1 Q b. I f cp(b)= 1 Q b = 0, then ab = 0 for some a E since a is a unit in S - R and B is an S - R-module, we have 0 = a-'(ob) = b. To see cp is epic, note that s-' Q b = cp(s-'b). The second statement is obvious from the identification s- 0b = b/s. Without identification, it says we have commutativity of

'

s;

'

'

.f

B

b

B'

The next theorem says S-' preserves nice modules.

Theorem 3.76: Let S be a subset of a commutative ring R, and let A be an R-module. (i) (ii) (iii) (iv) (v)

If If If If If

A is R-free, then S - ' A is S - 'R-,free. A isjg., then S - ' A i s f g . A isjnitely related, then S - A is jinitely related. A is R-projective, then S - ' A is S-'R-projective. A is R-injective, then S - A is S - 'R-injective.

Proof: (i) S-' preserves sums. (ii) If F + A + 0 is exact with F tg. free, then S - ' F + S - ' A + 0 is exact, with S - 'F f.g. free. (iii) If F , F , + A 0 is exact with F , and F, f.g. free, then S-'Fl + S - IF, -+ S - ' A -+0 has the same properties. (iv) S - preserves summands. (v) Consider the diagram of S - R-modules and maps -+

'

-+

'

S-'A

104

3 Projectives, Injectives, and Flats

where A is R-injective. By Lemma 3.75, this diagram may be viewed as S-'A

Since the diagram of R-modules and maps A

can be completed because A is R-injective, the map S-'g completes the original diagram; thus, S - ' A is S- 'R-injective. I It is true that S-' preserves flatness, but this requires a short lemma. Lemma 3.77: For R-modules B and A, there is a natural isomorphism S-'(B@,A)

%

S-'B@s-lRS-'A.

Proof: An elementary argument is to define an isomorphism by bringing an element of S - ' B oS1 R S - ' A into a normal form:

xbi/si 0 ai/ai = C(bi/siaJ0 ai =

(slip l/oi across tensor sign)

I l/siai)(bi ( 0 Ui).

A fancier argument is to show S-'(B 0 A ) satisfies the universal mapping problem for S - R-bilinear maps f : S- 'B

Define h:(b/s, a/.) reader. I

H

M (l/so)(b 0 a). Details of either version are left to the

Theorem 3.78: I f B is a flat R-module, then S - ' B is a jlat S-'R-module.

105

Localization

Proof: Let ~ : A + A ' be monic in IR%n; is l @ f : s - ' B @ s - I R A + S - ' B @ S - l R A' monic? By Lemma 3.75, we may ask whether

'

1 @ S - f :S - B OS- L RS- ' A -+ S

'B OS ~

S- 'A'

is monic; by Lemma 3.77, we may even ask whether

'

S - ( 1 @ f ): s '( B @ R A )

-+

'

S - ( B @R A')

is monic. The affirmative answer is now apparent: 1 @ f is monic because B is R-flat, and S-'(l o f )is monic because S - is an exact functor. I

'

This theorem may also be proved using Exercise 3.38. Corollary 3.79:

Let S , c S be subsets of R . For any R-module A, S-'A

= S - ' ( S ; ' A ) = S ; ' A @sl

i,S-'R.

Proof: In Exercise 3.48, weobserved that S - ' A = S-'(S; ' A ) = S; '(S- ' A ) (for all satisfy the same universal mapping problem). But S;'(S-'A)

= S;'(A@,S-'R) = S;'A

Since S,'(S-'R)

= S-'R,

OS,IRSI1(S-'R), by Lemma 3.77.

we have the desired formula.

I

The proper attitude is that S - I R is a less complicated ring than R. For example, if R is a domain, its quotient field is a localization (thus, Theorem 3.73 is a vast generalization of Corollary 3.48). More generally, if P is a prime ideal in R and S = R - P ( ( 0 ) is a prime ideal when R is a domain), S - 'R is a ring having only one maximal ideal. In this very important special case, one writes R,, M,, and f, instead of S ' R , S - I M , and S'f,where M is an R-module and f is an R-map. The next theorem illustrates how one may use localization to reduce a problem to a simpler setting. Theorem 3.80: (i) I f M is an R-module with M , = 0 for every maximal ideal P of R, then M = 0. (ii) I f f : M -+ N has f,: M , + N P monic. (or epic) for every maximal ideal P , then f is monic (or epic). Proof: (i) Assume M = 0, and choose m E M with m # 0. If I = { r E R:rm = 0}, then I is a proper ideal of R, hence is contained in some maximal ideal P. Now m/l E M , = 0, so Theorem 3.71 provides CJ E R - P, i.e., 0 $ P, with urn = 0. But 0 E I c P, a contradiction. (ii) Consider the exact sequence 0 -+ K -+ A4 1, N , where K = kerf. For each maximal ideal P, Corollary 3.74 gives an exact sequence

0 +K,

+M

/

P LX p .

By hypothesis, each f, is monic, whence A',, part (i). Therefore f is monic.

=0

for all P, and so K

= 0, by

106

3 Projectives, Injectives,and Flats

If each fp is epic, there is a similar argument using coker f.

I

A variation on the proofjust given is useful. Theorem 3.81: Let M be Jg. and S c R. Then S - ' M there exists o E S with oM = 0.

= 0 if

and only if

Proof: Let M = ( x l , . . . , x,). If S - ' M = 0, then each xi/l = 0 in S - ' M . By Theorem 3.71, there is oi E with oixi = 0. Note that o = n o i E and that oxi = 0 for all i. It follows that OM = 0. For the converse, oM = 0 implies o ( S - ' M ) = 0. O n the other hand, S - ' M is an S-'R-module, and o is invertible in S-'R. Therefore, multiplication by o is an automorphism of S - ' M (whose inverse is multiplication by l/o).It follows that M = 0. I

Our final desire in this chapter is to obtain an analogue of Lemma 3.77 for Hom in place of @ :we wish to show S - ' HomR(N,M ) Z Horn,- l R ( S - 'N, S - l M ) , but, in contrast to Lemma 3.77, some restriction must be imposed on N. To see that the formula is not generally true, take R = Z and S-'R = Q. If N = Q and M = Z, then Hom,(Q,Z) = 0, hence its localization is 0; on the other hand, @ Q, Q @ Z) = HOmQ(Q,Q) = Q. We first give two very general results. Lemma 3.82: Let A be an R-algebra, and N afinitely related R-module. For every A-module M , there is a natural isomorphism 8:HOmR(N,M ) 7 Hom,(N ORA, M ) given by f H y, where f"(x @ 1) = f ( x )for all x E N. Proof: We first establish an isomorphism when N is Eg. free. If N has a basis { e l , . . . ,e m ) , then N ORA has a basis _lei @ 1,. - .,em@ 11. Define 8:HomR(N,M ) + Hom,(N @ A , M ) by f H f , where f i e i @ 1) = f ( e i ) ,i = 1,. . . , n. That 8 is a well-defined isomorphism is a restatement of the definition of basis. Now assume N is finitely related, so there is an exact sequence R' -+ R" + N -+ 0.

There is a commutative diagram with exact rows: 0+

HOmR(N, M )

+

HOmR(RS,M ) + HOmR(R', kf)

107

Localization

As the vertical maps 8 are isomorphisms, the (dashed) induced map is also an isomorphism, and it has the desired formula. We leave the check of naturality to the reader. I Lemma 3.83: Let B b e p a t and Njinitely related. For any module M, there is a natural isomorphism $1

given by b 0g

B ORHOmR(N, M ) 2 Horn,( N , M ORB )

Hg b ,

where gb(n)= g(n)0b.

Proof: It is clear that $ arises from the R-bilinear function sending ( b , g )H g b , it is natural (in N ) , and it is an isomorphism when N is f.g. free. There is an exact sequence when N is finitely related:

R' + Rs+ N

+ 0.

Since B is flat, there is a commutative diagram with exact rows: 0 + B 0Hom(N, M ) + B 0Hom(R'. M ) + B 0Hom(R'. M )

0 + Hom(N, M 0B ) + Hom(R", M

0B ) -+ Hom(R'. M 0B )

Since the second two vertical maps are isomorphisms, so is the first.

I

Theorem 3.84: Let S c R, and let N be u,finitely related R-module. There is a natural isomorphism, for every R-module M ,

cp:s-' HornR(N,M)%Hom,.,R(S-'N,S-'M) given by g / 1 H g, where g ( x / l ) = y(x) 01 .for all x E N .

'

Proof: By definition, S - HomR(N,M ) = S- 'R ORHomR(N,M). Since S- R is flat (Theorem 3.73), Lemma 3.83 gives an isomorphism

'

s- ' HOmR(N,M) 2 HornR(N , s- 'M);

since S- R is an R-algebra, Lemma 3.82 gives an isomorphism HOmR(N,S-'M) 2 Horn,- IR(S-'N,S-'M). The composite is an isomorphism having the given formula.

4

Specific Rings

The time for examples has come. We pose twin problems: if conditions are imposed on projective, injective, or flat modules, how does this affect the ring of scalars R ; if we put conditions on R, how does this affect these special modules? We deviate from earlier style in that a number of results are stated without proof (though references are given). There are almost no exercises, for they would be more ring-theoretic than homological. We do, however, include a little ring theory to give the reader some feeling for the theorems, thereby removing some of theii- mystery. Of course, what is included and excluded is a matter of taste.

NOETHERIAN RINGS

Let us begin with a class of rings that arose several times in Chapter 3. Definition: A ring R is left noetherian if every left ideal is f.g. Examples: 1. Every principal ideal ring (possibly with zero divisors, e.g., Z / n Z , possibly noncommutative, e g , division rings). 2. If R is left noetherian, so is R [ x ] , polynomials in which x commutes with all constants in R (Hilbert Basis Theorem) [Lambek, 1966, p. 701. By induction on n, R left noetherian implies R [ x , , . . . ,x,] left noetherian.

108

109

Noetherian Rings

3. If R is commutative noetherian, so is any localization S - ' R [Kaplansky, 1970, p. 571. 4. Every quotient ring of a left noetherian ring is left noetherian. 5. If R is the ring of all matrices of the form ($ ,"), where a E Z and b, C E Q, then R is right noetherian but not left noetherian [Small, 19661. (Dieudonne gave an example of this phenomenon in the 1950s [CartanEilenberg, 1956, p. 161.) Theorem 4.1 : R is left noetherian if' und onlj, if ererj' submodule of a $g. left R-module M is also j g . Proof: Assume M = (xl,. . . , xJ, and S is a submodule of M . We do an induction on n that S is f.g. If n = 1, then M is cyclic, hence M z R/Z for some left ideal Z (Theorem 2.2). It follows that S 2 J / I , where J is some left ideal containing I . Since R is left noetherian, J , hence its image J/Z E S, is f.g. Suppose n > 1. If M' = R x , , there is an exact sequence

0 + M'

--+

M

+M/M'

-+

0,

and M / M ' can be generated by n - 1 elements. There is another exact sequence O+S n M ' + S + S / ( S n M')+O.

Now S n M' is f g , being a submodule of the cyclic module M', while S/(S n M ' ) E ( S + M')/M' c M / M ' is f.g. by induction. It follows that S is f.g. For the converse, the submodules of R , namely, the left ideals, are f.g., hence R is left noetherian. 1 Corollary 4.2: I f R is left noetherian, then ererj.J:g.module i s j n i t e l y related.

Proof: If 0 .+ K by Theorem 4.1.

+ F -,B -,0

1

is exact, where F is f.g. free, then K is f.g.

Corollary 4.3: Zf R is left noetherian, e r u y j : g . ,flat module is projectioe. Proof: Couple Corollary 4.2 with Corollary 3.58.

I

Note that we have identified the rings in the hypothesis of Corollary 3.34. module M has a free resolution Corollary 4.4: If R is left noetherian, erer.j~,fg.

. . . + F, in which each F , is $g.

+Fn-l+ . . . + F ,

+Fo+

M -+O

110

4

Specific Rings

Proof: Since M is fg., we may choose a f.g. free module F , and an exact sequence 0 + K O-+F , --* M -+ 0. Since R is left noetherian, K O is f.g., so we may choose a f.g. free module F , mapping onto K O and obtain an exact sequence 0 + K , + F , + K O+ 0. Moreover, K , is f.g., so we may iterate this process as in Theorem 3.8. I

Definition : A module M (over any ring) has ACC (ascending chain condition) if every ascending chain of submodules M , c M , c M3c”

stops, i.e., there is an integer n with M , = M,,

= M,+

Theorem 4.5 : A module M (over any ring) has ACC module of M is Jg.

,= . . . .

if and only if every sub-

Proof: Assume every submodule is tg. and let M , c M 2 c . . . be an ascending chain of submodules of M . If M* = u Z l M i , then M* = (x,, . . . , x k ) , by hypothesis. Each x j E Miu); let n = max i ( j ) . Then M* c M , c M*, so that the chain stops. Assume S is a submodule of M that is not f.g. Choose s1 E S and define M , = R s , . Assume, by induction, that we have chosen sl,. . . , s, E S so that if M i = (sl,. . . ,s,), then M , S M , 5 . - . 5 M,. Since S is not f.g., M , is a proper submodule of S ; we may thus choose s,+ E S with sn+ .$ M,. Define M,+, = M , R S , + ~so, that M , 5 M , + l . By the axiom of choice,’ there is an ascending sequence of submodules of M , namely, M c M , c . . . , that plainly does not stop. This contradicts ACC. I

,

+

,

A module with ACC is often called “noetherian” because of this theorem. Corollary 4.6: A ring R is left noetherian if and only ideals.

if it has ACC on left

Proof: If we consider R as a module, its submodules are the left ideals.

Corollary 4.7:

I

I f R is left noetherian, every f i g . module has ACC.

Proof: Theorems 4.1 and 4.5.

I

Lemma 4.8: Let A be a module (over any ring) and cp: A I f A has ACC, then cp is an isomorphism.

-+

A an epimorphism.

This is a subtle logical point. Induction allows us to choose a suitable s, for each integer n. The axiom of choice allows us to make all these choices simultaneously. Of course, one could phrase this more formally.

111

Noetherian Rings

Proof: Let

K , = kercp and let K z = ker(cpz), where cpz = cpq. Clearly K , c K , and cp2 is epic. There is a commutative diagram

The version of the Third Isomorphism Theorem in Exercise 2.10 shows that if K , # 0, then K 1 is a proper submodule of K , . This procedure may then be iterated, by induction, to obtain ker(cp")S ker(cp"+'). Since A has ACC, it must be that K , = kercp = 0, so that cp is an isomorphism. I Theorem 4.9: If R is left noetherian, then R has IBN. Moreover, i f A is free of rank n, then any generating set with n elements is a basis.

Proof: Assume A is free and A 2 R" 2 R", where m 2 n (R" is the sum of m copies of R). If rn > n, there is an epimorphism cp : A + A having a nonzero kernel. But A has ACC, by Corollary 4.7. Therefore cp is an isomorphism, by Lemma 4.8, contradicting ker cp # 0. Let A have a basis {xl,. . . , x n } and let A = ( u l , . . . , a,,). Define an epimorphism cp :A --* A by x i w a,. If ker cp # 0, we reach a contradiction as above. I Theorem 4.10:

T h efollowing are equivalent jor a ring R.

(i) R is left noetherian; (ii) every direct limit (directed index set) oj'injective modules is injective; (iii) every sum of injective modules is injectiile. Proof: (i) e=(ii) Let (Ei, cp;] be a direct system of injective modules over a directed index set. By Theorem 2.17, Ei consists of elements [ei],ei E Ei,

where [ei] = ,liei+ S and li:E i

1 5

4

E iis the ith injection.

112

4

Specific Rings

Consider the diagram

where J is a left ideal. Since R is left noetherian, J = (a,,. . . ,a,). If F is free on {xl, . . . ,x,} and we map F + J by x i H a,, there is an exact sequence 0 + K + F + J + 0. Moreover, Theorem 4.1 gives K = ( y , , . . . , y,); each y, = c r p j x j ,where rpj E R . The idea is to force f to take values in some one E , . In detail, faj = [ei(j,].Since the index set is directed, there is an index k 2 i(j), all j . Define & E E,, j = 1, . . . , n, by d = q$"eicj), so thatfaj = [ei(j,]= [a].Each y, in K yields [ z r p j e J ]= 0. By Theorem 2.17, there is an index q ( p ) 2 k with ' p ~ ( p ~=~0. pFinally, j d choose an index m 2 q ( p ) , all p , and define bi E Em by b-' = pie'. The end result of these choices is that E Em,j = I, . . . ,n, faj = [g], and xrpjg= 0, all p . Define f ' : J Em by aj H .'b Note that f ' is well defined, for we have taken care that all the relations are sent to 0. Since Em is injective, there is a map g ' : R + Em extending f'.Finally, define g : R + I@ Ei by r H [g'r]. The map g does extend f,for ~

--f

guj

= [ S ' U j ] = [f'Uj]

=

[bj] = faj.

Therefore, 1 5Ei is injective. (ii)*(iii) A finite sum of injectives is injective and, by Exercise 2.33, every sum is a direct limit, with directed index set, of its finite partial sums. Thus HEi is injective. (iii) * (i) (This implication is due to Bass.) We show that if R is not noetherian, there is an ideal I and a map from I to a sum of injectives that cannot be extended to R . If R is not noetherian, there is a strictly increasing . . . Let I = I,; note that I / I , # 0 sequence of left ideals I , f I? for all n. Imbed I / I , in an mjective module En. We claim that U E , is not injective. Let n,:l-+ I / I , be the natural map.-'For each a E I , n,u = 0 for large n, so the map f:I + n E , given by

2..

a H ( q a , nza,

u,"=

. . . , n,u, . . .) = (n,a)

does have its image in LIE.. We regard f as a map I + H E , . Suppose there were a map g : R H E n extendingf. Write g ( 1 ) = (x,). Choose m and choose --f

113

Noetherian Rings

a E I , a 4 I,. Now n,a # 0, so that ga = f a has nonzero rnth coordinate n,a. But g(a) = ag(1) = a(x,) = (ax,). Therefore n,u = ax,, so that x, # 0. Since rn is arbitrary, we have contradicted the fact that almost all the coordinates of an element in a sum are zero. Let us give a direct proof that R left noetherian implies HEi is injective when each Ei is injective. Consider the diagram

O

-

l

L

R

where I is a left ideal. Since R is left noetherian, I = ( a l , . . . , a,). For each j , the “vector” faj in H E i has only finitely many nonzero coordinates; since there are only finitely many a j , { fu, , . . . ,.fun} collectively involve only finitely many Ei, say, Ei,, . . . , E i m . It follows that imf c E , , 0 . . . @ Eim, which is injective, being a finite sum of injectives. There is thus a map g : R -,Ei, 0 .. . @ Eim extending ,f, which we may regard as having its image in the larger module L I E i . We quote a theorem of Chase analogous to Theorem 4.10. Definition: A ring R is left coherent if every f.g. left ideal is finitely related. Examples: 6. Every left noetherian ring is left coherent. 7. If k is a field, then the ring of polynomials over k in infinitely many variables is coherent (but not noetherian). Theorem A (Chase): A ring R is k f t coherent offlat lefl R-modules is Pat.

if and only if every product

Proof: [Anderson-Fuller, 1974, p. 2291.

Another important class of rings is dual (in the lattice-theoretic sense) to noetherian rings. Definition: A ring R is left artinian if it has DCC (descendingchain condition) on left ideals (i.e., every descending chain of left ideals I EI I, 3 . . . stops). Examples: 8. If k is a field, every finite-dimensional k-algebra R is left (and right) artinian.

114

4

Specific Rings

9. A product of finitely many left artinian rings is left artinian (in a product of rings R = n R i , each Ri is a two-sided ideal of R). 10. Every finite ring is left (and right) artinian. 11. There exist rings that are left artinian but not right artinian. (See [Faith, 1973, p. 3371 for an example; it consists of certain suitably restricted 2 x 2 triangular matrices.) 12. Every left artinian ring is left noetherian (Hopkins-Levitzki Theorem) [Anderson-Fuller, 1974, p. 1721. Definition: A ring R is left perfect ifit satisfies DCC on principal right ideals.

(This curious mixing of left and right is explained by the theorems to come.) Examples: 13. IfR is left or right artinian, then R is left perfect [AndersonFuller, 1974, p. 3181. 14. A ring may be left perfect but not right perfect [Anderson-Fuller, 1974, p. 3221. Theorem B (Bass): A ring R is left perfect if and only R-module is projective.

if every p a t left

Proof: [Anderson-Fuller, 1974, p. 3151. Theorem C (Chase): Every product of projective left R-modules is projective if and only if R is left perfect and right coherent. Proof: [Chase, 19601. TheoremD (Chase): If R is commutative, every product of projective R-modules is projective if and only (f R i s artinian. Proof: [Chase, 19601. Theorem E (Bass): A ring R is left perfect ifand only ifevery left R-module has a projective cover. Proof: [Anderson-Fuller, 1974, p. 3151.

We shall return to projective covers later in this chapter (Theorems 4.46 and 4.50). Exercises: 4.1. and injective.

If R is a division ring, every left R-module is projective

115

Semisimple Rings

4.2. If R is a domain but not a field, an R-module that is simultaneously projective and injective must be 0. 4.3. If R is left artinian, then R contains a minimal left ideal S, i.e., S # 0 and S contains no smaller nonzero left ideal.

SEMISIMPLE RINGS We are going to characterize those rings for which every module is projective. Definition: A module A is simple (or irreducible) if A # 0 and A has no proper submodules; A is semisimple if it is a sum of(possib1y infinitely many) simple modules. Theorem 4.11 : A module A is semisimple if and only ifevery submodule of A is a summand.

Proof: Suppose A is semisimple: A = uk E K sk,where S k is simple. Given a subset I c K , define S, = Uk.1 s k . If B is a submodule of A, choose, by Zorn’s lemma, a subset I of K maximal with S , n B = 0. We claim A = B @ S , ; to prove this, it suffices to show S k c S , + B for every k E K . Clearly this is so for k E I. If k $ I, then maximality gives ( S , + S,) n B # 0. Writing what this means in terms of elements, one sees ( S , B) n Sk # 0. Since Sk is simple, ( S , B) n s k = S,, i.e., S k c S , + B. Suppose, conversely, that every submodule of A is a summand. We first show each nonzero submodule B contains a simple submodule. If b E B, b # 0, Zorn’s lemma gives a submodule C of B maximal with b $ C. Since every submodule of B is a summand of A, hence of B (Exercise 2.23), B = C @ D for some submodule D. We claim D is simple, for if D’ is a proper submodule of D, then D = D’ @ D“ and B = C @ D = C @ D’ @ D”. Either C 0 D’ or C 0 D“ does not contain b, contradicting the maximality of C . Second, we show A is a sum of simple modules. By Zorn’s lemma, there is a family of simple submodules of A , jS,:k E K ) , maximal such that the submodule M they generate is their sum D S k . By hypothesis, A = M @ B for some B. If B = 0, we are done. If B # 0, then B = S @ B’ for some simple S, by our first argument. The family { S , S , : k E K j violates the maximality of { & : k E K } , whence A is semisimple.

+

+

Corollary 4.12: Every submodule and eirery quotient of u semisimple module A is semisimple.

116

4

Specific Rings

Proof: If B is a submodule of A, Exercise 2.23 shows the criterion of Theorem 4.11 applies to B. If A / B = C , then B is a summand of A, so that C is isomorphic to a submodule of A, hence is semisimple.

Definition : A ring R is left semisimple if, as a left R-module, it is semisimple. Since the submodules of R are its left ideals, R is a sum of simple left ideals. Moreover, since R has a unit, it is easy to see R is a sum of a finite number of simple left ideals. Of course, a simple left ideal is just a minimal left ideal. Example: 15. Wedderburn’s theorem asserts that every semisimple ring R is a finite product of full matrix rings over division rings: R E ni= End,,( where End,$(Vi) = Horn,#( and vl. is a left vector space over a division ring Ai (say, of dimension ni).Moreover, the numbers r and niand the division rings Ai are uniquely determined by R.

v),

v, v),

It follows from Wedderburn’s theorem that R is left semisimple if and only if it is right semisimple. Proofs of this theorem abound in the literature. In particular, it is shown in [Anderson-Fuller, 1974, p. 1541 that R is semisimple if and only if R is a finite product of simple artinian rings (where “simple” means no nontrivial two-sided ideals). Here is a proof, due to Rieffel,2 that R simple left artinian implies R E End,(V) for a division ring A and a left vector space V over A. If V is a minimal left ideal of R (which exists, by Exercise 4.3), then A = End,( V ) is a division ring (an elementary fact called Schur’s lemma). Now V is a left A-module if we write endomorphisms on the left, and I/ is finite-dimensional (since R is left artinian). Define cp: R + End,( V ) by cp,(u)

= ru,

r E R, u E V .

One checks easily that cp is a ring map. Since cp is not identically 0 (it preserves the identity), ker cp # R ; since R is simple, cp is one-one. To see that cp is onto, we first show that cp( V )is a left ideal in End,( V ) :indeed, we show fcp, = cprcu, for all f E End,( V ) and all u E V . If u E V , then u H uu E End,( V ) = A, so that f a A-map says

f ( U 4 = (f(4)u (i.e.,fcommutes with the right multiplication by u). This last equation may all u E V , as claimed. be restated as (fcp,)u = (~,-(~p, Now VR is a nonzero two-sided ideal in R (for 1 E R), so that R = VR. Hence q(R) = cp(V)cp(R) is a left ideal in EndJV), which must be the whole I am grateful to I. Kaplansky for calling my attention to this proof.

Semisimple Rings

ring since cp(R) contains the identity of EndA(R).Therefore cp: R is an isomorphism of rings.

117 + EndA(V)

Definition: Let G be a (multiplicative) group and k a commutative ring. The group ring kG is the k-algebra whose additive structure is that of a free k-module with basis the elements of G , and whose multiplication is given by the multiplication in G and the distributive law. Examples: 16. Maschke's theorem [Curtis-Reiner, 1962, p. 411 says that if G is a finite group and k is a field whose characteristic does not divide the order of G, then kG is semisimple. (See Theorem 10.28.) 17. If k is a field of characteristic 0, then R = k[x]/(x" - 1) is semisimple (for R is just kG, where G is cyclic of order n). 18. R = Z/nZ is semisimple if and only if n is square free.

Here is the reason we have been discussing semisimple rings. Theorem 4.13: The following are equivalent for a ring R :

(i) (ii) (iii) (iv) (v)

R is semisimple; every left R-module is semisimple; every left R-module is injective; every short exact sequence of leji R-modules splits; every left R-module is projective.

-

Proof: (i) (ii) Since R is semisimple as a module, every free module is semisimple. By Corollary 4.12, every quotient of a free module is semisimple. But this says every module is semisimple. (ii) * (iii) If M is a module, then M may be imbedded in an injective module E. By Theorem 4.11, M is a summand of E, and hence is injective. (iii) (iv) Theorem 3.19. (iv) * (v) If M is a module, there is a short exact sequence 0 -+ K -+ F -,M -,0 with F free. As this sequence splits, M is a summand of F , hence is projective. (v)+(i) If I is a left ideal of R , then the module R/I is projective, by hypothesis. By Corollary 3.13, I is a summand of R . By Theorem 4.1 1, R is semisimple as an R-module, whence R is a semisimple ring. I Remark: It now follows from Corollary 3.42 that if R is semisimple, so is R,, the ring of all n x n matrices over R . This may also be proved directly from Wedderburn's theorem.

This is an appropriate place to show that projective modules really do occur naturally. The next theorem describes separability of field extensions

118

4

Specific Rings

in such a way that Galois theory may be done for commutative rings, not just for fields. Theorem 4.14: Let L and k be Jields, with L a jinite separable extension of k. Then L is a projective L @k L-module. Proof: First of all, Exercise 1.13 shows that L @ k L is a k-algebra, where multiplication is given by (II @ lz)(r10 r2)= 1,1; @ 41;.It is obvious that L is an (L - L)-bimodule, so that Exercise 1.14 shows that L is a module over L r& L (note that the exercise does apply, for L commutative allows us to identify L with Lop).To prove the theorem, it suffices to prove L @k L is a product of fields, for then it is semisimple and every module is projective. Since L is a finite separable extension of k, the theorem of the primitive element [Lang, 1965, p. 1851 provides an element a E L with L = k(a). Let f(x) E k[x] be the irreducible polynomial of a. There is an exact sequence of k-modules

O+(f)-+k[x]~L+O, where (f) is the principal ideal generated by f(x); moreover, 7t is a k-algebra map. Since k is a field, every k-module is flat, so we have exactness of 0

(f)

@k

L + k[x]

@k

L

aL

@k

L + 0.

Now n @ 1 is a ring map, even a k-algebra map, so that (f) in k[x] @k L . Therefore, there is a k-algebra isomorphism

@k

L is an ideal

L @k L Z k[x] @ L/(f) 0 L.

If L[y] is the polynomial ring with indeterminate y, then there is a k-algebra isomorphism 8:k[x] @k L + L[ y] given by g(x) 0 1 H M Y ) (we identify x @ 1 with y); moreover 8 takes (f) @ L onto (f(y)). Therefore, 2 L[Y]/(f(Y)).

Now f , though irreducible over k, may factor over L. However, separability tells us there are no repeated factors:

ff~ = npi(Y) )

in

‘CYI~

where the pi(y) are distinct irreducible polynomials. The principal ideals (pi(y ) ) are thus distinct maximal ideals in the principal ideal domain L[ y ] ,

119

Von Neumann Regular Rings

so the Chinese Remainder Theorem3 gives a k-algebra isomorphism

n

L [ Y ] / ( ~ ( Y ) )( L [ Y ] / ’ ( P ~ ( Y ) ) ) i

This concludes the proof, for each L [ y ] / ( p i )is a field.

I

The converse of Theorem 4.14 is true [DeMeyer-Ingraham, 1971, p. 491, so finite separable field extensions may be characterized in terms of projectivity. Notice that the characterization “ L is a projective L @k L-module” makes sense for k any commutative ring and -Lany commutative k-algebra.

VON NEUMANN REGULAR RINGS The next class of rings deals with flatness. Definition: A ring R is von Neumann regular if, for each a E R , there is an element a’ E R with au’a = 0. Examples: 19. Boolean rings ( r Z = r for every r E R). 20. R = Endk(V),where V is a (possibly infinite-dimensional) vector space over a field k. 21. Semisimple rings are von Neumann regular (this follows easily from Example 20 and the Wedderburn theorem). Lemma 4.15: I f R is von Neurnann regular, erery J g . left ideal is principal, generated by an idempotent (an element e with e2 = e).

Proof: First of all, let us show every principal left ideal Ra is generated by an idempotent. There is an element a‘ E R with ua‘a = a. It follows that e = a‘a is idempotent. Moreover, Ra = Re, for a E Re and e E Ra. To show that an arbitrary f.g. left ideal is principal, it suffices, by induction, to prove Ra Rb is principal. There is an idempotent e with Ra = Re. Also, Ra Rb = Re Rb(1 - e): both a and b are in the right side; both e and b( 1 - e) are in the left side. As in the first paragraph, there is an idempotent f with Rb( 1 - e) = Rf ;moreover, f = rb(1 - e)for some r E R. It follows that f e = 0. We do not know whether ef = 0, so we adjust f .

+

+

+

If R is a ring and I , , . . . , I , are “pairwise comaximal” two-sided ideals, i.e., I j + I , = R for all j # k, then R / n I j 2 n ( R / f j )as rings. In our case. i t is easy to see n i ( p i ( y ) = ) (/I!)).

120

4

Specific Rings

Define g = (1 - elf. Then g2

= (1 - e)f(l -

e)f = (1 - e)(f - fe)f ge = 0 = eg, and

= (1 -

elf’ Rg = Rf.

= (1 -

elf

= g,

Therefore, Ra + Rb = Re + Rg. We claim that Re + Rg = R(e + 9). Clearly Re + Rg 3 R(e + 9). For the reverse inclusion, note that re sg = (re + sg)(e + g ) E R(e + g ) (since eg = 0 = ge). I

+

Theorem 4.16: A ring R is von Neumann regular R-module is flat.

if and only if every right

Proof: Assume R is von Neumann regular, and let B be a right R-module; let-0-+ K -+ F + B 0 be exact, where F is free. By Theorem 3.55, it suffices to prove K I = K n F I for every f.g. left ideal I . By Lemma 4.15, I = Ra for some a E R. We need only show that if k E K and k =fa E Fa, then k E Ka. But --f

k

= fa = faa’a = ka‘a E Ka.

Therefore B is flat. For the converse, assume a E R . By hypothesis, the cyclic right R-module RJaR is flat. Theorem 3.55 applied to the exact sequence 0 -,aR -+ R -+ RJaR -+ 0 gives (aR)I = aR n RI = aR n I

for every left ideal I . In particular, if I = Ra, then aRa = aR n Ra and a E a R n Ra=aRa.

This is thus some a’ E R with a = aa’a, and R is von Neumann regular.

1

Here is a stupid argument that R semisimple implies R von Neumann regular. If R is semisimple, every module is projective, hence flat, and so R is von Neumann regular.

HEREDITARY AND DEDEKIND RINGS

We have seen how the assumption that every module is “special” forces constraints on R . Let us now assume every ideal (or every f.g. ideal) is special. Definition: A ring R is left hereditary if every left ideal is projective. A Dedekmd ring is a hereditary d ~ m a i n . ~ Recall that “domain” means “commutative integral domain”

121

Hereditary and Dedekind Rings

Examples: 22. Every semisimple ring is left hereditary. 23. Every principal ideal domain R is hereditary, hence is a Dedekind ring (for every nonzero ideal is isomorphic to R). 24. Small’s ring of triangular matrices (Example 5 ) is left hereditary but not right hereditary. For the first example of this phenomenon, see [Kaplansky, 1958bl. There are more examples after Corollary 4.26. Theorem 4.17 (Kaplansky) t#R is lefi hereditary, then every submodule of a free module is isomorphic to a sum of left ideals.

Proof: Let F be a free module with basis [x,: k set K is well ordered. For each k E K , define

Fk

=

U

E

K } , and suppose the index

R.Xi;

ispan S/pS, which will suffice. If s = (miei)E S, then almost all its coordinates mi are divisible by p . There is thus an integer N so that

+

N

s-

1 miei = ps'

i= 1

for some s' E S. In SIPS, the coset of s is thus a finite linear combination of cosets of e , . I

123

Hereditary and Dedekind Rings

Using the continuum hypothesis, Specker proved that B

=

((miei)E G:there exists N with lmil I N for all i},

the subgroup ofall bounded sequences, is free. Nobeling [Fuchs, 1973,p. 1731 was able to generalize this result in two ways; first, the continuum hypothesis need not be assumed; second, the index set of the product need not be countable. The cited proof is an elegant one due to Bergman. Corollary 4.18 characterizes left hereditary rings. In order to see this, we first prove a lemma. Lemma 4.22: A module P is projective g’ and only $ every diagram with Q P

injective can be completed to a commutaticr diuyrum. The dual is also true. Proof: If P is projective, the diagram can be completed with no hypothesis on Q . For the converse, consider the diagram P

we want a map P + A making the diagram commute. There is an injective Q and a monk 0:A + Q. Imbed the dpgram in P

O+A‘

cri

+Q-

Q”

0

124

4

Specific Rings

where Q” = coker oi,71 is the natural map, and p exists by diagram chasing. By hypothesis, there is a map y: P -+ Q giving commutativity. Another diagram chase gives im y c im o. Theorem 4.23 (Cartan-Eilenberg): ring R :

The joflowing are equivalent for a

(i) R is left hereditary; (ii) every submodule of a projective module is projective; (iii) every quotient of an injective module is injective. Proof: (i)

(ii) Corollary 4.1 8. (ii) * (i) R itself is projective, being free, and hence its submodules, the left ideals, are projective. (iii)+.(ii) Consider the diagram with exact rows

Q

-

Q”

0

where P is projective and Q is injective. By Lemma 4.22, it suffices to find a map P’ Q making the diagram commute. Now Q” is injective, by hypothesis, so there exists a map P -+ Q” giving commutativity. Since P is projective, there is a map P Q giving commutativity. The composite P‘ -+ P -+ Q is the desired map. (ii) 3 (iii) Dualize the argument just given, using the dual of Lemma 4.22. I -+

-+

Recall Theorem 3.23 : every injective module is divisible; recall Exercise 3.14: every quotient of a divisible module is divisible. We conclude that divisible modules may not be injective, i.e., the converse of Theorem 3.23 is false. If R is not left hereditary, there exists an injective having a noninjective quotient. Another consequence of Theorem 4.23, using Corollary 3.42: if R is left hereditary, so is the matrix ring R,. (This result may be proved without the categorical Corollary 3.42.) Let us show that the definition of Dedekind ring above coincides with a more classical definition. Definition: Let R be a domain with quotient field Q. An ideal I in R is

125

Hereditary and Dedekind Rings

invertible if there are elements a , , . . . , a,

E

I , q , , . . . , q,

E

Q with

(i) q i l c R , i = 1, . . . , n ; (ii) 1 = qiai. Remarks: 1. Every nonzero principal ideal Ra is invertible: choose a, = a and q1 = l/a. 2. If Z is invertible, then I is generated by a,, . . . , a,. 3. Define I - ' to be the R-submodule of Q generated by q , , . . . , 4,; theniflisinvertible,II-' = R = l ~ l l , w h e r e l l= ~ '( ~ a j y j : a j E Z , y j E I - l ) . 4. A fractional ideal is a f.g. nonzero R-submodule of Q. All the fractional ideals form a commutative monoid under multiplication, with unit R. We shall see, in Corollary 4.25, that every nonzero ideal in a Dedekind ring R is invertible. It follows that the family of all fractional ideals forms an abelian group (which turns out to be free with basis all prime ideals). One defines the class group of R as the quotient group of this group by the subgroup of all principal ideals. 5. One can show I - z HomR(l,R). Theorem 4.24: If R is a domain, a nonzero ideal I is projective if and only if it is invertible. Proof: If I is projective, it has a projective basis (Theorem 3.15): there are elements (ak:k E K ) c I and maps ( P k : I + R such that

(1) if a E I , almost all qka = 0; (2) ifa E I, then a = x(qka)ak. If b E I and b # 0, define qk E Q by qk

= qk(b)/h.

Note that qk does not depend on the choice of b : if b'

E

I and b # 0, then

b'Vk(b)= Vk(b'b)= Pk(bb')= bq,(b'),

so that q,(b)lb = cp,(h')/b'. It follows that qkl c R for all k : if b E I and b # 0, then qkb = [qk(b)/b]b= qk(b)E R . By condition (l), if b E I and b # 0, then almost all qk(b)= 0. Since qk = qk(b)/b,there are only finitely many nonzero qk. Finally, condition (2) gives, for b E I ,

126

4

Specific Rings

If we discard all ak for which q k = 0, there remain finitely many a, E I . Moreover, if b # 0, we may cancel b from both sides of the above equation to obtain 1 = &a,. Therefore, I is invertible. Conversely, assume I is invertible, and let a,, . . . , a, E I , q l , . . . ,q,, E Q be as in the definition. Define pk:I -+R by a H qka (which lies in R because c R). If a E I , then c(qka)ak

= xqkaak = a x q k a k = a.

Therefore, I has a projective basis, hence is a projective module.

I

Corollary 4.25: A domain is a Dedekind ring if and only if every nonzero ideal is invertible.

This corollary is the link to the classical notion of Dedekind ring [Zariski-Samuel, 1958, p. 2751. Corollary 4.26: Every Dedekind ring is noetherian. Proof: Invertible ideals are always f.g.

I

Remarks: 1. It can be shown [Zariski-Samuel, 1958, p. 2791 that every ideal in a Dedekind ring can be generated by two elements. 2. The ring of integers in an algebraic number field is a Dedekind ring [Zariski-Samuel, 1958, p. 2831. In particular, there are Dedekind rings that are not principal ideal domains; the easiest example is R = {u b m : a ,b E Z } . 3. The polynomial ring R in noncommuting variables over a field is left and right hereditary; it can be shown [Cohn, 1971, p. 801 that every ideal in R is free. There are thus hereditary rings that are not noetherian.

+

We now generalize Theorem 3.24. Theorem 4.27: A domain R is a Dedekind ring if and only module is injective.

if every divisible

Proof: If every divisible module is injective, then every quotient of an injective is divisible, hence injective.Theorem 4.22 shows that R is hereditary, and it is Dedekind since it is a domain. Assume R is Dedekind. If D is divisible, we show that, for every nonzero ideal I and every map f : I + D, there exists an extension g off to R . Since I is invertible, there are elements al, . . . ,a,, E I , q l , . . . , q,, E Q with Cqiai = 1 and qiI c R. Since D is divisible, there are elements di E D with fai = aidi. IfaE I ,

f a = f(xqiaia) = c(qia)fai = C(qia)aidi = ax(qiai)di-

Semihereditary and Priifer Rings

127

If we set d = C(qiai)di,then d E D and f a = ad for all a E I . Define g :R -+ D by r H rd. Therefore D is injective, by Theorem 3.20. I As long as we are discussing projective ideals, let us note the following fact.

Theorem 4.28: I f R is a unique factorization domain (UFD = factorial ring), an ideal I is projective if and only if it is principal. Proof: In a domain R, every nonzero principal ideal is isomorphic to R, hence is projective. Conversely, assume I is projective and nonzero, hence invertible. Choose elements a,, . . . , a, E I , q 1,. . . , q, E Q with 1 = &ai and qil c R. Write qi = bi/ci, and assume, by unique factorization, that bi and ci have no nonunit factors in common. Since (hi/ci)uj E R, we must have ci dividing a j , all i, j . Let c = lcm{ci}. We claim I = Rc. First, c = CCbiUi/Ci = C ( h i C / C i ) U i E I ,

for bic/ci E R ; therefore, Rc c I . For the reverse inclusion, ci divides aj for all i, j implies c divides a j , all j , whence u j E Rc for all j . I

SEMIHEREDITARY AND PRUFER RINGS Definition: A ring R is left semihereditary if every f.g. left ideal is projective. A semihereditary domain is called a Priifer ring.s Examples: 25. Every left hereditary ring R is left semihereditary (of course, if R is left noetherian, the distinction vanishes). 26. Every valuation ring(i.e., a domain such that, given any two elements, one divides the other) is a Priifer ring. (In such a ring, every f.g. ideal is principal; these special Priifer rings are called Bezout rings.) 27. Let X be a noncompact Riemann surface, and let R be the ring of all complex-valued analytic functions on X . Helmer [1940] proved that R is a Bezout ring. 28. Every von Neumann regular ring R is left and right semihereditary (this follows quickly from Lemma 4.15). 29. The ring of all algebraic integers is a Btzout ring [Kaplansky, 1970, p. 721. 30. Chase, [1961] gives an example of a left semihereditary ring that is not right semihereditary. Recall that “domain” means “commutative integral domain”.

128

4

Specific Rings

We remark that left semihereditary rings are left coherent, for f.g. projectives are finitely related. Theorem 4.29: I f R is leji semihereditary, then every j : g . submodule A of a free module F is a sum of$nitely many f . g . left ideals.

Proof: Let F have basis { x k :k E K } . Since A is f.g. and each generator of A is a finite linear combination of x i s , we see that A is contained in a free summand of F generated by finitely many x i s . We may, therefore, assume F is free with basis {xl, . . . , x,}. We do an induction on n that A is a finite sum of f.g. ideals. If n = 1, then A is isomorphic to a f.g. ideal. If n 1, define B = A n (Rx, 0 * . 0 Rx,- J. Each a E A has a unique expression a = b + rx,, where b E B and r E R. If cp:A + R is defined by a c* r, then there is an exact sequence

=-

O + B + A % I -+0,

where I = im cp is a f.g. ideal of R. Since I is projective, this sequence splits . . @ Rx,- the inductive and A 2 B 0I . Since B is contained in Rx, 0. hypothesis gives B, hence A, a finite sum of f.g. ideals. I

,

The reader has undoubtedly observed that the proofjust given is Theorem 4.17 stripped of its transfinite apparel. Albrecht [1961] proves that every (perhaps not f.g.) projective module over a left semihereditary ring is a sum of f.g. left ideals. Theorem 4.30: A ring R is left semihereditary if and only submodule of a projective module is projective.

if every f . g .

Proof: Let A be a f.g. submodule of a projective module P. Now ' P is a submodule of a free module F (even a summand). By Theorem 4.29, A is a sum of Eg. ideals, each of which is projective since R is semihereditary. Therefore, A is projective. For the converse, R itself is projective, hence its f.g. submodules, the f.g. left ideals, are projective. Therefore, R is left semihereditary. 1

Let us now say a few words about Priifer rings. Note first that if R is any domain and M an R-module, then its torsion submodule is defined by t M = { m E M:rm = 0 for some r e R, r # O } .

Because R is a domain, tM is a submodule. Moreover, MItM is torsion-free, i.e., its torsion submodule is 0.

129

Semihereditary and Priifer Rings

Lemma 4.31: Zf R is a domain with quotient field Q, then every torsion-free R-module A can be imbedded in a vector spaee over Q. I f A is torsion free andJg., then A can be imbedded in cr,f.g.frreR-module. Proof: Imbed A in an injective module E . Since A is torsion free, A is also imbedded in EItE, which is torsion free and divisible. By Exercise 3.20, EItE is a vector space over Q . Assume A is f.g. If one chooses a basis of E/rE, then each of the generators of A is a linear combination of only finitely many basis vectors. Therefore, we may assume A imbedded in a finite-dimensionalvector space V , namely, the subspace of EItE spanned by the finitely many basis vectors just mentioned. Let V have basis { u l , . . . , v,). If A is generated by { a l , .. . ,a,,}, then , r i j , s i j E R . If s is the product of all s i j , then each a, = c j ( r i j / s i j ) u jwhere Is-'ul, . . . , s-lu,,,} is independent in V , and the R-submodule B of V generated by this set is free. Clearly A is imbedded in B. I Theorem 4.32: A domain R is a Priijkr ring free module A is projective.

if

and only

if every J g torsion-

Proof: If R is a Priifer ring, it is a domain, and so Lemma 4.31 shows A may be imbedded in a free module. By Theorem 4.30, A is projective. Conversely, every submodule of R is torsion free, so every ideal is torsion free and every f.g. ideal is projective. I

Theorem 4.33: Zf R is a Pruter ring, then a module B is p a t if and only if it is torsion free. Proof: Suppose B is flat, and 0 -+ K -, F % B --* 0 is exact, where F is free. By Theorem 3.55, K n F I = KI for every ideal I . In particular, this holds when I is principal. Suppose b E B and rh = 0 for some r # 0. If x E F and cpx = b, then rx E K n rF = r K ; hence r x = rk for some k E K . But F is torsion free and r(x - k ) = 0, so that x = k E K . Therefore b = cpx = 0, and B is torsion-free. Suppose B is torsion-free. By Corollary 3.48, it suffices to show every f.g. submodule A of B is flat. Since R is Priifer. Theorem 4.32 shows A is projective, so it is surely flat. I Remark: The first half of the proof shows that flat modules are torsion-free for every domain R. Thus, projective generalizes free, injective generalizes divisible, and flat generalizes torsion-free.

130

4

Specific Rings

Instead of assuming ideals are projective, what happens if we assume they are flat? Flatness gives us nothing new if we assume rings are noetherian. Theorem 4.34: A leji noetherian ring R is leji hereditary leji ideal is p a t .

if and only if every

Proof: If R is left hereditary, every left ideal is projective, hence flat. Conversely, every left ideal is f.g. since R is left noetherian. By Corollary 4.3, f.g. flat modules are projective, so every left ideal is projective and R is left hereditary. I

In Theorems 9.24 and 9.25,we shall classify all (not necessarily noetherian) rings in which every ideal is flat. It will be seen that this class contains all left semihereditary rings and all right semihereditary rings.

QUASI-FROBENIUS RINGS If we assume every left ideal of R is injective, it is easy to see that R is semisimple. Let us now consider R itself as a module. It is no restriction at all to assume R projective or flat; it is obviously a stringent condition to assume R is injective (= seIf-injective). Definition: A ring R is quasi-Frobenius if it is left and right noetherian and R is an injective left R-module. It can be shown that the asymmetry of the definition is only virtual: R must be injective as a right R-module [Jans, 1964, p. 781. It is also true [Jans, 1964, p. SO] that quasi-Frobenius rings are left and right artinian. Clearly semisimple rings are quasi-Frobenius; in particular, kG is quasiFrobenius when G is a finite group and k is a field whose characteristic does not divide the order of G . Although there are other examples, as we shall see, the most important example of a quasi-Frobenius ring is kG with no restriction on characteristic k. This ring is important in the theory of modular group representations. Theorem 4.35: If R is a principal ideal domain and I = Ra is a nonzero ideal, then RII is quasi-Frobenius.

Proof: As an R-module, RII is torsion and cyclic, say, with generator x. An ideal of RII has the form JII, where J is an ideal containing I . If J = Rb, then bc = a for some c E R (since Ra = I c J); moreover, bx is a generator

131

Quasi-Frobenius Rings

of J / I c R/I (for x

=

1 + Ra). Consider the diagram RII

t

0

-

I

R/l

J/I

Now f ( b x ) = sx for some s E R. Since cbx

= ax = 0, we

have

0 = cf(bx)= C S X , so that cs E Ra. Therefore cs = ru = rbc, for some r E R ; cancelling c gives s = rb, so that f ( b x ) = sx = rbx. Define g : R/I -+ R/I to be multiplication by r. Our calculation shows that g extends j , whence R/Z is self-injective. Finally, R/I is quasi-Frobenius, for it is clearly noetherian. I

Remark:

Exercise 9.20 generalizes this by allowing R Dedekind.

Corollary 4.36:

The following rings are yuasi-Frobenius:

(i) ZJnZ, where n # 0 ; (ii) k [ x ] / Z , where k is a jield and I is u nonzero ideal. Remark:

Compare this Corollary with Examples 17 and 18.

Theorem 4.37: Let R be lefi and right noetherian. Then R is quasi-Frobenius if and only if every projective module is injectire.

Proof: Assume R is quasi-Frobenius, and P is projective. Now P is a summand of a free module F = D R . Since R is noetherian, Theorem 4.10 shows F is injective (since R itself is injective); thus P is injective. For the converse, R itself is projective and hence injective, by hypothesis. Thus R is quasi-Frobenius, since we are assuming R is noetherian. I Definition: A module is indecomposable if it is not the sum of two proper submodules. Definition: If R = u L i , where each Liis indecomposable, then the Liare called principal indecomposable modules.

It follows easily from the definition that every left artinian ring R has such a decomposition, and hence has principal indecomposable modules.

132

4

Specific Rings

Theorem 4.38: If R is quasi-Frobenius, there is a bijection between minimal leji ideals and principal indecomposable modules.

Proof: First of all, we know that R is left artinian, so R does have a decomposition into indecomposables. If I is a minimal left ideal, let E ( I ) be its injective envelope in the injective module R. Being injective, E ( I ) is a summand of R. If E ( I ) = A @ B, then minimality shows either I n A = 0 or I n B = 0, which contradicts E ( I ) being an essential extension of I . Therefore E ( I ) is a principal indecomposable module. If E is a principal indecomposable module, then E contains a minimal left ideal I (Exercise 4.3). Now E is injective, being a summand of R. Let E ( I ) be the injective envelope of I contained in E. Since E is a summand of E ( I ) and E ( I ) is indecomposable, we have E = E ( I ) . Therefore, E is an essential extension of I . If I’ is another minimal left ideal in E, then I n I’ = 0, contradicting E being essential. We conclude that I H E ( I ) is a bijection. I This result takes on added interest when we observe that every simple module over a quasi-Frobenius ring is isomorphic to a minimal left ideal [Curtis-Reiner, 1962, p. 4011. This fact can be used to show that f.g. injective modules E over a quasi-Frobenius ring are projective (it is true that all injectives here are projectives). The module E is a sum of indecomposables E,, each of which contains a simple submodule. One may now repeat the argument above to show each E, is isomorphic to a principal indecomposable module, hence is projective. Definition: Let R be a finite-dimensional algebra over a field k. Then R is a Frobenius algebra if R z Homk(RR,k ) as left R-modules. Observe that the dual space Homk(RR,k ) is a left R-module as in Theorem 1.1S(i). Theorem 4.39:

Every Frobenius algebra R is quasi-Frobenius.

Proof: Every left or right ideal of R is a vector space over k, so finitedimensionality of R shows R is left and right noetherian. Theorem 3.44 applies to show HOmk(R,k ) Z RR is injective, so R is quasi-Frobenius. I

Lemma 4.40: Let R be a jinite-dimensional k-algebra over a jield k. If there is a k-map f : R -+k whose kernel contains no nonzero lejit ideals, then R is a Frobenius algebra. Proof: Define 6:R + Homk(R,k ) by 0,(x) = f ( x r ) . It is easy to check that each 0, is a k-map and that 8 is an R-map: r’e, = Next, we claim 6 is

133

Local Rings and Artinian Rings

monic. If 8,

= 0, then

0 = d,(x)

= f(xr)

for all x E R,

and Rr c kerf. By hypothesis, r = 0. Finally, if dimkR = n , then dim, Hom(R, k ) = n, being the dual space. Therefore, 8 must be epic, hence an isomorphism. I We are now able to give the important example. Theorem 4.41: If k is a field and G a f i n i t e group, then kG is a Frobenius algebra, hence is quasi-Frobenius. Remark: There is no restriction on the characteristic of k.

Proof: By Lemma 4.40, it suffices to exhibit a linear functional f : R -, k whose kernel contains no nonzero left ideals. Each r E R has a unique expression r

m,x,

=

where

m,E

k.

XEG

Definef:R -+k by r- m,, the coefficient of 1. Ifr = zm,x, thenf(x-'r) so that Rr c kerf gives m, = 0 for all x E G, whence Y = 0. I

= m,,

LOCAL RINGS AND ARTINIAN RINGS Definition: A ring R is local if it has a unique maximal left ideal.

One must be aware of the context when reading about local rings, for many authors assume local rings are commutative and noetherian as well. Examples: 31. Every division ring is local. 32. If p is prime, then Z/p"Z is local, for any n. 33. I f k is commutative local, the ring of formal power series k[[x]] over k is local. 34. If M is an injective indecomposable R-module, then End,(M) is a local ring [Lambek, 1966, exercise on p. 1041. 35. Ifp is a prime and Q is the rationals, then Z, = {a/b E Q:(b,p) = 1) is a local ring. 36. If R is a commutative ring and P is a prime ideal in R, then the localization R, = S-'R (where S = R - P ) is a local ring [Kaplansky, 1970, p. 241. Note that Example 35 is a special case of this example: take R = Z and P = (p).

134

4

Specific Rings

Lemma 4.42: Let R be a local ring with maximal left ideal J .

(i) (ii) (iii) (iv) (v)

r E R is invertible (i.e., is a unit) if and only if r 4 J ; i f r E J , then 1 - r is invertible; if A is a f . g . R-module with J A = A, then A = 0 ; J is a two-sided ideal; R/J is a division ring.

Proof: (i) In any ring R, Zorn’s lemma shows that every proper left ideal is contained in a maximal left ideal; in particular, if x E R is not invertible, then x E R x is contained in the unique maximal left ideal J . The converse is trivial: if x E J , then x cannot be invertible, for J is a proper ideal. (ii) If r E J , then 1 - r 4 J and hence is invertible. (iii) Let {al, . . . , a,,} be a minimal generating set for A in the sense that no ai may be deleted leaving a generating set. Since J A = A, we have n

a, =

1 riai,

i= 1

ri E J , all i.

Since 1 - r , is invertible, a, is a linear combination of the remaining ai, violating minimality. (iv) Since J is a left ideal, J R is a two-sided ideal. As J c J R (for 1 E R ) and J is a maximal left ideal, either J = J R (and we are done) or J R = R. This last equation violates (iii). (v) R/J is a ring because J is a two-sided ideal. Since every r outside J is already invertible in R, by (i), every nonzero element in R/J is invertible, i.e., R/J is a division ring. I Remark: Lemma 4.42(iii) is a special case of Nakayama’s lemma (see Theorem 4.47 for a better version). It is essential that the R-module A be f.g. For example, let R = Z, (Example 35) and let A = Q. Here J = ( p ) , the principal ideal generated by p, and it is easy to see p Q = Q. Lemma 4.43: Let R be a local ring with maximal left ideal J ; let A be an R-module with minimal generating set {a,, . . . ,an>.I f F is free on { x l , . . . , x n } , cp:F + A is dejined by cpxi = ai, and K = kercp, then K c JF. Proof: If K # JF, there is an element l r i x i E Knot in JF. By Lemma 4.42(i), one of the coefficients, say, r , , must be invertible. But x r i a i = 0, so that

contradicting the minimality of {al,. .. ,an).

I

135

Local Rings and Artinian Rings

Theorem 4.44: If R is a local ring, ewrj3,f.g. projective module A is free. Proof: Construct an exact sequence

as in Lemma 4.43, so that K c J F . Now projectivity of A gives F = K 0 A’, where A‘ 2 A. Hence JF = J K 0 JA‘. Now J K c K c JF, so Exercise 2.23 gives

K

=J

K @ ( K n JA’).

Since K n JA‘ c K n A’ = 0, we have K = J K ; since K is f.g., being a summand (hence an image) of the f.g. module F, we have K = 0 (Lemma 4.42(iii)). Therefore, A is free. I It follows from Exercise 3.4 and Lemma 4.42(v) that local rings have IBN, i.e., the rank of a free module is well defined. Endo [1961] proves that every f.g. flat module over a local ring is free; Endo [19621proves that every f.g. flat module over a (commutative) domain is projective (he attributes this result to Cartier). Kaplansky [1958a] proves that every summand of U M u , where each M u is a countably generated module over an arbitrary ring, is again of the same form (“countably generated means there is a set of generators of cardinality INo). It follows immediately that, for any ring, every projective module is a sum of countably generated projectives. If a ring R has every countably generated projective actually free, therefore, then every projective R-module is free. In particular, Kaplansky proves in the paper cited above that all R-projectives are free when R is local or when R is Btzout. We now show that f.g. modules over a local ring R have projective covers. This is not true for arbitrary R-modules, for Bass’s Theorem E above would imply R left perfect, contradicting the existence of local rings that do not have DCC on principal right ideals (Example 35). Rings for which every f.g. module has a projective cover are called semiperfect. Lemma 4.45: If R is local with maximal left ideal J and module, then J A is a superfluous submodule.

if A

is a f . g . R-

Remark : See Exercise 3.31. Proof: If B is a submodule of A with B + J A = A , we must show that B = A . NOWA / B = ( B + J A ) / B = J ( A / B ) .Since A / B is f.g., Lemma 4.42(iii) gives AIB = 0, whence B = A. I

136

4 Specific Rings

Theorem 4.46: I f R is a local ring, every J g . module A has a projective cover. Proof: Let us call an epimorphism essential if its kernel is superfluous. We must find a projective P and an essential epimorphism cp: P + A. This is precisely what is done in Lemma 4.43,for kercp c J P and Lemma 4.45 shows that JP, hence any submodule of JP, is a superfluous submodule of P. I

We now give a second, more complicated, proof of Theorem 4.46, because this second proof will allow a generalization for artinian rings. Second proof: By Lemma 4.45,the natural map n : A + A/JA is an essential epimorphism. Now A / J A is a f.g. R/J-module, i.e., a finite-dimensional left vector space over the division ring R / J ; hence A/J E (R/J)",the sum of n copies of R/J. Consider the diagram

A

J

1

*A/JA-0

where cp:R"+ (R/J)"is the natural map on each summand. Each coordinate map of cp is an essential epimorphism (Lemma 4.45 applied to R itself), so Exercise 3.33 shows that cp:R" -,A / J A is a projective cover. Since R" is projective (even free), there is a map E:R"+ A making the diagram comE rp. Now ker E c ker cp, so ker E is a superfluous submodule of R" mute: ~ C = (since ker cp is). Finally, if a E A, there exists x E R" with na = cpx = n ~ x ; thus, EX - a E kern = JA. Hence A = im E + JA, so that E is epic because J A is a superfluous submodule. Therefore E:R"+ A is an essential epimorphism, i.e., a projective cover. I One can axiomatize this proof if one can find an ideal to play the role of J . Once we do this, we will be able to prove the existence of projective covers when R is left artinian. Definition: If R is a ring, its (Jacobson) radical J is the intersection of all its maximal left ideals.

In a local ring, the radical is the unique maximal left ideal. Theorem F: Let R be a ring with radical J .

137

Local Rings and Artinian Rings

(i) i f r E J , then 1 - r is invertible; (ii) J is a two-sided ideal in R. If R is left artinian, then

(iii) RIJ is a semisimple ring; (iv) idempotents can be lifted: i f 7 an idempotent e E R with e + J = 7.

E

RIJ is an idempotent, then there is

Proof: All proofs may be found in [Anderson-Fuller 19741: (i) and (ii) on p. 166; (iii) on p. 170; (iv) on p. 301 (in order to apply the result there, one needs to know J is nil; even more is true: it is nilpotent, as proved on p. 172). I

Theorem 4.47 (Nakayama’s Lemma): l f R is u ring with radical J and A is a f.g. module with J A = A, then A = 0. Proof: Exactly as the proof of Lemma 4.42(iii).

I

Corollary 4.48: If R is a ring with rudirul J urid A is a J g . module, then J A is superfluous. Proof: Exactly as the proof of Lemma 4.44.

I

Lemma 4.49: Let R be a left artiniun rincq with radical J. If 7 is an idempotent in RIJ, there is a projective R-module P rind a projective cover E : P + (R/J)P. Proof: By Theorem F, we may lift the idempotent Z to an idempotent e E R. There is thus an exact sequence

O+Je+Re$(RiJ)F-+O, where n:re H rZ (one readily checks that ker F = Je). Setting A = Re in Corollary 4.48, we see that J e is a superfluous submodule of Re. Finally, e idempotent in R gives Re a summand of R , hence R e is R-projective. I

Theorem 4.50: I f R is left artinian, erel-1. f.q. module A has u projective cover. Proof: The proof proceeds exactly as that for local rings. Let J be the radical of R, and note that the natural map n : A --t A/JA is an essential epimorphism (Corollary 4.48). Since R I J is semisimple, the RIJ-module A/J has a decomposition into simple modules: A,/J 2 U S i , and Siis isomorphic to a minimal left ideal of R/J. Every ideal of R I J is a summand, and there is an idempotent ZiE R/J with Si2 (R/J)u,. By Lemma 4.49, there are

138

4

Specific Rings

projective covers c i : P i+ S i , hence a projective cover &:UPi + A / J A (Exercise 3.33). The proof is completed as in Theorem 4.46. I Theorems 4.46 and 4.50 show that local rings and artinian rings are semiperfect.

POLYNOMIAL RINGS

In the mid-l950s, Serre proved that if A = R[tl, . . . ,fk], where R is a field, then every f.g. projective A-module P has a f.g. free complement: there is a f.g. free module F with P 0 F free. Serre posed the problem: are projective A-modules free? If k = 1, then A = R [ t ] is a principal ideal domain, so projectives are free, by Corollary 4.20. In 1958, Seshadri proved that if B is a principal ideal domain, then f.g. projective B[x]-modules are free; in particular, setting B = R[t] gives an affirmative answer to Serre’s problem when k = 2. There was much interest in this problem for k > 2; indeed, it was one of the main reasons for the development of algebraic K-theory. Remarkably, the problem was solved simultaneously, in January 1976, by Quillen in the United States and Suslin in the Soviet Union. We first present Vaserstein’s version of Suslin’s proof; afterwards, we shall outline Quillen’s solution. We remark that Bass [19631 proved that “big projectives” (i.e., non-f.g. projectives) are free over these rings. Definition: Let A be a commutative ring and let A” be the free A-module of rank n. A unimodular column is an element CI = (al, . . . , a,) E A” such that there exist bi E A , i = 1,. . . , n, with cl=l a&, = 1. Definition: A commutative ring A has the unimodular column property if, for every n, every unimodular column u is the first column of some n x n invertible matrix over A . This definition may be rephrased. Let GL(n, A ) denote the multiplicative group of invertible matrices over A , and let c1 denote the column vector having first coordinate 1 and 0’s elsewhere. Then u is the first column of an invertible matrix over A if and only if CI

= Mcl

for some M E GL(n, A ) . That a column u = ( a l , . . . ,a,) be unimodular is a necessary condition that it be the first column of some M E GL(n,A),for M is invertible if and

139

Polynomial Rings

only if its determinant, det M , is a unit in A, and Lagrange expansion shows det M is a linear combination of the entries in the first column of M . For the remainder of this chapter, all rings are commutative. Theorem 4.51: If every f.g. projective A-module is free, then A has the unimodular column property. Proof: If a = ( a l , . . . , a,) E A" is a unimodular column, there exist bi E A with x a i b i = l . Define cp:A"+A by ( rl . . . . , r n ) - x r i b i . Since cp(a)=l, there is an exact sequence

o+ where K

= ker 9.As

K - + A"% A + O ,

A is projective, this sequence splits, and

A" = K @ (3). By hypothesis, K is free (of rank n - 1). If ( a 2 ,. . . , an) is a basis of K , then adjoining c( gives a basis of A". Let ( E ~ ., . . , E,: be the standard basis of A", i.e., E~ has all coordinates 0 except for 1 in the ith place (this agrees with our earlier notation for E ~ ) .The transformation T:A"+ A" with T E=~a and T E=~ mi for i 2 2 is invertible, and the matrix of T with respect to the standard basis has first column a. I In general, the converse of Theorem 4.51 is false. For example, the reader may show that A = 2/ 62 has the unimodular column property, yet we know A has nonfree projectives. Definition: A f.g. module P is stably free if there exists a f.g. free module F such that P 0 F is free. Remarks: 1 . Every stably free module is projective, being a summand of a free module. 2. With this terminology, Serre's theorem states that f.g. projective A-modules are stably free when A = R [ t , , . . . , t k ] with R a field. 3. Kaplansky exhibited a stably free module that is not free over the ring of all continuous real-valued functions on the 2-sphere {(x, y , z) E R3: x2 + y 2 + z2 = 1 ) (see [Swan, 19621). This example may be modified to exhibit the same phenomenon over A = Z[x, y , z]/(x2 + y 2 + z2 - 1). The first algebraic proof of this is given by Kong [ 19771.

Eilenberg's observation (Exercise 3.1 3 ) shows that we must insist that the complement F be f.g. The next result shows that the notion of stably free is of interest only when all modules in sight are f.g.

140

4

Theorem 4.52 (Gabel): Assume P 0A" f .g., then P is free.

= F,

Specific Rings

where F is free. I f P is not

Proof: Let {ei:i E I} be a basis of F. Since P is an image of F, the index set I must be infinite. Consider the exact sequence O+ P+ F

3 A"+o.

If ( e l , . . . ,E"} is the standard basis of A", choose xj E F with cp(xj)= ej, 1 5 j < m. As each x j has only finitely many nonzero coordinates, they Aei, collectively involve only a finite subset J c I . If we define F' then cp' = cp[ F' is epic and F = P + F'. Setting P' = P n F', there are exact sequences

=uisJ

and

O+P-rF'%A"+O

O-+P'-+P+P/P-+O.

Clearly the first sequence splits: F' = P' @ A". The second sequence also splits, for Aei. PIP' = PJP n F' z P + F / F ' = F J F Thus, P z P @ P / P . Since I - J is infinite, there is a free module F" with P / P z A" 0F". Therefore P z P @ P/P 2 P 0(A" @ F )z (P'0A") 0F

and P is free.

z F' @ F ,

4

Theorem 4.53: I f A has the unimodular column property, then every stably free A-module P is free. Proof: By induction on the rank of a free complement of P, it suffices to prove P is free assuming P 0 A is free: P @ A = A". Let n be the projection of A" on A having kernel P, and let {el, . . . , en} be the standard basis of A". There exists a = ( a l , . . . , a,,) E A" with 1 = n(a)= Eain(ei); thus, u is a unimodular column. By hypothesis, a = Me1 for some M E GL(n,A). Let T :A" -+A" be the corresponding linear transformation: T E ,= u and T E=~a, for i 2 2 (where az, . . . , an are the other columns of M). If Tei E P for all i 2 2, we claim T' = TI ( E ~ ., . . ,e n ) : ( E ~. ,. . , E,) P is an isomorphism, proving P is free. Visibly, T' is monic. If B E A", then /? = Ty for some y = E r i e i ; hence B = T(rIel 6), where 6 = C;=z ri.zi. In particular, if p E P, then

+

/3 - T6 = rlTel = r l a E P n ( a ) = 0,

whence p E im T and T' is epic. It thus suffices to show TE,E P for i 2 2; in matrix terms, it suffices to show the columns a 2 , . . . , a,, may be chosen to lie in P = kern. Suppose

141

Polynomial Rings

n(ai)= lifor i 2 2. By elementary column operations, for each a’ 2 2, replace ai by cri = a, - Lia. The new matrix is invertible, has first column a, and, if i 2 2, then n(c(1)= n(ai)- L,n(a)= 0 and a: E P . I Theorem (Serre): If R is a j e l d , then t g . projectiile R [ t l , .. . , t,]-modules are stably free. Proof: Theorem 9.45.

I

Corollary 4.54: If A = R [ t , , . . . , t k ] , where R is a3eld, has the unimodular column property, then t g . projective A-modules are free. Proof: Serre’s theorem and Theorem 4.53.

1

Definition: If B is a commutative ring, a polynomial f(y) E B [ y ] has virtual degree s if f ( y ) = boy”

+ b,yS-‘ + . . . + h,,

bi E B.

Since we do not demand that the leading coefficient bo # 0, the degree of f(y) may be smaller than its virtual degree (of course, f ( y ) has many virtual degrees). Theorem 4.55 (Suslin Lemma): Assume B is u commutative ring. Let s 2 1 and consider polynomials in B [ y ] : f ( y ) = y S + u a , y ” - ’+ . . . + u s ; b , - f - ’ + . . . + b,. g(y) = Then, for each j , 1 Ij Is, the ideal (.f( j*),g( y ) ) in B [ y ] contains a polynomial of virtual degree s - 1 und leuding coeficient b j . Proof: Define I

=

{leading coefficients of those h( y ) E ( f ,g) having virtual degree J - 1 ;

it is clear that I is an ideal in B containing b , . We prove by induction on j that I contains bl , . . . , bj, j I s. Define q’(y) E ( f g ) by g’(y) = yg(Y) - b , f ( ~=)

C:= (hi+ 1

1

- b,ai)u’-’.

By induction, I contains the first j - 1 coefficients of g’(y), the last of which is bj - b l a j - 1. It follows that bj E I . I

Observe that performing elementary row operations on a matrix M E G L ( n ,A ) corresponds to left multiplication by a matrix N E GL(n,A).

142

4

Specific Rings

Thus, if a column a can, after suitable row operations, be completed to an invertible matrix, then a itself can be so completed: N u = M c l implies O! = N - ' M E l . Theorem 4.56 (Horrocks): Let B be local, and let a = (al, . . . , a,) be a unimodular column over A = B [ y ] . If some ai is monic, i.e., has leading coeficient 1, then a is thejrst column of an invertible matrix over A . Proof (Suslin): If n = 1 or 2, the theorem holds for any commutative ring A ; therefore, we may assume n 2 3. We do an induction on s, the degree of the monic polynomial ai, noting that the case s = 0 is trivial. By our preceding remark about row operations, we may assume al is monic of degree s > 0 and the other polynomials a 2 , . . . , a, have degrees 5 s - 1. Let m be the maximal ideal in B. Thus mA consists of those polynomials each of whose coefficients lies in m.The column B E A"/mA"is unimodular (over B/m)[y], so that not all ai, i 2 2, lie in mA; assume a2 $ mA. Thus, a2 = b l y s - ' + . . . + b,, and some bj $ m; since B is local, bj is a unit. By Suslin's Lemma 4.55, the ideal (a1,a2)in A contains a monic polynomial of degree s - 1. Therefore, the elementary row operation of adding a linear combination of a, and a2 to a3 produces a monic polynomial in position 3 of degree s - 1. One may now apply the inductive hypothesis. I Notation: If A = B [ y ] , where B is a commutative ring, and if a( y) = (ai(y ) ) is an n-rowed column over A, then

GL(n, A)a(y ) = { M a (y ) :M E GL(n, A ) ) .

As the entries of M E GL(n,A ) lie in A = B [ y ] ,they are polynomials in y ; one may thus write M = M ( y ) . Theorem 4.57: Let B be a domain, A = B [ y ] , and a ( y ) = ( a i ( y ) )a unimodular column ouer A, one of whose coordinates is monic. Then a(Y) = W Y ) B , where M ( y ) E GL(n,A ) and Proof: Define

I

= {bE

B is a unimodular column ouer B.

B:GL(n, A)a(w + bx) = GL(n, A)a(w),all w, x E A } .

It is easy to see that I is an ideal in B : in particular, if b, b' E I , then GL(n,A)a(w

whence b

+ b' E I .

+ bx + b'x) = GL(n,A)a(w + bx) = GL(n,A)a(w),

143

Polynomial Rings

Suppose I

= B,

so that 1

E

I . Set w

= J',

b = 1, and x

=

- y to obtain

G L ( n ,A)LY(y ) = G L ( n ,A)a(O). Thus, ~ ( y=) Mcr(0) for some M E G L ( n ,A ) ; since p = a(0) is a unimodular column over B, the theorem would be proved. We may therefore assume I is a proper ideal of E, so that I c J for some maximal ideal J . Since B is a domain, E is contained in the localization B,. As B, is a local ring and a ( y ) is a unimodular column over B J [ y ] having a monic coordinate, Horrocks' Lemma 4.56 applies to give 4 Y )=M(yk,

for some M ( y ) = ( m i j ( y )E) G L ( n ,B,[y]). Adjoin a new indeterminate z to B j [ y ] and define a matrix N ( Y , z )= M ( Y ) [ M ( Y

+

Z)]-'

E

GL(REj[Y,z])

(the matrix M ( y z) is obtained from M ( y) by replacing each of its poly, it is easy to nomial entries mij(y)by mij(y z ) ; if M ( y - ' = ( h i j ( y ) )then see that ( h i j ( y z ) ) is the inverse of M ( y + z ) ) .Observe that the definition of N ( y, z ) gives

+

+

N(y,O) = l,,

the n x n identity matrix.

Since ~ ( y=) M ( y ) c l , it follows that a ( y + z ) = M ( y + z)cl. Therefore N(Y,Zb(Y + 4

(*I

=N(y,z)M(y

+ I)&, = M ( Y h = 4 Y ) .

Each entry of N ( y , z ) is a polynomial in B,[J,z], hence may be written as f ( y ) g( y, z), where each monomial in g( J; z ) involves a positive power of z. Since N ( y, 0) = l,, we must have f'( y ) = 0 or 1, and we conclude that the entries of N ( y, z) are polynomials in B,[ J,, z ] containing no nonzero monomials of the form Ayi with i > 0 and A E B j . Let b be the product of all denominators occurring in coefficients of the polynomial entries of N ( y , z ) ; by definition of B j , we have b 4 J and hence b 4 I. Further, N ( y ,bz) E G L ( n , B [ y , z ] ) for we have just seen that replacing z by bz eliminates all denominators. Equation (*) gives

+

GL(~,B[Y,Z])+ C ~bz) Y = GL(n,B[y,z])cc(y). For fixed w, x E A = B [ y ] , define a B-algebra map q:B[y, z ] -,A by q ( y ) = w and q ( z ) = x. Applying cp to the equation above gives G L ( n ,A)LY(W + bx) = GL(n.A ) ~ ( w ) , and this contradicts b $1.

I

144

4 Specific Rings

The reader surely appreciates how ingenious this elementary argument is! The coefficient ring is seriously restricted in the next technical lemma, which is the computation at the heart of the “Noether Normalization Lemma”.

Lemma 4.58 (Noether): Let A = R[t,, . . . , tk],where R is ajield, and let a E A have (total)degree 6 ; set m = 6 + 1. Define Y

= tk

and, for 1 I i < k - 1, define yi = ti - tr

k-i

Then a = ra‘, where r E R and a’ is a monic polynomial over the polynomial ring R [ y l , . . . Yk-l]. 7

Proof: First of all, R [ y,, . . . ,yk- 1] is a polynomial ring (i.e., { y , , . . .,y k - ,} is an independent set of transcendentals), for the defining equations for the y’s give an automorphism of A (with inverse t, H t k and ti H ti + trk-’ for 15 i I k - 1). Denote the k-tuple (rnk-’, mk-’ , . . . , m , l ) by p, and denote k-tuples (jl,. . . ,j,) with each ji 2 0 by ( j ) . The dot product p . ( j ) is, of course, ,,,k-ljl

+ . ..

+

rnjk-1

+ j,.

The polynomial a may be written

a = C r(,@ . . . t k , (j)

where ru) are nonzero elements of R. Replacing each ti as in the statement gives a

=

C r(j)(yl+

. . . (yk-,

pk-’)Jl

(i)

+ y”)jk-’yJk

Expand each monomial for fixed ( j ) and separate the “pure” power of y from the rest: a=

C r(j)y” (j)

(j)

+CJj)(Y,, ..

. 7

Yk-19

Y),

(j)

where each mixed term An involves some yi to a positive power; moreover, for each ( j ) ,the highest power of y is p . ( j ) . Since m is greater than the (total) degree of a, each indexing k-tuple ( j )= (jl,. . . ,j,) has 0 I j i m for all i. It follows that ( j ) # ( j ’ ) implies p . ( j )# p . ( j ’ ) ,for this is just the unique representation of a positive integer written in base m. All the exponents p . ( j ) distinct implies all the terms ru)y”(j) distinct. If d is the largest p * ( j ) , then d is the degree of a as a polynomial

-=

145

Polynomial Rings

in y with coefficients in R[y,, . . . ,J'k-

a

=

ry

d

,I. Therefore,

+ g ( y , , . . . . r k - I , v),

where r E R and g has degree in y strictly less than d. As r # 0 and R is a field, we may define a' = yd r - 'y( y , , . . . , yL - y). I

+

,,

Theorem 4.59 (Quillen-Suslin): If A = R [ t , , . . . , tk], where R is a j e l d , then every &g. projective A-module is free.

Proof: We do an induction on k , the case I, = 1 holding because R[t] is a principal ideal domain. By Corollary 4.54, it suffices to prove A has the unimodular column property. Let CI = (ai(t,, . . . , tk)) be a unimodular column over A ; we may assume a, # 0. By the Noether Lemma 4.58, a , = ra;, where r E R and a; E R[yl, . . . , y,- ,] is a monic polynomial (J,, defined as in the Noether lemma). Since r is a unit (lest a, = 0),there is no loss in generality in assuming a , = a;, i.e., a l is monic. Theorem 4.57 thus applies to give CI =

Ma,

where M E GL(n, A ) and is a unimodular column over B = R [ y,, . . . ,yk- ,]. By induction, B has the unimodular column property, so that

fi for some N in GL(n,B). Hence M N S(

=

= NE1 E

GL(n, A) and

MNE,.

I

Generalization of Noether's lemma to more general rings enables one to relax the hypotheses on R. Quillen's approach is more module-theoretic, and makes use of an elementary idea used by Seshadri in his solution of Serre's problem for k = 2. Definition: If A is an R-algebra, an A-module P is extended from R if there is an R-module Po and an isomorphism P 2 P,, ORA. Examples: 37. Every free A-module is extended from R. 38. If an A-module P is extended from R and S is a subset of A, then S - ' P is extended from R (if P 2 P0@,A. then S - ' P = P @ , S - ' A , and associativity gives

s-'P

2

(Po @ R A ) @ A S - ' A 2

@R

S-'A.

39. If P 2 Po ORA , the R-module Po may not be unique (if R = Z and then the Q-module Q is extended from Z : Q 2 Z OzQ. However, for any torsion group T , we also have Q = ( Z 0 T )@,Q.

A

= Q,

146

4

Specific Rings

40. There exist A-modules that are not extended (if R is a field and A is an R-algebra, then every R-module is free (being a vector space) and hence every extended A-module must also be free.) The uniqueness question: “If P is extended from R, is the module Po unique?”, is easily settled for polynomial rings. Definition: An R-algebra A (not necessarily commutative) is augmented if there is an R-algebra map E :A -,R.

Since ~ ( 1= ) 1, the augmentation E must be epic. The kernel of the augmentation ideal, is a two-sided ideal in A.

E,

called

Examples: 41. If A = R[ U], polynomials in a set of variables U , then A is augmented with augmentation E : A -,R given by f H constant term off. 42. If G is a group, the group ring RG has an augmentation: 4Crgg) =

1%.

Lemma 4.60: Let A be a commutative augmented R-algebra with augmentation ideal I ; let P be an A-module. (i) If P z Po ORA, then Po 2 P / I P . (ii) I f , in addition, P is projective or t g . , then so is Po.

Proof: (i) Tensor the exact sequence 0 + I an exact sequence Po & I

+

Po OR A

+A +R

-,0 by Po to obtain

+ Po Q R R + 0.

The last term is P o , the middle term, by hypothesis, is P, and the image of the first term is IP. We conclude that Po 2 P / l P . (ii) If P is f.g., the formula P / I P z Po shows Po is also f g ; if P is Aprojective, then Exercise 3.11 applies with R 2 A/I to show that Po is R-projective. I The existence question: “Given an A-module P , when is P extended from R?”, is quite difficult. In his solution of Serre’s problem, Quillen discovered the following criterion. Theorem (Quillen): Let R be u commutative ring, and let P be a Jinitely related R[t]-module. I f the R,,,[t]-module P,, = P 0R,,[t] is extendedfrom R,, for every maximal ideal m of R, then P is extended from R.

Proof:

[Lam, 1978, pp. 125-1281.

As in Suslin’s solution, one prefers polynomials with coefficients in a local ring.

147

Polynomial Rings

Definition: If S is the set of all monic polynomials in R [ f ] ,write

R(t) = S - ' R [ l ] . The elements of R(t) may be regarded as rational functions

+ ' . ' +ao)/(t"+ h,- i t " - i + . . . + bo).

(a,t"+u,-lP-'

Definition: If P is an R[t]-module, define P ( t ) = S- ' P = P OR[,] R(t).

There is an analogue of Horrocks' lemma. Theorem H : I f R is local and P is LI f:g. projri,tire R[t]-module such that P ( t ) is a free R(t)-module,then P is pee. Proof: [Lam, 1978, pp. 114-1161.

I

Assuming Quillen's theorem and Theorem H, let us see how Quillen's solution reads. Theorem 4.61 (Quillen):

Let \Lt be a class o/' rings such that

(i) if R E 2I, then R ( t ) E 2I; (ii) If R E 2I and m is u maximal ideul of R , then R,,,[t]-projectivesare free.

Then,for all k 2 1 and for all R are extended from R .

E

91, f:g. projectice R [ t , , . . . , tk]-modules

Proof: We do an induction on k . Assume k = 1 and P is f.g. R[t]-projective. By condition (ii), R,,,[t]projectives are free for every maximal ideal m of R . In particular, P,,,= P @ RJt] is free, hence is extended from R , , , (Example 37). By Quillen's theorem, P is extended from R . Assume k > 1 and P is f.g. R [ t , , . . . , t,]-projective. We establish notation:

Po = P/(tl,.. . , t/oP; P , = P i ( t 2 , . . . , t/JP; S, = {all monic polynomials in R [ t I] 1, which we view as a (multiplicatively closed) subset of R [ t l , . . . , tk]. Now define ~=S;lp=P@R[rl.

=

@R[rl,.

.. ,rk]

...,r*]S;lR[tl,...,tk] R(t)[t2,

...

9

fk],

where we have written t in place of f l . Since R ( t )E 'u, by (i), induction says the projective R ( t ) [ t 2 ,. . . , tk]-module P' is extended from R(t). As poly-

148

4

Specific Rings

nomial rings are augmented,

. . , tk)P' @~(~)R(t)[r,, . . . , tk]. Using Exercise 1.12 (with I = ( t , , . . . , tk)) and the definition P' = P @ R(t)[t,, . . . , tk], we have P'

P'/(fZ,.

P'/(t2

From the initial step k

=

9

. . . tk)P 9

pl @R[t]

R(t).

1, we have P , z Po ORR[t].

Combining these isomorphisms gives

(*I

P'

Po @R R(t)[t,, . . . ,tk].

Now define B = R[t,, . . . , tk] and S = {all monic polynomials in B[t]}; note that S-lB[t] = B(t).Since S , c S, Exercise 3.48 gives P(t) = S - I P = S-l(S;lP) = S-lP'.

But, by definition of S - ' , P(t)=S-'P'=

P O R ( f ) [ I , ,....f,]S-'R(r)[tZ,...,tk] =

@R(f)[tz,

= Po &B(t), ==

@If

. . . ,f k ]B(t)

by equation (*),

B, @B B(t).

Let rn be a maximal ideal in B. Defining P,,,(t)= P ( t ) O B B I , ,we , apply Lemma 3.77 to obtain p,,,(t)= (Po ORB),,,OB,,,B,,,(t).

Now (Pog RB),,,is B,-projective because Po is R-projective; it is free because B,,, is local; therefore, P,,(t) is free. By Theorem H, PI,,is free, and so PI,,is extended from B,,,for every m (Example 37). Quillen's theorem shows P is extended from B = R[t2,. . . , tk], and induction shows P is extended from R.

I

Theorem 4.62 (Quillen-Suslin): R[t,, . . . , tk]-module is free.

If R is a field, then euery fig. projective

Proof: Let % be the class of all fields. If R E %, then it is easily seen that R(t) is the quotient field of R[t] consisting of all rational functions (for the leading coefficient of a nonzero polynomial is a unit). To verify condition (ii) of Theorem 4.61, note that R = R,, and that R[t] is a principal ideal domain. Theorem 4.61 says that f.g. projective R [ t l , . . . , tk]-modules are extended

Polynomial Rings

149

from R. As every R-module is a vector space, hence free, extended modules are free. I Theorem 4.63 (Quillen-Suslin): I f R is a principal ideal domain, then everyfig. projective R[t,, . . . , tk]-module is,free. Proof: Let 2l be the class of all principal ideal domains. Condition (i) of Theorem 4.61 needs some standard commutative algebra: one shows that R(t) E 2l by showing it is a unique factorization domain in which nonzero prime ideals are maximal. Condition (ii) is not standard. Assume P is R,,,[t]-projective, where rn is a maximal ideal in R. Now P ( t ) = P ORrn[,, R,,,(t) is R,,(t)-projective, hence R,,,(t)-free, for R , E '$ (standard I commutative algebra) and R,,,(t)is a principal ideal domain (condition (i)). Since R,,, is local, Theorem H gives P free. Now apply Theorem 4.61 to conclude that projective R[t,, . . . , tk]-modules are extended from R; since R[t,, . . . , t k ] is augmented, such modules are extended from projective, hence free, R-modules, and so are themselves free. I Remarks: 1. Seshadri's theorem would also prove condition (ii) in Theorem 4.63. 2. Theorem 4.61 applies to the class 2l of all Dedekind rings. In particular, our remarks after Corollary 4.20 show that if R is Dedekind, then every f.g. projective R[t,, . . . , t,]-module has the form A 0 B, where A is free and B is extended from an ideal of R. 3. Brewer and Costa [19781 have proved that the class 2l of all Bezout rings in which nonzero prime ideals are maximal satisfies the hypotheses of Theorem 4.61 (the ring of all algebraic integers is such a ring), thereby proving f.g. R[t,, . . . , t,]-projectives are free for these rings. 4. Quillen applied Theorem 4.61 to the class 2l of all regular rings (i.e., rings of finite global dimension; see Chapter 9 for the definition) of Krull dimension 1 2 (i.e., the longest chain of prime ideals has two steps: Po c P , c P 2 ) [Lam, 1978, p. 1381. 5. Other examples of rings R for which projective R[t,, . . . , tk]modules are free are formal power series rings K[[x,, . . . , x.]], where K is a field [Lam, 1978, pp. 150-1621. This result was proved, independently, by Mohan-Kumar and by Lindel and Liitkebohmert. 6. Ojanguren and Sridharan [1971] exhibit a division ring D and a nonfree projective D[x, y]-module. Thus, the noncommutative version of the Quillen-Suslin theorem is false.

5

Extensions of Groups

With this chapter we begin the proper study of homological algebra. The group theoretic computations that we shall do will yield constructions analogous to those for the singular homology and cohomology groups of a topological space that we sketched in Chapter 1. Let E be a (possibly nonabelian) group, A an abelian normal subgroup, and G = EfA. We write E and its subgroup A additively, but we write the quotient group G multiplicatively. Definition: An extension of A by G is an exact sequence

O + A + E - + G + 1. Remark: The definition of extension makes perfectly good sense for any group A, but we assume A abelian throughout. Lemma 5.1: An extension of A by G determines a homomorphism 8:G + Aut(A), where Aut(A) is the group of all automorphisms of A. Proof: If x E G , let Ax E E be a lifting of x (i.e., Ax H x). Define 8,: A + A by O,(a) = Ax a - Ax, noting that im 8, c A since A is normal in E. The function 8:G + Aut(A) sending x H 0, is well defined; this follows from a short computation using the facts: (i) A is abelian; (ii)if A'x is a second lifting of x, then I'x - Ax E A. Finally, 8 is a homomorphism, for since the value of 8 is independent of choice of liftings, we may choose A(xy) = Ax Ly, I

+

+

150

151

Extensions of Groups

We recall a definition from the previous chapter. Definition: If G is a multiplicative group (perhaps infinite), the integral group ring ZG is the ring whose additive group is the free abelian group with basis G, and whose multiplication is determined by the multiplication of G and the distributive laws.

A typical element of ZG is EXEC m,x, where rn, E Z and almost all m,

Theorem 5.2: If 0 -+ A ZG-module.

+

E

-+

G

+

= 0.

1 i s un extension, then A is a jefr

Proof: In brief, G acts on A by conjugation. In detail, if x E G and a E A, define xu = e,(a) = Ax a - Ax. Since H is a homomorphism, la = a and (xy ) a = ~ ( y a )We . now define the action of an arbitrary element of ZG by (&,x)a = Cmx(xa). I

+

It should now be clear why we write A additively and G multiplicatively. There is a second way of writing conjugation. Suppose we had defined O:(a) = -Ax + a + Ax. The reader may check that O':G -,Aut(A) given by x w exis an antihomomorphism:O'(xy) = 11'( y)O'(x). This version ofconjugation makes A into a right ZG-module, where ax = @:(a). Exercises: 5.1. I f x E G, then H: = O X - , . 5.2. Let A be a left ZG-module. Show that A can be made into a right ZG-module by defining ax = x - ' u . (If G is not abelian, A need not be a (ZG - ZG) bimodule.) Conclude that ZG-modules can, at our pleasure, be considered as either left or right modules. It follows that if A and B are ZG-modules, we can always arrange matters so that A OZcBmakes sense. 5.3. Give an example in which G is abelian (so that ZG is commutative) and a ZG-module A in which xu # ax (i.e.,the left and right structures do not coincide). [If R is a commutative ring, one usually makes a left R-module A into a right module by defining ar = ru; if A happened to be a right Rmodule, however, there is no reason to expect that these two right actions of R should be the same.] 5.4. The integral group ring functor is adjoint to the unit group functor: there is a natural bijection

HOmRings(ZG7R )

Hom(;roups(G,U ( R )1 3

where G is a group, R a ring, and U ( R )the group of units (=invertible elements) in R. This last exercise is the formal reason A is a left ZG-module in Theorem 5.2, for Aut(A) is the group of units in End(A).

152

5 Extensions of Groups

5.5. If G and H are groups, then there is a ring isomorphism

Z(G x H) 2 ZG &ZH. Definition: A left ZG-module A is trivial if x a = a for all x E G and all a E A. Exercise: 5.6. The following are equivalent for the left ZG-module A, where 0 + A + E -+ G -+ 1: (i) A is trivial; (ii) A is contained in the center of E; (iii) 8:G + Aut(A) is trivial, i.e., 8, = 1 for all x E G. Suppose A is a left ZG-module, and 0 + A -,E -+ G + 1 is an extension. It may happen that the extension equips A with a left ZG-module structure distinct from its original structure. Definition: A left ZG-module A realizes the operators of an extension 0 + A + E + G + 1 if the two left ZG-actions on A coincide: x a = 8,(a) = I x

+ a - Ax,

all a E A, x E G.

The extension problem is: Given a left ZG-module A, determine all extensions of A by G that realize the operators. From now on, we shall say “G-module” instead of left ZG-module. Definition: An extension 0 + A + E 5 G -+ 1 is split if there is a homomorphism I : G + E with nI = 1,; the middle group E is then called a semidirect product. Theorem 5.3: An extension 0 + A + E 5 G -+ 1 is split if and only if E contains a subgroup C 2 G (not necessarily normal) with A + C = E and AnC=O.

Remark: Such a subgroup C is called a complement of A in E. Proof: If the extension splits, then define C = imI. Since nl = l,, I is monic, so it is an isomorphism between G and C. Now im I is a transversal of A in E (i.e., a complete set of coset representatives), so that A C = E. Finally, if a E A, then nu = 1; if a E C, then a = I x for some x E G, and na = l rl x = x . Therefore if a E A n C, then a = I 1 = 0. Conversely, each e E E can be written e = a c, a E A, c E C ; moreover, this expression is unique because A n C = 0. Note that K I C: C + G is an isomorphism, and define I:G + E as (nI C ) - l . I

+

+

153

Extensions of Groups

Definition: Let 0 + A + E 5 G 1 be an extension. A lifting is a function A:G -+ E with 7cA = 1, and A1 = 0. -+

Exercises: 5.7. An extension is split if and only if it has a lifting that is a homomorphism. 5.8. Show there is a split exact sequence

0 + 2/32

-+

s, + 2/22 + 1,

where S , is the symmetric group on three letters. Show there is no projection of S , onto 2 / 3 2 (compare Exercise 2.22). 5.9. If an extension 0 -+ A -+ E + G -+ 1 is split and if A is G-trivial, then E is isomorphic to the direct product of A and G. Theorem 5.4: A n extension 0 -,A -+ E -, G 1 is split ifand only if (i) there is a group E whose elements are all ordered pairs (a,x ) E A x G with addition given by -+

(a,x )

+ (u‘,y) = (a + xa’, xy);

(ii) there is a lifting A so that the function cp: E (a,x ) is an isomorphism.

+E

dejined by a + Ax

H

Proof: Suppose the extension splits, and A: G + E is a lifting that is a homomorphism. As in Theorem 5.3, every element of E has a unique expression of the form a + Ax. We compute addition: (a + Ax) + (a’ + Ay) = a =a =a

+ ilx + a’ - Ax + ilx + Ay + ( A s + a’ - Ax) + ilxy + xa’ + Axy.

It follows that E’ is a group and cp: E + E’ is an isomorphism. Conversely, define a function ill :G -+ E’ by x H (0, x), and note A1 is a homomorphism. But Al = cpl, so that A = cp-l A, is a homomorphism and the extension splits. We now solve the extension problem in the sense that we construct all possible addition tables for “middle groups” E. Suppose an extension 0 + A + E -+ G + 1 is given. If A: G + E is a lifting (A need not be a homomorphism!), then im l is a transversal of A in E. Since A1 = 0, every element of E can be uniquely expressed as a + Ax. Both l ( x y )and Ax + Ay represent the same coset of A, so we have

(*I

Ax

for some element [x, y ] E A.

+ Ay = [x, y ] + 4 x y )

154

5 Extensions of Groups

Definition: The function [ 1: G x G -+ A defined by formula (*) is called a factor set.

Further, A realizes the operators (there is no other G-structure in sight), so there is a second formula

(**I

xu

+ Ax = Ax + a.

Theorem 5.5: Let A be a G-module. A function [ ] : G x G -+ A is a factor set if and only if

(i) [x, 11 = 0 = [ l , x ] ; (ii) X [ Y , z ] - [ X Y , z ] + [x, Y Z ] - [ x , Y ] for all x , y, z

E

= 0,

G.

Proof: Suppose [ ] is a factor set. The first equation follows from (*) and the assumption 11 = 0; the second follows from the associative law in E. For the converse, we must construct an extension of A by G (in which A realizes the operators), and we must then choose a lifting 1 so that [ ] satisfies (*). Let E be the set of all ordered pairs (a,x ) E A x G with addition defined by

+ (a',Y ) = ( a + xa' + [ x , Y ] , X Y ) . (Compare this with the addition in the semidirect product.) Formula (ii) gives associativity, the identity is (0,l), and -(a,x)

=(-x-'a

-X - ~ [ X , X - ~ ] , X - ~ ) ,

so that E is a group. Define K : E -, G by (a, x ) e,x ; clearly K is onto with kernel {(a, 1 ) : aE A } that we identify with A. Thus 0 + A -+ E 5 G -+ 1 is an extension. If we define iLx = (0,x), then it is easy to check that A does realize the operators, viz, formula (**) holds, and [ ] satisfies formula (*). I Definition: Z2(G,A ) is the abelian group of all factor sets under pointwise addition.

Observe that the zero of Z2(G,A ) is the factor set that is identically zero, and it corresponds to the semidirect product. The definition of factor set arose from considering an extension and it depends on a choice of lifting 1. Changing 1 should still give the same extension. Theorem 5.6: Let 0 -+ A -+ E + G -,1 be an extension, and let , I and A': G -+ E be lijlings. If [ ] and ( ) are the corresponding factor sets, then there

155

Extensions of Groups

is a function ( ) :G + A satisfying (i) (1) = 0; (ii) (x, y ) - [x, y ]

=x( y) -

(xy)

+ (x)

for all x , y E G.

Proof: For any x E G, both Ax and A’x lie in the same coset of A, so there is an element ( x ) E A with ( x ) = A‘x - A x , Since A 1

= 0 = 2’1,

we have (1

> = 0. The main formula is derived as follows:

+ A‘y = ( x ) + Ax + ( y ) + Ay ( A realizes the operators) = ( x ) + x ( y ) + i x + 3,y = ( x ) + x ( y ) + [ x ,y ] + I x y = { ( x ) + x ( y ) + [ x , y ] - ( X L . ) ) + A’xy. It follows that (x,y ) = ( x ) + ( x y) + [x,j-1- ( x y ) , and this is formula (ii) I’x

since each term lies in the abelian group A .

I

Definition: B2(G,A ) is the set of all functions f ;G x G + A for which there is a function ( ): G --t A with (1) = 0 such that

f(x,Y ) = X ( Y >

-

(-.v>

+ (x>.

The elements of B 2( G, A ) are called coboundaries. Theorem 5.7: B2(G,A ) is a subgroup of Z 2 ( G ,A).

Proof: A computation is again left to the reader. In particular, one shows that coboundaries satisfy the two properties of Theorem 5.5 and hence are factor sets. I Definition: e(G, A ) = Z Z ( G , A ) / B 2 ( G , A ) .

We summarize our results. Corollary 5.8: Two factor sets arising from un extension via two choices of liftings determine the same element of e( G, A).

We have been led to the following equivalence relation. Definition: Two extensions 0 + A .+ E 4 G -+ I and 0 --t A + E’ + G + 1 are equivalent if there are factor sets [ ] and ( ) of each with [ ] - ( ) E B2(G, A ) .

156

5 Extensions of Groups

A semantic point: equivalence is a relation among extensions and not merely among factor sets: if [ 3’ and ( )’ are two other factor sets of these extensions, then [ 1’- ( )’ E B2(G,A ) also, as the reader should check. Theorem 5.9: Two extensions of A by G are equivalent f i t in a commutative diagram’ :

if and only if they

0-A-E-G-1

O-A-E’----+G-l

Proof: If the extensions are equivalent, then the construction of Theorem 5.5 allows us to regard E as all a ;Ix with addition

+

+ Ax) + (a’ + Ay) = a + xu’ + [ x , y ] + Axy. Similarly, we may regard E’ as all a + A’x (denote the factor set here by (a

( x ,y)). We are told that

[x, Y ]

- (x, y ) = x( I.> - ( X Y >

+ (x>.

Define q:E -+ E’ by a + Ax H a + ( x ) + A‘x. It is obvious the diagram commutes, and a dull computation shows that rp is a homomorphism. Conversely, suppose 13.: G -+E is a lifting determining a factor set [ 1. By commutativity of the first square, [x, y ] E A gives rp[x, y ] = [x, y ] , so that rp[ ] = [ 1. By commutativity of the second square, qA:G + E’ is also a lifting, and, applying cp to the defining equation

Ax

+ AJ = [x, y ] + Axy

shows that rp[ ] is the factor set determined by ~13.. If ( ) is any factor set of E’, however, we know that ( ) is equivalent to rp[ 1, by Corollary 5.8. Since rp[ ] = [ 1, the two extensions are equivalent. I Remark: Diagram chasing shows that cp must be an isomorphism. Although the equivalence relation is defined in a natural manner, it can happen that there are inequivalent extensions of A by G with isomorphic middle groups. If p is a prime, we shall see later that e ( Z / p Z ,Z / p Z ) z Z / p Z . If 0 + Z / p Z -+ E -+ Z / p Z + 1 is an extension, then E has order p z , hence is abelian [Rotman, 1973, p. 841. It follows easily (one ought not invoke the

’

One usually defines equivalence of extensions In terms of this diagram; we have chosen to give the definition in terms of factor sets because it arises directly from calculations.

157

Extensions of Groups

Fundamental Theorem of Finite Abelian Groups) that E E Z / p z Z or E z Z / p Z 0 Z / p Z . There are thus only two middle groups E, but there are p equivalence classes of extensions. Here is an explicit construction of two such inequivalent extensions. Let p be an odd prime, let A be cyclic of order p with generator a, and let E be cyclic of order p z with generator x . Define i :A -,E by ia = px; define j :A + E by j a = 2px. Suppose these definitions lead to equivalent extensions, i.e. there is a commutative diagram 0AA

0AA

-i

.I

E

E

0

E/iA-

-

E/jA-

0

Note that iA = j A , so we may take the same map n : E E/iA in both extensions. Now commutativity of the first square shows that /3 is multiplication by 2, but this is not compatible with commutativity of the second square. -+

Corollary 5.10:

If e(G, A ) = 0, then every extension

is split, and E is a semidirect product.

Proof: By Corollary 5.8, any given extension is equivalent to the “obvious” extension 0 -* A + E’ + G + 1 in which E’ is a semidirect product. By Theorem 5.9, there is an isomorphism cp: E -+ E’ showing that E ? E and that the given extension is split. I

The Schur-Zassenhaus lemma (Theorem 10.27) gives a condition guaranteeing that e(G,A ) = 0 : if G and A are finite groups whose orders are relatively prime. The theorem goes on to say that if C and C’are complements of A in E, then C and C’ are conjugate. Let us examine this. Definition: A derivation (or crossed homomorphism)is a function ( ): G -+ A with

(XY) = X ( Y > + (x). The set of all derivations, Der(G, A ) , is an abelian group under pointwise addition. If A is a trivial G-module, then Der(G. A ) = Hom(G, A).

158

5 Extensions of Groups

Exercise: 5.10. Let A be a semidirect product of A by G, and let A: G + E be a lifting. (i) I x = ( ( x ) , x ) for some (x) E A ; (ii) I is a homomorphism if and only if ( ) is a derivation. Definition: An automorphism cp of E stabilizes an extension 0 + A + E G + 1 if the following diagram commutes:

-,

0-A-E-G-l

A G E 10-Exercise: 5.11. Given an extension 0 + A + E + G -P .l, the set of all stabilizing automorphisms of E, denoted s(E), is a subgroup of Aut(E). (Of course, s ( E ) depends on the choice of extension having middle group E.) Theorem 5.11: Let 0 + A + E isomorphism s ( E ) z Der(G, A).

+ G -+

1 be an extension. There is an

Proof: Let A:G -+ E be a lifting. Then E consists of all a + Ax, so that if cp E Aut(E) is stabilizing, then cp(a + Ax) = a + ( x ) + Ax for some ( ): G + A. A short computation (using the fact that cp is a homomorphism) shows that ( ) is a derivation. We claim that the derivation ( ) does not depend on the choice of lifting I . If A‘:G + E is another lifting, then cp(a + I’x) = a + [XI + A’x, and [XI = cpA‘x - L’x E A . Therefore, -(x)

+ [XI = I x - cplx + cpA‘x - I‘x = I x + cp( - Ax + A’x) - I’x = 0

+

since -Ax L’x E A, hence is fixed by cp. The function 1, there is a diagram a EXt"( B) +EXt"(UPk, B) Ext"-'(UPk,B) Ext"- '(uLk,B)

n

Ext" - '(Pk, B )

dn

li

Ext" - '(Lk, B

)

y

n

EXtn('4k,B)

-n

Ext"(Pk, B)

The first and last terms in each row are 0 (Theorem 7.6), so exactness gives a and 8' isomorphisms. By induction, there is an isomorphism $, and d'$a-':Ext"(UA,, B) Ext"(A,, B) is an isomorphism. 1 -+

n

Theorem 7.14: For all n, Ext"(A,

n&) nExt"(A, S

Bk).

Proof: Dual to the proof of Theorem 7.13. In particular, begin with exact sequences 0 -+ B, -+ Ek -+ Q, 0, where E, is injective, and then apply the long exact sequence for Ext(A, ) to 0 -+ n B k -+ n E k -+ 0. -+

nQk -+

Given a complex of R-modules and R-maps, the complex obtained after applying Hom, is a complex of abelian groups and Z-maps. If R is commutative, however, we may consider R-modules as (R - R)-bimodules and equip each Hom, with an R-module structure (Theorem 1.15). Also, the maps between the Hom's are then R-maps.

Theorem 7.15: If R is commutative, then Ext:(A, B ) is an R-module. Proof: Ext" is, by definition, a quotient of two R-submodules of HomR(P,, B), where P , is the nth term of a projective resolution of A. I

We generalize this result. Let r E Z(R), the center of R, and let A be an R-module. Then p r :A + A defined by a H ra is an R-map, called multiplication by r(or homothety). Ifr 4 Z(R), then p r may not be an R-map (e.g., take A = R).

Theorem 7.16: If p : A - + A is multiplication by r, where r E Z(R), then p*:Extt(A,B) -+ Exti(A,B) is also multiplication by r. The same is true in the second variable.

Proof:

Consider the diagram

. . --+PI+ P O - + A - + O

201

Elementary Properties

where the rows are identical projective resolutions of A . Recall the definition of p*: first fill in a chain map g over p, say g = { g n : P ,+ P,},then apply Horn,( ,B), and p* is the induced map p*(zn

+ boundaries) = g;z. + boundaries.

We also know that any choice of chain map over p gives the same p*. In particular, define g by letting g,:P, -+ P , be multiplication by r. This is a chain map over p, and p*(z,

+ boundaries) = rz, + boundaries = r(z, + boundaries).

1

Exercises: 7.2. Use Theorem 7.16 to show that Ext",A,B) is a Z(R)module. 7.3. Let R be a domain and A an R-module. Prove that A is torsion-free if and only if every multiplication p,:A + A (with r # 0) is monic; A is divisible if and only if every such multiplication is epic. 7.4. Let R be a domain and A a divisible R-module. If rA = 0 for some r E R, r # 0, then A = 0. 7.5. Let Tbe a functor that preserves multiplications:whenever p r :A -, A is multiplication by r E Z(R), then Tpr:TA ---* T A is also multiplication by r. If R is a domain and A is a torsion-free, divisible R-module, then T A is also torsion-free divisible. 7.6. Let R be commutative and A an R-module with r A = 0 for some r E R. If T preserves multiplications, then rTA = 0. 7.7. Every additive functor T :Ab + Ab preserves multiplications. More , generally, if R is a subring of Q, then every additive functor 7':JJJI + m preserves multiplications.

The theorems we have just proved are useful in computing Ext. We give the next result for abelian groups, although it obviously generalizes to modules over any domain. Theorem 7.17: If B is an ahelian group, then

Exti(Z/rnZ, B) 2 B/rnB. Proof: Apply HornZ(,B ) to 0 + Z 5 Z 4 Z/mZ -+ 0, where the first map is multiplication by rn, to obtain exactness of

Hom,(Z, B) 1 :Hom,(Z, B ) + Exti(Z/rnZ, B) -,Exti(Z, B). Since Z is projective, even free, the last Ext = 0; also, the first map is still multiplication by m, while Hom,(Z, B ) may be identified with B (Theorem 1.16). It follows that Exti(Z/mZ, B) z B/rnB. I

202

7 Ext

If A and B are f.g. abelian groups, recall the Fundamental Theorem asserting each of them is a sum of cyclic groups; the results above permit explicit computation of Extd(A, B). Exercises: 7.8. IfA and Bare arbitrary abelian groups, then Ext$(A, B) = 0 for all n 2 2. (Hint: Use Corollary 4.19: every subgroup of a free abelian group is free abelian.) 7.9. If A is an abelian group with mA = A for some m E Z, then every exact sequence 0 + A + E -+Z/mZ -+ 0 splits. 7.10. If A and C are abelian groups with mA = 0 = nC, where (m,n) = 1, then every extension of A by C is split. 7.11. If A is a torsion abelian group, then Exti(A, Z) z Horn(A,R/Z), where R/Z is the circle group. 7.12. (i) If G is a nonzero divisible group, then Hom(Q, G ) # 0. (ii) If p is prime, then Z/pmZ)/(Urn Z/prnZ)is divisible. 7.13. Using Exercise 7.12, prove there is a nonsplit extension

(nrn

O+UZ/prnZ+E-+Q+O. rn

(We shall see later (Theorem 9.3) that this result is false if the index set of the sum is finite.)

EXT' AND EXTENSIONS

We seek an interpretation of Extk(C, A ) , and we begin with a definition motivated by Theorem 5.9.

Definition: Two extensions commutative diagram

r and r' of A by C are equivalent if there is a

5': 0 - A - E ' A C - O It is easy to check that equivalence is, in fact, an equivalence relation (to prove symmetry, use the Five Lemma to show that cp must be an iso-

203

Ext’ and Extensions

morphism). The equivalence class of an extension 5 : O -+ A -+B + C + 0 is denoted [ 0. There is an exact sequence of R*-modules 0 -+ K *

+ F* + A*

4

0,

246

9 Son of Specific Rings

where F* is R*-free. By Lemma 9.26, @,(A*) I 1 + max{pdR(K*),pdR(F*))I 1

+ max{pd,(K*),pd,(R*)).

Since A* is not R*-projective, Exercise 9.7 gives pd,.(K*) = n - 1, so that induction gives pd,(K*) 5 n - 1

+ pdR(R*).

Combining these inequalities: pd,(A*) I 1

+ mU{n - 1 + pdR(R*),pdR(R*)}I n + pdR(R*).

Finally, suppose we have equality whenever A* # 0 and pd,dA*) I 1. Assume @,,(A*) = n 2 2, and let pd,(R*) = r. By Exercise 9.7, the exact sequence of R*-modules 0 -+ K* + F*

+ A* + 0,

with F* an R*-free module, has pdR.(K*) = n - 1 2 1. In particular, K* # 0 and so induction gives pdR(K*) = n

- 1 4-1.

Denoting pdR(A*)by a, Exercise 9.12 gives

r = pd,(F*) I max{n - 1 + r,a}.

This inequality is strict, for

max{n - 1 + r , a } 2 n - 1 + r > r.

The remainder of Exercise 9.12 gives

pdR(A*) = pdR(K*) -k 1 = (n - 1 + I) + 1 = n -k r.

I

Theorem 9.33: Let x E R be a central element that is neither a unit nor a zero divisor. l f R* = R/Rx, and if lD(R*) < co,then ID(R) 2 ID(R*) + 1.

Before proving this result, let us draw an instant consequence. Theorem 9.34: For any ring R,

ID(R[r]) = lD(R) + 1.

Proof: Corollary 9.30 gives the inequality 1.Corollary 9.28 allows us to assume ID(R) < 0 0 ; since R[t]/tR[t] r R as rings, Theorem 9.33 applies to give the reverse inequality. I

247

Hilbert’sSyzygy Theorem

Corollary 9.35

(Hilbert’s Theorem on Syzygies): If k is a Jield,

D(k[t,, . . . , r,])

= n.

Proof: Induction on n, after observing D(k) = 0.

Corollary 9.36: If k is afield and R has a free resolution

= k [ t , , . . . , t,],

0 - F, -+. . . -+ F ,

-+

A

-+

I then every R-module A

0.

Proof: Corollary 9.35 and the Quillen-Suslin theorem.

Note that the hypothesis lD(R*) < LX in Theorem 9.33 is essential. An example is provided by R = Z and x = 4 : D(Z)= 1 and D(Z/4Z)= co. Proof of Theorem 9.33: It suffices to show that if A* is a nonzero R*module with pd,,(A*) < 00, then

pdR(A*) = pdp(A*) f 1 (for then take sup of both sides, realizing that not every R-module need be of the form A*). Since x is central, multiplication by x is an R-map R -+ R ; moreover, x not a zero divisor gives an exact sequence of R-modules

o-+ R:R+

R*+o.

1; in fact, pd,(R*) = 1, otherwise this sequence It fo~lowsthat pd,(R*) I would split, forcing Rx to be a summand of R, and this would imply x is a zero divisor. By Theorem 9.32,

pd,(A*) I pdp(A*) + 1;

moreover, if we can prove equality whenever A* # 0 and pd,,(A*) I 1, we are done. It suffices to eliminate strict inequality. If pd,,(A*) = 0, then pd,(A*) I 1 and strict inequality means pd,(A*) = 0, i.e., A* is R-projective. But xA* = 0, while X P # 0 if P is any nonzero projective ( P is a submodule of a free R-module, hence each element of P has coordinates in R; since x is not a zero divisor, x cannot annihilate any nonzero element of P). Since A* # 0, pd,.(A*) # 0. If pd,,(A*) = 1, then pd,(A*) I 2 and strict inequality means pd,(A*) I 1. We have already shown pd,(A*) # 0 ; we claim pd,(A*) # 1. Otherwise there is an exact sequence of R-modules 0 -+ K

-+

F

+ A*

-+

0,

where F is R-free and K is R-projective. Since xA* = 0, there is an exact

248

9

Son of Specific Rings

sequence of R*-modules

0 + K / x F + F/xF + A*

-+

0.

Now F/xF is R*-free, so that pd,.(A*) = 1 shows that K / x F is R*-projective. The exact sequence of R*-modules 0 + xF/xK

+K / x K +K/xF

--*

0

must split, whence xF/xK E FJK z A* is a summand of K I x K . As K is R-projective, K / x K is R*-projective, and thus A* is R*-projective, contradicting pd,.(A*) = 1. I We now offer a second proof of Theorem 9.33. After all, the "moreover" clause of the crucial Theorem 9.32 does remind one of the slow starting inductions so characteristic of dimension shifting. An obvious way to prove lD(R[t]) 2 m for some (finite) m is to exhibit a pair of R[t]-modules A and B with Extm(A,B) # 0; we shall be able to do this once we have proved the next result.

Theorem 9.31 (Rees): Assume x E R is a central element that is neither a unit nor a zero divisor, and let R* = R/xR; assume B is an R-modulefor which multiplication by x, p = p x : B-+B, is monic. Then, for euery R*-module A, there are isomorphisms Ext",(A, B/xB) z Exti+ ' ( A ,B). Proof: First of all, the natural map R + R* enables us, by change of rings, to regard A as an R-module; since an R*-module is merely an R-module annihilated by x, the quotient B/xB is an R*-module; thus, both sides make sense. Next, we claim HOm,(A, B ) = 0. If a E A, then xa = 0, for A is an R*module. Consequently, if f :A + B, then xf(a) = f ( x a ) =f(0) = 0,

and the hypothesis on B forces f = 0. To prove the theorem, we use the axioms characterizing the contravariant Ext functors (Exercise 7.27, proved by dimension shifting). Define contravariant functors G": + Ab, n 2 0, by G"(A) = Ext;'

'(A, B).

Clearly { G"} is a strongly connected sequence of contravariant functors. Exactness of 0 + B 1:B + B/xB + 0 gives exactness of HOmR(A,B) + HOm,(A, B/xB) + Exti(A, B) 5 ExtA(A,B).

249

Hilbert's Syzygy Theorem

We have already seen that Hom,(A,B) = 0. O n the other hand, p* is also multiplication by x (Theorem 7.16); since A is annihilated by x, we have imp* = 0. We conclude that the connecting homomorphism is an isomorphism HOm,(A, B / x B ) 1Extk(A, B). However, R -,R* onto allows us to use Exercise 9.13 to conclude Hom,,(A, B/xB) = Hom,(A, B/xB). There is thus a natural equivalence Go Horn,( ,B/xB). It remains to prove that Cn(P*) = 0 for all free R*-modules P* and all n 2 1. Choose a basis of P*, and let Q be the free R-module with the same basis. There is an R-exact sequence O+ Q : Q

-+

P* - 0

(again we have used x a central non-zero-divisor). Exactness of

. . . + EXt;(Q, B ) + Ext;' '(P*,B ) + Ext:'

'(Q, B ) + . .

shows that G"(P*)= Ext;'

' ( P * ,B ) = 0,

n 2 1.

1

The axioms have been verified and the theorem is proved.

Corollary 9.38 ( = Theorem 9.33): Assume x E R is a central element that is neither a unit not a zero d i i i w r , and l r ~ rR* = R j x R . I f lD(R*) = n < GO, then ID(R) 2 ID(R*) + 1. Proof: Assume M is an R*-module with pd,.(M) = n. By Exercise 9.6, there is a free R*-module F* with Ext;(M,F*) # 0. If we define B as the free R-module on the same basis as F* (precisely as in the previous proof), then B / x B z F*. By the theorem of Rees,

Exti' ' ( M ,B ) 2 Ext;*,(M, F*) # 0, and this says pd,(M) 2 n + 1. Therefore, ID(R) 2 n + 1

=

ID(R*)

+ 1.

1

There is another comparison theorem that is often useful. Definition: Let R be a (not necessarily commutative) ring. A right R-module M is faithfully flat if M is flat and whenever M ORN = 0, then N = 0. Examples: 5. R[t] is faithfully flat, for it is a free R-module. 6. If R is the ring of integers in an algebraic number field, then R is a faithfully flat Z-module (for R is Z-free). 7. Q is a flat Z-module, but it is not faithfully flat.

250

9 Son of Specific Rings

Exercise: 9.14. Assume R is a subring of A with AR flat. Prove that AR is faithfully flat if and only if A QR(RII) # 0 for every proper left ideal Z of R. Conclude that A is faithfully flat if and only if AZ # A for every proper left ideal I of R. Theorem 9.39 (McConnell-Roos): Assume R is a ring with lD(R) = n < 00, and that R is a subring of A, where A, is faithfully p a t . Assume either

(i) (ii)

RA RA

is projective, or isJlat and R is left noetherian.

Then 1D(R)I ID(A). Remark: Goodearl [I19741 gives an example showing one must assume R has finite global dimension. Proof: For any left R-module M, there is a map cp:M -,A Q M given by m H 1 Q m; we claim cp is monic. If K = ker cp, then K = ( m E M: 1 Q m = O}. Since A R is flat, exactness of 0 + K f M gives exactness of

O+AQK-

l@i

AQM.

But the definition of K gives im(1 @i)

=

(1 @ m

E

A @ M : m E K } = 0;

there is thus an exact sequence 0 + A @ K + 0. Therefore A Q K = 0, so that A faithfully flat implies K = 0. Assume (i): R A is projective. Choose RM with pdR(M)= n. There is an R-exact sequence 0 - M - , A @ M -+C-,O.

Now pd,(C) I n, since ID(R) I n, so Exercise 9.9 gives pdR(A@ M ) = n. It suffices to prove pd,(A Q M) = n, for then lD(A) 2 n, as claimed. Take an R-projective resolution of M : 0 + P , -+. . . + Po + M -+ 0.

Since A R is flat, there is an A-exact sequence 0 + A @ P , + . . . -+A Q P o + A @ M -+ 0.

By Exercise 3.10, each A Q P i is A-projective. Hence pd,(A Q M) I n. For the reverse inequality, choose an A-projective resolution of A Q M:

. . .+ Q,+

+ Q,

4 . .* +

Qo --t A @ M + 0.

251

Serre’s Theorem

Since R A is projective, it is easy to see that each Qi is also R-projective. Therefore, pdR(AQ M ) = n implies (Theorem 9.5(iv)) that we may assume O + K , - + Q , - l +.‘.-+(IO-+ A Q M + O

is an R-projective resolution of A 0 M (where K, = ker(Q.- + Qn-2) considered as an R-module). Hence, n 5 pd,(A Q M ) . Assume (ii): R A is flat and R is left noetherian. The above argument may be repeated, yielding pd,(A Q M) = n and pd,(A Q M ) I n. For the reverse inequality, we use R left noetherian to enable us to switch to flat resolutions (Theorem 9.22). Assume an A-flat resolution of A 0 M :

. . . + F,+l + F , + ’ . .-+ F ,

+A

@ M -0.

Since A R is flat, each F i is R-flat (for Fi OR = (FiQ, A ) QR , and use Exercise 3.38). But now Theorem 9.13(iv) applies, and the proof is completed as in case (i). I

SERRE’S THEOREM We now give an elementary proof of the theorem of Serre mentioned in Chapter 4; the basic idea of this proof is due to Borel, Serre, and Swan and the exposition provides details of the sketch given by Kaplansky [1970, p. 1341. Definition: A module M has FFR (finite free resolution) of length I n if there is an exact sequence O-+F,+F,-l

+.-.+Fo+M+O

in which each Fi is f.g. free. If M has FFR, then M is Eg., even finitely related. The reader may easily provide examples of such modules. Exercises: 9.15. A projective module has FFR if and only if it is stably free. (Hint: Induction on length.) 9.16. Let R be left noetherian and let M be a left R-module with FFR of length I n . Every free resolution of M , each of whose terms is f.g., has a stably free nth syzygy. (Hint: Generalized Schanuel lemma, Exercise 3.37.) Lemma 9.40: I f M has a projective resolution

0 -+ P, -+. . . + P , f P , 2 Po 5 M d

-+ 0

in which each Pi is stably free, then M hirs FFR qf length I n

+ 1.

252

9

Son of Specific Rings

Proof: We do an induction on n. If n = 0, then E is an isomorphism Po G M . Since Po is stably free, there are f.g. free modules F , and F,with F , E Po OF,, i.e., there is an exact sequence 0 -+ F, + F , -+ M --f 0. If n > 0, let K O = ker E. If F is a f.g. free module with Po 0 F free, then we have exactness of O+P, +-.

dr@lp -,p2 fd' P, 0 F -P,O

F

5 M -0,

where d ; : p 2 H ( d , p 2 , 0 ) and ~ ' : ( p ,f ,) H ~ ( p , ) .Since kerd = K O0 F has a stably free resolution with n - 1 terms, the proof is completed by induction. I Theorem 9.41 : Let R be left noetherian and0 + M' -+ M + M" -+ 0 an exact sequence of left R-modules. I f two of the modules have FFR, then so does the third one. Proof: Since two of the modules have FFR, they are f.g.; since R is noetherian, the third module is also f.g. That R is noetherian also provides free resolutions of M' and M" each of whose terms is f.g. (Corollary 4.4). Use the Horseshoe Lemma 6.20 to insert a resolution with f.g. free modules in the middle. For each n, there is an exact sequence of syzygies

0 -+Kb -+ K , -+ K :

-+

0.

Assuming two of { M ' ,M , M " } have FFR of length 5 n, Exercise 9.16 shows two of { Kb, K , , K i ) are stably free. If one of these is KZ, then the sequence of syzygies splits and the third syzygy is also stably free; Lemma 9.40 now gives the result. If KL and K , are stably free, then 0 + KL -+ K , -+ F,'-

4..

. + FG -+ M" -+ 0

is a resolution of M" by stably free modules, and Lemma 9.40 gives the result in this case as well. I Definition: A family 8 is a nonempty subclass of R?Dl such that whenever 0 -+ M -+ M + M" + 0 is exact and two of the modules lie in 8, then the third one lies in 8 as well.

Theorem 9.41 states that the class of all FFR modules over a left noetherian ring is a family. There are some easy observations. Every family 5 contains the zero module: if M E 8, then 0 -+ M -% M 0 -+ 0 is exact. From this it follows that M' z M and M E 3 implies M' E 3: families are closed under isomorphism. It is easy to see that any intersection of families is again a family. Thus, every subclass 6 of RYJl generates a family. --f

253

Serre's Theorem

If 6 is a subclass of $X containing the zero module, define E ( 6 ) as the subclass of all modules contained in a short exact sequence having two terms in G. Clearly E(G) is a subclass containing 6. Define an ascending chain of subclasses :

E o ( 6 )= 6 ;

E n +'( Z) = E ( E " ( 6 ) ) .

Lemma9.42: I f G is a subclass of R'iJJnl containing the zero module, then u2=o E " ( 6 ) is the family generated by 6.

Proof: It is clear that any family containing 6 must contain each E"(G), hence the union of them. On the other hand, u E " ( 6 ) is a family containing G : if 0 -+ M' + M .+ M" + 0 is exact with two terms in UE"(G),then these two terms lie in some E " ( 6 ) ,and so the third lies in E ( E " ( 6 ) )= E""(6). I Corollary 9.43: Let G be a class of modules over a left noetherian ring, each member of which has FFR. Then ezlery module in 8, the family generated by (5,has FFR.

Proof: If M E 8, there is a least n with M E E"(G);we do an induction on n. If n = 0, then M E 6 and has FFR by hypothesis. If n > 0, then M is contained in a short exact sequence whose other two terms lie in E " - ' ( 6 ) and hence have FFR, by induction. By Theorem 9.41, M has FFR. Theorem 9.44: Let R be commutative northerian and assume every t g . R-module has FFR; then every Jq. R[t]-module has FFR. Before proving Theorem 9.44, let us deduce Serre's theorem from it. Theorem 9.45 (Serre): module is stably free.

If k i s a.field, then eoeryjig. projective k [ t , , . ..,t,]-

Proof: We first do an induction on n that every f.g. k [ t , , . . . , t,]-module has FFR. If n = 1, then k [ t ) is a principal ideal domain and every f.g. module 1. If n > 1, let R = k [ t , . . . . , t n - l ] ,which is noetherian has FFR oflength I by the Hilbert Basis Theorem. By induction, every f.g. R-module has FFR, so that Theorem 9.44 gives every f.g. R[t,]-module has FFR. In particular, every f.g. projective module over R [ f n ]= k [ t , , . . . , t,] has FFR, hence is stably free by Exercise 9.15. I

To prove Theorem 9.44, we use two simple lemmas, the first of which is a form of Zorn's lemma available in noetherian rings. Lemma 9.46: I f R is a left noetherian ring, then every nonempty set 9l of left ideals of R contains a maximal member.

254

9

Son of Specific Rings

Proof: Assume 9l has no maximal member. Choose I , E 9l; since I , is not maximal, there is I , E 9l with I , S I , . This procedure may be iterated indefinitely, so the axiom of choice provides a strictly ascending chain that violates ACC. I

Remark: One may show that the property in Lemma 9.46 characterizes left noetherian rings. Definition: If M is a left R-module, then its annihilator is ann(M) = { r E R:rM = O}. One defines the annihilator of an element x E M by ann(x) = ann(Rx). It is clear that ann(M) is a left ideal in R. Lemma 9.47: Let R be commutative noetherian and M a nonzero R-module. If 1 is an ideal maximal among {ann(x):x E M,x # 0}, then I is a prime ideal.

Proof: Choose x E M with I = ann(x). Let ab E I and b 4 I , i.e., abx = 0 and bx # 0. Then ann(bx) =I I

+ Ra 2 I .

If a 4 I , the maximality of I is violated. Therefore, I is prime.

I

Proof of 9.44: Consider the statement: the family 8 in RLt,!Ul generated by all (extended) modules R[t] @ N , where N is a f.g. R-module, consists of all f.g. R[t]-modules. If this is true, then we may invoke Corollary 9.43, for N has FFR implies R[t] @ N has FFR (because R[t] is R-flat). Assume there are f.g. R[t]-modules not in 5. By Lemma 9.46, there is an ideal I maximal among {ann(M) n R:M is a f.g. R[t]-module not in 8). Choose such a module M with I = ann(M) n R. We normalize by replacing R by R/I and R[t] by (R/I)[t]. If 8‘ is the family generated by all modules extended from f.g. R/l-modules, then it is easy to see that A E 8 implies A / I A E 8’. If M‘ is a f.g. (R/I)[t]-module with J / I = ann(M’) n (R/I) # 0, then J M ’ = 0. Construing M’ as an R[t]module, ann(M‘) n R = J 3 1 ; maximality of I shows M‘ E 8,hence M’E 5’. The force of the normalization is that we may assume that if M’ is a f.g. R[t]-module with ann(M’) n R # 0, then M‘ E 8. In the chosen module M (over the noetherian ring R[t]), we may apply Lemma 9.46 to find a nonzero x1 E M with ann(x,) maximal among {ann(x):x E M,x # 0). Similarly, there is x2 E M with ann(x, + (x,)) maximal among {(ann(x + (xl)):x + (xl) E M/(x,),x + (xl) # O}. Iterating,

255

Serre's Theorem

there are elements x l , x2,.. . in M and submodules M , ( x l , x z , . . . , x i ) with

=0

and M i=

0 = M , c M , c M2 c ....

By construction, M i / M i - 2 R[t]/P,, where Pi= ann(x, + M i - 1) is a prime ideal (Lemma 9.47). By Corollary 4.7, R[t] noetherian and M fg. imply M has ACC; therefore, M = M , for some r. By induction on r, using the fact that 8 is a family, M E 8 if R[t]/P E 6 for every prime ideal P in R[t] (and this is the desired contradiction). Now P n R c ann(R[t]/P) n R, so the normalization gives R[t]/P E if P n R # 0. We may thus assume P n R = 0; as P n R is always a prime ideal in R, we may assume R is a domain; let Q be its quotient field. Let P = (g . . . ,g,,) and choose f E P a nonzero polynomial of minimal degree. We claim ann(P/(f)) n R # 0. For each i, 1 s i In, the division algorithm in Q [ t ] yields gi = fqi

+ ri,

where q i and ri E Q [ t ] and degree ri < degree f . Clearing denominators, there are nonzero a, E R with aigi = fqi

+ r;.

where q: and r: E R[t] and degree ri < degree f . By minimality of degree f, aigi = fq: E ( f ) for all i. If a = na,, then a # 0 and a P / ( f ) = 0. Hence a E ann(P/(f)) n R, and the normalization gives P / ( f ) E 8. Exactness of 0

together with ( f )

+

E

+

p

+

P / ( , f )--* 0

E R[t] E 8 imply PE3,for 8 is a family. Finally, exactness of

0 +P shows R[t]/P

(f)

+ R[t] -+

R[t]/P

+0

8, as desired. I

We mention a useful tool; for some applications, see [Kaplansky, 1970, 44.31. Exercises: 9.17. Assume R is a ring with IBN and O-+F,+. . . -+F,-+M-+O is R-exact, with each F if.g. free. Prove that x ( M ) = ~ ~ = , ( 1)' rank(F,) is independent of the choice of FFR. (Hint: Schanuel.) The integer x ( M )is called the Euler characteristicof M . Compare Exercise 3.6. 9.18. Assume R has IBN and 0 M ' M + M" + 0 is an exact sequence of left R-modules, two of which have an Euler characteristic. Then the third module also has an Euler characteristic, and -+

-+

x ( M )= z ( M 7 + x ( M " ) .

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9 Son of Specific Rings

MIXED IDENTITIES Various natural isomorphisms involving tensor and Hom can be extended to isomorphisms involving Tor and Ext. Theorem 9.48: Let R and S be rings, and consider the situation ( A R ,RBS, J). If A is p a t , there are isomorphisms

Toc(A @R B, C )

A

@R

To$(B, C).

Proof: Let PBbe a deleted projective resolution of B. Naturality of the associativity isomorphisms for tensor imply an isomorphism of complexes

( A 0P B ) @sC 2 A O R ( P B @ C), and hence isomorphisms, for all n 2 0,

@ c,1. Since A is flat, A @ PBis a (deleted) flat resolution of A ORB ; the left hand side is thus Toe(A ORB, C). To evaluate the right hand side, flatness of A means the functor A OR is exact. By Exercise 6.4, Hn( ( A @ P B ) @S

Hn(A @R

(PB

@ c)) 2 A

c )2 Hn(A @R

@R Hn(PB

(PB

@ c)) = A @R Toe(B, c).

[

There is a very useful variant of this result. Theorem 9.49: Let R be commutative, S a subset of R, and A, B R-modules. There are isomorphisms, for all n 2 0,

S-'Tor:(A,B) 2 TO 0, Theorem 10.26 gives mH"(G,A)= 0. On the other hand, kA = 0 implies kH"(G,A) = 0 (Exercise 7.6). Since (m,k) = 1, it follows that H"(G,A)= 0 for all n > 0. When n = 2, this says that E is a semidirect product of A by G (Corollary 5.10); when n = 1, this says that any two complements of A in E are conjugate (Theorem 5.13). I

Remark: The hypothesis that A is abelian may be removed ([Rotman, 1973, p. 1491 for the first statement; conjugacy is true also, but much harder). Theorem 10.28 (Maschke): Let G be a .finite group of order m, and let k be a Jield whose characteristic does not divide m ; then the group ring kG is semisimple. Proof: By Theorem 5.14, it suffices to prove H'(G,A) = 0 for every kGmodule A. Now l/m E k, since k is a field whose characteristic does not divide m. Therefore, multiplication by m is an automorphism of A (whose inverse is multiplication by l/m) which induces an automorphism of H'(G, A) that is also multiplication by m (Theorem 7.16). It follows that mH'(G, A) = H'(G, A). On the other hand, mH'(G, A) = 0, by Theorem 10.26, whence H'(G, A) = 0.

I

Theorem 10.29: l f G is ajnire group and A is a,$g. G-module, then H"(G, A ) and HJG, A ) are$nitefor all n > 0. Proof: By Exercise 10.4, H"(G,A) and H,(G, A) are f.g. abelian groups for all n 2 0. If n > 0, these groups are annihilated by ]GI, whence the result. I

There is a remarkable connection between cohomology and homology groups of G. For a G-module B,we write B* for the character module (with G-action as in Theorem 1.15):

B*

= Horn,(

B, Q/Z).

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10 The Return of Cohornology of Groups

Theorem 10.30 (Duality Theorem): For every Jinite group G and every G-module B, there are isomorphisms for all n 2 0

(H"(G,B))*z H,(G,B*). Proof: Recall the mixed identity, Theorem 9.51: Let R and S be rings with R left noetherian, and consider the situation (RA,R B S , C,) in which A is f.g. and C is injective. Then there are isomorphisms for all n 2 0,

Tor:(Hom,(B,

C ) ,A ) 2 Hom,(Extl(A, B), C ) .

Set S = Z and R = ZG (which is left noetherian, by Exercise 10.4); take A = Z and C = Q/Z. The isomorphism becomes Tor,ZG(B*,Z) z (Ext",(Z, B))*. By Exercise 10.3, Tor,ZC(B*,Z) z Tor,ZG(Z,B*) = H,(G, B*), while the right hand side is (H"(G,B))*,as desired. Remark: Using the remark after the proof of Theorem 9.51, one may prove the Duality Theorem 10.30 for any group G for which the trivial Gmodule Z has a G-projective resolution each of whose terms is f.g.

The next theorem explains why H,(G,Z) is called the multiplier of G when one recalls the discussion of projective representations. Theorem 10.31 : Let G be a Jinite group, and let k be an algebraically closed field of characteristic 0. If k # is the multiplicative group of nonzero elements of k considered as a trivial G-module, then

H,(G,Z) 2 H 2 ( G , k # ) . Proof: Since k is algebraically closed, every element in k # has an nth root for every n > 0. This says that the multiplicative abelian group k" is divisible. The torsion subgroup of k', consisting of the roots of unity, is isomorphic to Q/Z because k has characteristic 0 (consider k = C to convince yourself). Divisibility, hence injectivity, of Q/Z gives

k#

= Q/Z 0 V ,

where V is a vector space over Q (Exercise 3.20); thus, H2(G,k#) z H2(G,Q/Z)0 H 2 ( G ,V ) .

However, the argument of Theorem 10.28 gives H2(G, V) = 0, whence

H2(G,k')

r H Z ( G Q/Z). ,

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Computations and Applications

Exactness of 0 -+ Z + Q + Q/Z -+ 0 gives exactness of

H2(G,Q)

-+

H2(G,Q/Z)

+

H3(G,Z)

+

H3(G,Q),

and the outside terms are 0 by the Theorem 10.28argument. We conclude that 2 H 2 ( G ,QiZ) z H 3 ( G ,Z).

H2(G,)'k

A similar argument gives an isomorphism in homology H3(G, Q/Z) 2 H,JG. Z). Applying the Duality Theorem 10.30 yields an isomorphism (H3(G,Z))*z H,(G,Z*). By Theorem 10.29, H 3 ( G ,Z) is finite, whence ( H 3 ( G ,Z))* z H 3 ( G , Z). Also, Z* 2 Q/Z, so that H3(G,Z) 2 H,(G, Q/Z). Assembling all the isomorphisms gives H 2 ( G ,k') z H2(G,Z), as desired. I Observe that H , and H 3 were useful to us, even though we have not given them down-to-earth interpretations. Definition: A group G has cohomological dimension s n, denoted cd(G) I n, if Hk(G,A ) = 0 for all G-modules A and all li > n. If n is the least such integer, one defines cd(G) = n ; if no such integer n exists, define cd(G) = 00.

Exercises: 10.16. Prove that cd(G) = 0 if and only if G = (1). 10.17. If G is free of rank >O, then cd(G) = 1. 10.18. If H " + ' ( G , A )= 0 for all G-modules A , then H k ( G , A )= 0 for all k > n and all G-modules A . 10.19. Let G be free abelian with basis S = ( x I , .. . , x,}. Prove that ZG z S-'Z[x,, . . . , x,]; use Hilbert's Syzygy Theorem to prove cd(G) I n + 1. Remarks: 1. The ring of Laurent polynomials in consists of all formal sums

C mixi,

.Y,

coefficients in Z,

fl

m iE Z,

i=k

k, n (possibly negative) integers, with obvious addition and multiplication. One may easily generalize this definition to several (commuting) variables, and observe, using Exercise 10.19, that Laurent polynomials in n variables, coefficients in Z, is ZG, where G is free abelian of rank n. 2. If G is free abelian of rank n, then cd(G) = n (Corollary 11.50). The reason the bound in Exercise 10.19 is too high is that global dimension I n

296

10 The Return of Cohomology of Groups

demands vanishing of ExtiG(B,A),all k > n and all pairs of G-modules B, A, whereas cd(G) In only demands such vanishing in the special case B = Z.

Is there a relation between cd(S) and cd(G) when S is a subgroup of G? Theorem 10.32 (Shapiro's Lemma): Zf S is a subgroup of G and A is an S-module, then, for all n 2 0,

H"(S, A) 2 H"(G,Hom,(ZG, A)). Proof: First of all, the right side makes sense, for Homs(ZG, A) may be regarded as a G-module (as in Theorem 1.15, i.e., as in the adjoint isomorphism). A mixed identity, Exercise 9.23, arising from the adjoint isomorphism, gives Ext",(ZG @G Z, A) z EXtiG(Z,Homs(ZG,A)). Since ZG & Z z Z, this is the desired isomorphism H"(S, A) 2 H"(G,Hom,(ZG, A)).

I

Corollary 10.33: If S is a subgroup of G, then cd(S) 5 cd(G).

Proof: Without loss of generality, we may assume cd(G) = n < 00. I f k > n and A is an S-module, then Shapiro's lemma gives I l k ( S, A )

where A*

= Hom,(ZG,A).

z

H k ( G, A*) =

0,

I

Lemma 10.34: Let G be a j n i t e cyclic group of order k and with generator x. Dejne elements of ZG: D =x - 1

and

N

=

1+ x

+ x2 + . . . + x k - ' .

Then the following is a G-free resolution of Z:

. . .+ ZG 5 ZG

ZG 3 ZG -$ Z 4 0 ,

where D and N denote multiplication by D and N , respectively. Proof: First, ZG is commutative, so multiplications are G-maps. Since ND = DN = xk - 1 = 0, we do have a complex. Since ker E = g = im D (Theorem 5.19), there is exactness at the first step. Suppose u = CfZ; mixi E ker D. Computing gives mo = m, = . . . = mkso that u = moN E im N . Finally, if u E ker N, then 0 = uN = ~~1~ rni)xj, so that c m i = 0, and u = -D[mol + (mo + m l ) x + . . . + (mo + . . . + m k - l ) ~ k - lE] imD. I

(ZZ;

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Computations and Applications

Exercises: 10.20. Prove Lemma 10.34 using Gruenberg's resolution with respect to the presentation (x I xk)of G. 10.21. If G is finite cyclic, and A is a G-module, then D A = gA. Theorem 10.35: Let G be a j n i t e cyclic group of order k. If A is a G-module, dejine NA = { a E A : N a = O}. Then H 2 " - l (G, A ) = NA/DA,

Ho(G,A ) = A',

H2"(G,A)= A G / N A .

Proof: Apply HornG(, A ) to the resolution of Lemma 10.34. Corollary 10.36: Let G be a jinite cyclic group of order k and let A be a trivial G-module. Then Ho(G, A ) = A, H 2 n - 1( G , A ) = , A = { a E A : k a = O } , H2"(G,A ) = A/kA.

In particular, HO(G,Z)= z,

H2"-I (G, Z) = 0,

and

H2"(G,Z) = Z/kZ.

Corollary 10.37: If G isjnite cyclic (of order # l), then cd(G) = co.

Proof: Take A = Z considered as a trivial G-module; by Corollary 10.36, H2"(G,Z) # 0 for all n. Corollary 10.38: If cd(G) < co,then G is torsion-free.

Proof: If G contains an element x # 1 of finite order, then Corollary 10.33 shows cd((x)) Icd(G), contradicting Corollary 10.37. I Exercises: 10.22. If G is finite cyclic, then H,(G,A) E H"+'(G,A) for all G-modules A and all n 2 1. (Compare with the Duality Theorem 10.30.) 10.23. If G is finite cyclic, then pd(Z) = l x and D(ZG) = 00. 10.24. If R = Z[x]/(xk - l), then D(R) = co. 10.25. (Barr-Rinehart) Define k ( G , A ) = Ext;,(g,A). Prove that

B o ( G , A )z Der(G,A)

and

Z?'(G, A ) 2 H"+'(G, A),

n 2 1.

(Hint: Theorem 5.21.) What are the groups of cohomological dimension l ? Stallings [1968] proves the following results. Theorem. A jinitely related torsion-free group G is a nontrivial free product if and only if H1(G,(Z/2Z)G)has more than two elements.

298

10 The Return of Cohomology of Groups

The hypothesis on G can be considerably relaxed [Cohen, 19721; the important point here is that Stallings discovered a homological criterion for certain groups to be free products; he used this criterion to prove the next two theorems. Theorem. I f G is j:g. then cd(G) I1 ifand only ifG is free. This theorem was a conjecture of Eilenberg and Ganea. Observe that this generalizes the Schreier-Nielsen theorem. Lemma. If S is a subgroup of finite index in G, then H'(S,(Z/2Z)S) s H1(G,(Z/2Z)G). This is an easy result which we quote to make the next result plausible. Theorem. I f G is $g. torsion-free, and if G hus a free subgroup of finite index, then G is free. This theorem was a conjecture of Serre who proved that if G is torsion-free and S is a subgroup of finite index, then cd(G) = cd(S). Swan [1969], using Stallings's results, was able to remove the hypothesis that G be Eg. in the last two theorems. We end this chapter by voicing the reader's discontent; one must compute homology in order to use it, and there is still difficulty even in determining whether a homology group is nonzero. A technique is needed, and, in fact, a technique exists: spectral sequences. While it does not solve all problems, it is a powerful tool.

11

Spectral Sequences

Spectral sequences (invented by Leray and Koszul) are certain types of sequences of homology modules; however, one often abuses language by saying spectral sequences is a technique of computing homology. Actually, there are two main techniques. The first gives a useful exact sequence, the “five-term exact sequence of terms of low degree”. The second technique involves a notion of convergence of a spectral sequence to a limit module H . We shall be more precise later, but we wish to indicate now to what extent the limit H is determined by the approximating sequence. It is a familiar procedure in group theory to consider a normal series of a group G, G

=

Go 3 G I

1G, 3

...,

and its corresponding factor groups G J G , , It is also a familiar procedure to consider a series of submodules of a module H ,

H = H o ~ H 1 ~ H z ~ . . . , now called a “filtration” instead of a normal series, and its factor modules H i / H i + l . The limit of a spectral sequence is H only in the sense that one obtains the factor modules of some filtration of H . Of course, this usually does not determine H to isomorphism, but it does provide information.

EXACT COUPLES AND FIVE-TERM SEQUENCES We begin with some simple definitions.

299

300

11 Spectral Sequences

Definition: A graded module is a sequence of modules M = { M , : p E Z}. If M = { M , ) and N = {N,} are graded modules and a is a fixed integer, then a sequence of homomorphisms f = { f,: M , -,N,,,} is a map of degree a. One writes f :M + N. P A complex C = . . . -+ C , d-+ C,- I +. * . determines a graded module C = {C,:p E Z] if one ignores the differentiation d = { d , : p E Z). The map d : C + C has degree - 1. If (C, d ’ ) is another complex, a chain map f :C + C gives a map of degree 0 (with fd = df ’), while a homotopy is a certain type of map of degree + 1. A second example of a graded module is the homology of a complex C : H , ( C ) = ( H , ( C ) : p E Z ) . One may reverse this procedure: given a graded module {A,:p E Z}, define a complex

A =.’ .+A,

3 A,-,

4’.

.

in which each d, = 0; one says A is a complex with zero differentiation. Exercises: 11.1. Degrees add under composition: iff :A4+ N has degree + b. 11.2. All graded modules (over a fixed ring) and all maps having a degree comprise a category. Note that

a and g: N + K has degree b, then gf: M + K is a map of degree a

Hom(M, N ) =

U

osz

(n p

Hom(Mp,Np+.)

)

Recall a mnemonic introduced when we first saw long exact sequences: exact triangles. If 0 + A B 5 C -+ 0 is a short exact sequence of complexes, then the long exact sequence may be written .1;

Regarding the vertices as graded modules, the maps f, and g* have degree 0 and a has degree - 1. Conversely, given any exact triangle, one may write down a long exact sequence if he knows the degrees of the maps. Definition: A bigradedmodule is a doubly indexed family of modules are bigraded M = { M p , q : ( p , 4E) Z x Z}. If M = { M p , q } and N = modules and if (a,b) is a fixed ordered pair of integers, then a family of

301

Exact Couples and Five-Term Sequences

homomorphisms f = { f,,,:M P , ,-+ N , + a. One writes f :M N .

,+

is a map of bidegree (a,b).

b}

-+

Exercises: 11.3. Bidegrees add under composition: if f : M -+ N has bidegree (a, b) and g :N -+ K has bidegree (a’,67, then gf: M K has bidegree (a a’, b b‘). 11.4. All bigraded modules (over a fixed ring) and all maps having a bidegree comprise a category.

+

+

-+

In the category of bigraded modules, there are subobjects and quotient objects. IfM,,, c N,,, for all p , q, then M = { M p , q ;is a (bigraded)submodule N has bidegree (0,O). Deof N = { N , , ¶ ) ; visibly, the inclusion map M ; natural map fine the (bigraded) quotient module N / M as { N p , q / M p , q }the N N / M also has bidegree (0,O). There is a consequence of this elementary definition. Given f : M -+ N with bidegree (a,@, imf should be a (bigraded) submodule of N ; what is (im f),,,? Since (im f),,,c N,,,, we are forced to define -+

-+

(imf)p.q

=fp-o,q-b(Mp-o,q-b)

= im(fp-o,q-b)

Np,q.

(Thus, (imf),,, is not im(f,,,), which lies in N , + , , ¶ + & . On the other hand, there is no problem with indices of kerf: if one defines (kerf),,,

= Wf,,J,

then kerf is a (bigraded) submodule of M . It is now clear how to define exactness of a sequence of bigraded modules. We defer giving examples of bigraded modules; later, we shall define a “bicomplex” which will determine bigraded modules in much the same way that a complex determines graded modules (ignore differentiation; homology). Definition: An exact couple is a pair of bigraded modules D and E, and maps u, B, y (each of some bidegree) such that there is exactness at each vertex of the triangle D

3!

*D

E

Obviously, an exact couple generalizes the notion of exact triangle (since one does not generalize merely for the sake of generalization, there are not

302

11 Spectral Sequences

three distinct bigraded modules at the vertices; in practice, an exact couple is what one encounters). Given an exact couple D = (Dp,q),E = {Ep,q),and maps a, p, y of bidegrees (a,a'), (b,b'), (c,c'), respectively, one may write down a long exact sequence for each fixed q : ' . . - P+ E P - c , q - c . Dp,q'

Dp+a,q+a'

b +

Ep+o+b.q+a'+b'

'''.

.

Conversely, given infinitely many long exact sequences as above, they may be assembled into one exact couple. Let us now use this barrage of notation (actually, a clever organization of a maze of data, due to Massey) to obtain concrete results. The next theorem is due to Kreimer, while the notion of acyclicity when dealing with composite functors was first formulated by Grothendieck. We maintain our (too restrictive) hypothesis that all functors are additive functors between categories of modules (one may replace module categories by certain more general categories). Definition: Let F:d -+ (5 be a functor of either variance. A module B in

B is right F-acyclic if (RPF)B= 0 for all p 2 1, where RPF is the pth right derived functor of F; a module B in B is left F-acyclic if (LpF)B= 0 for

all p 2 1, where L,F is the pth left derived functor of F.

If F is covariant, recall that (RPF)B= HP(FEB), where EB is a deleted injective resolution of B. It follows from Theorem 7.6 that every injective module is right F-acyclic. Exercises: 11.5. Every projective module is left F-acyclic for any covariant functor F, and is right F-acyclic for any contravariant functor F. 11.6. If F = A OR , then every flat module RB is left F-acyclic.

-

A composite of functors may give an exact couple. Observe first that if dn G:2I + 23,then(R"G)AE 23 for every module A in 2I (if. * .+ X, + X , - --*' * * Gd, is a deleted resolution in %, then . . . + GX, GX,- -+ . . . is a complex in 23, whence ker Gd,/im Gd,. E 93).

Theorem 11.1: Let G:2I + d and F: 23- (5 be functors such that F is left exact and whenever E is injective in 2I, then GE is right F-acyclic. For each module A in '3, choose an injective resolution 0 -P A -+ Eo -+ E' -+ ' . . and define

Zq = ker(GEq+ GEqi').

303

Exact Couples and Five-Term Sequences

Then there exists an exact couple with ( R p F ) ( R q G ( A ) ) if p 2 0 , 4 2 0 , otherwise, if p 2 0, q 2 1, if p = - 1 , 421, othrrwise,

and maps a : D + D OJ bidegrer ( - 1, l), p : D y : E -+ D of bidegree (1,O).

E of bidegree (1,

. +

-

l), and

Remarks: 1. Visualize a bigraded module as a family of modules, one sitting on each lattice point in the p-q plane. Thus, E lives in the first quadrant and D lives above the line q = 1 and to the right of p = - 1. 2. The basic idea is just to assemble the long exact sequences arising from the obvious short exact sequences [ 11 and [4] below.

Proof: Abbreviate RPG, RPF, R P ( F G )to G P , FP, ( F G ) P ,respectively. Our task is to exhibit, for each q 2 0, an exact sequence E - L q + 1.+DO,q+l + D - l . , + 2 + E o , q + l + ~ ~ . +

Dp,q + 1

-+

Dp- I . q+2

+

Ep3q + 1 +. . . .

By definition of E and D, we want exact sequences

o+FIZq+(FG)q+lA4 F ( G q + ' A ) . + . . . F P + 124 FPZq+ I FP( G4+ 1 A) + . . . +

+

~

Now GqAisjust the qth homologyofthecomplex~~~+GEq+GEq+l +.... We have already denoted the q-cycles of this complex by 2 4 ; denote the q-boundaries, im(GE4-I + CEq), by BY.There are short exact sequences

PI

0 + 2 4 + GEq

Bq+'

-+ 0

which give rise to exact sequences (since F is left exact and GEq is right F-acyclic)

c21

0 + FZq

.+

FGEq + F B q +

+ F1Z4+ 0,

all q 2 0,

and isomorphisms

[31

FPBq+' 2 F p f 1 Z q , all p 2 1, q 2 0. The definition of homology gives short exact sequences

[41

0 .+ Bq+ 5 2 4 '

A

Gq+ '

A + 0,

304

11 Spectral Sequences

which yields long exact sequences

o+

F@+' + FZq+' + F(GQ+'A)+ F'B4+' + F'Z"'

+ F'(Gq+'A)+-

a .

Using [3] to replace each term FPB4+' by its isomorphic copy Fp+'Zq, for p 2 1, almost leaves us with the desired exact sequences; only the first two terms are not correct. Consider the diagram FGEq

=

FGEq

1

1

0

0

1

II

I

where the first column is the end of the exact sequence [2], the map k is the composite FpFI = F(GEq + BQ+ 4 Zq+'), and W = coker k. Commutativity of the top square and exactness of the columns provide a unique map F'Zq -+ W (Exercise 2.7) making the bottom square commute; that Fv o k = 0 implies the existence of a map W F(Gq+'A) which makes the remaining square commute. Exactness of the middle row implies, by diagramchasing, exactness of the bottom row. It remains to identify W with (FG)¶+' A . Now Zq+' = ker(GEq+' + GEq+'), so that left exactness of F gives FZq+' = ker(FGE4" + FGEq+'). Therefore

'

W = coker(FGE4 -P F Z q + ' ) = FZq+'/im(FGE4 + FZq+') = ker(FGE4" + FGEq+')/im(FGE4 + FZq+'). Using left exactness of F once again, Zq+' 4 GEq+' implies FZq+' 4 FGEq", and thus im(FGEq + FZq+') = im(FGEq-+ FGEq+'). Therefore, W = Hq+'(FGEA)= (FG)¶+'A. I Exercise: 11.7. Give explicit formulas for the maps F'Zq + W and W -+ F(Gq+'A); give explicit formulas for these maps once W has been identified with (FG)q+'A. Theorem 11.2 (Cohomology Five-Term Sequence): Let G :U -+ 23 and F :23 -+ 6 be leji exact functors such that E injective in 2I implies GE is right

305

Exact Couples and Five-TermSequences

F-acyclic. Then there is an exact sequence

0 + (R'F)(GA)+ R'(FG)A + F ( R ' G A )+ (R'F)(CA)+ R'(FG)A for every module A in 2I. Proof: We return to the abbreviated notation above. Consider the exact sequences in Theorem 11.1 for q = 0 and q = 1 : 0 + F'ZO + (FG)'A + F ( G ' A ) -P F'GA + F'Z' + . . .

and 0 + F'Z'

+ (FG)'A + .

..

Just splice these two sequences together at F'Z', and remember that Z o = ker(GEo + GE') = GA, for G is left exact. I Remark: If G is not left exact, one must replace G A by (RoG)A in the sequence.

The following variant of Theorem 1 1.2 is useful. Theorem 11.3: Let G:2I -+ 8 and F : 8 + Q be left exact functors such that E injective in 2I implies GE is right F-acyclic. I f A is a module in 2I with (R'G)A= 0 for 1 5 i c q, then there is an exuct sequence O+(R4F)(GA)+Rq(FG)A+ F(RqGA)-+(Rq+'F)(GA)+R4+'(FG)A. Proof: Let us return once again to the abbreviated notation for derived functors. We shall only prove the special (though most important) case when G ' A = 0 (the rest being an exercise for the reader). Consider the exact sequences in Theorem 11.1 for q = 0, 1, and 2: 0 + F'ZO + (FG)'A -+ F(G'A) + F Z Z o+ F'Z' 0 + F'Z' + (FG)'A + F(G2A)+ F2Z' -P F'Z'

and

+ F'(G'A)

+. ..,

*. ,

4.

o + F I Z Z+ ( F G ) ~ +A . . . .

Splice the last two sequences together at FIZZ to obtain exactness of 0 + F'Z'

Since G ' A

= 0, the

+ (FG)'A + F ( G 2 A )+ F'Z'

4

(FG)3A.

first sequence ( q = 0) gives an isomorphism F2Zo 2 F'Z'.

Recalling that Z o = GA, the sequence now begins with F2(GA).Let us deal with the fourth term F'Z'. Since G ' A = Z ' / B ' , the hypothesis gives Z' = B'

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11 Spectral Sequences

and hence F2Z’ = F’B’. But isomorphism [ 3 ] in the proofofTheorem 11.1 gives F2B’ z iF3Z0 = F3GA. I An important use of this last result occurs in the cohomology of algebras, involving Hilbert’s “Theorem 90” and the Brauer group. It should be clear that the proofs just given dualize; we merely state the results for right exact covariant functors. Theorem 11.4 (Homology Five-Term Sequence): Let G:Iu -+ B and F:B -+ (5. be right exact functors such that P projective in Iu implies G P is left F-acyclic. Then there is an exact sequence

L2(FG)A--* (L2F)(GA)-+ F ( L I G A )+ L,(FG)A -+ (L,F)(GA)-+ 0 for every module A in Iu. Moreover, i f (LiG)A= 0 for 1 I i < q, then there is an exact sequence

L,, ,(FG)A --* (L,+, F ) ( G A )-+ F(L,GA) + L,(FG)A + (L,F)(GA)-+ 0. Let us illustrate these general results. Assume I3 is a group with normal subgroup N . Clearly, every II-module A may be regarded as an N-module, so that HornN(&A ) is defined. Recall Hom,(Z,A) = A N = { a E A : n .a

= afor

all n E N } .

For a ll-module A, the module AN is actually a n/N-module: if x E Il has coset x E n/N, define R.a=x.a, aEAN (one checks this is well defined). Therefore Hom,(Z, ), and its derived functors: Ex&,(& ) = H i ( N , ) are functors from n-modules to n/N-modules. What is the action of I3/N on H i ( N , A )for a ll-module A? Take a llprojective resolution of Z p = . . . + P, -+Po-+z-+o

(which is automatically N-projective, by Exercise 10.9), and let us compute H’(N, A ) by examining the complex HomN(P,, A). Exercise 10.7 tells us a reasonable way to make each HomN(Pi,A)into a ll/N-module: (Xf)(b) = x .f(x-’b),

Js

E

n/N,

b E Pi.

Hence, H’(N, A ) becomes a ll/N-module by K(Zi

+ Bi)= R . zi + B;,

where zi E HomN(Pi,A ) is a cycle and Biis the submodule of boundaries. (Observe that the action of Exercise 10.7 coincides with the action of n / N on Hom,(Z, A ) given above.)

307

Exact Couples and Five-Term Sequences

A similar discussion, now using Exercise 10.8, shows how I I / N acts on the homology groups H , ( N ,A ) when A is FI-module: X(b @ a) = b x - ' @ xu,

.tE

H/N, b E Pi.

Remark: One may prove [Gruenberg. 1970. p. 1511 that if N c Z ( n )and A is II-trivial, then H p ( N ,A ) and H,(N. A ) are H/N-trivial for all p . The proof of this is not difficult, but involves a longish digression. This fact

will be used several times in the sequel.

Theorem 11.5: If N is u normcil subgroup of there is an exuct sequence

o + H ' ( ~ / NA") , -,~ ' ( n A ),

+

n

ma' A is a II-module, then

H ' ( N , A ) " ' . ~-,H ~ ( I I / NA") , -+

~ ~ A() .n ,

Moreover, if Hi(N,A ) = 0 for I I i < y, there is un exuct sequence

0 -,H q ( n / N ,A")

+

W(n,A )

-+

H y N , A)"

-+ H4+ ' ( r I / N ,A") + H4+ yrI, A).

Proof: Let 'u be the category of n-modules and 23 the category of H/N-modules; define G:'u 23 by G = Hom,(Z, ) and F : % Ab by F = Hom,,N(Z, ). It is clear that G and f-. being Horn's, are left exact. If E is an injective n-module, we claim GE = EN is n/N-injective (hence right F-acyclic). Consider the diagram --f

-+

O-E"-E

3.

0-MI-M where f and i are IIlN-maps. By change of rings ZII + Z(II/N), every n/N-module may be regarded as a ll-module and every rI/N-map may be regarded as a II-map. Since E is II-injective, there is a ll-map f " : M -+ E extending f . But imf" c E N :n . fix) = , f i n . .u) = ,f"(x),since each x E M is fixed by every n E N . It follows that 7:M + EN is a rI/N-map, and EN is n/N-injective. The hypotheses of Theorem 11.2 (or t 1.3)hold. To apply these theorems, one must evaluate the composite FG. It is a simple matter, however, to see that (A")"" = A", i.e., FG = Hom,(Z, 1, and we are done. I

Remark: One may identify the map H ' ( n / N ,A") + H'(II, A ) as "inflation" and themap H'(ll, A)+H'(N, A)"" as "restriction", both ofwhich are familiar to specialists. The map H ' ( N , A)"" -+ H Z ( H , N ,A") is called "transgression".

308

11 Spectral Sequences

There are formulas for these maps in terms of the bar resolution B; recall that Bi is the free l7-module on all symbols [x,, . . . ,x i ] , xiE II, xi # 1, and cocycles are certain functions f with domain B i . It may be shown [Weiss, 1969, $2.51 that if clsf E Hi(II/N,A N ) , then inflation sends c l s f w clsg, where

. . ., xi] = f[x,,. . ., x i ] , in n/N,and cls f denotes the homology class off. If g[x,,

xj is the coset of x j

clsfE H'(ll, A ) , then restriction sends clsf-cls

h [ x , , . . . ,xi] = f[x,, . . . , xi],

h, where

xj E N , j

= 1,2,. . . , i.

Finally, if A is a TI-module, and clsj' E H i ( N , A), then1 E l l / N acts by

(f.f)[x,, . . . , x,]=t.f[t-'x,t , . . . , t-'xit],

x j c N , j = l , 2 ,..., i.

In particular, this last formula shows that the action of l l / N on H'(N,A) is trivial when A is ll-trivial and N c Z(ll). Here is the corresponding result for homology. Theorem 11.6: If N is a normal subgroup ofll and A is a ll-module, then there is an exact sequence

Hz(l-4 A ) Hz(II/N, A,) H , ( N , Ah,, H,(n,A ) 4H , ( n / N ,-4,) Moreover, i f Hi(N,A ) = 0 for 1 I i < q, there is an exact sequence -+

+

Hq+

A ) -b

Hq+

l(TI/N?A N )

-+

Hq(N,

Hq(n, A)

@/lv,

+

AN)

0. O.

Proof: Recall that A , = Z 0, A = A/nA, where n is the augmentation ideal of N. Let 2I be the category of ll-modules and 23 the category of ll/N-modules. Define G :9I -+ 23 by G = Z 0, . Note that Z 8, A is a ll/N-module if one defines g . ( m Q a ) = m Q g a, where g E l l / N is the coset of g E ll, a lies in the ll-module A, and m E Z. Define F:23 + Ab by F = Z . It is clear that G and F , being tensors, are right exact. If P is a projective ll-module, we claim G P = Z ONP is n/N-projective, hence left F-acyclic. If P 2 Zll, then Z ONP 2 P/nP s Zn/nZll E Z(n/N). It follows that if P is ll-free or II-projective, then P, = Z ONPis ITlN-free or n/N-projective. The hypotheses of Theorem 11.4 hold. To apply it, one must evaluate the composite FG. It is a simple matter to see that (A,)=/, = A,, i.e., F G = Z On ,and the theorem is proved. 1 Theorem 11.7 (=Theorem 10.12): For any group G, HAG, Z) where F is free and FIR z G.

( R n F')/[F,R],

309

Exact Couples and Five-TermSequences

Proof: Since R is a normal subgroup of F , Theorem 11.6 applies to give exactness of H , ( F , Z ) + H,(G,ZR) + H , ( R ,Z ) G

+H,(F,Z

) -+. . . .

Now Z R = Z , since Z is F-trivial, hence R-trivial, and Corollary 10.5 gives H,(F, Z ) = 0 since F is free. Thus, H2(G,Z ) = ker(Hl(R,Z),

-+

H , ( F , Z) ).

Corollary 10.3 shows H , ( F , Z ) z F / F and H , ( R , Z ) E'R/R'; moreover, R/R' E r/rf as G-modules, by Lemma 10.10 and the remark immediately thereafter. With the further aid of Lemma 10.11, we have

Hi(R, Z)G = Z

OG

R/R'

= (r/rf)/(fr

Z O G r/rf

(r/rf)/f(r/rf)

+ rf)/rf 2 r/fr + rf E R/[F, R ] .

Therefore, H,(G, Z) = ker(R/[F, R]

+ F/F').

We now cheat, asserting the map is just enlargement of coset, from which the result follows. I Our little cheating is just that; it is not difficult to identify the map. Accepting this, the reader may begin to appreciate the power of the five-term sequence. Here is another application of the five-term sequence. Let Q be the quaternion group of order 8 : Q = (x,1' I X *

= J', WJ.W = y ) .

We recall the following properties of Q: there is a normal subgroup N of order 2, namely, (x'), with Q / N 2 G , where G is the 4-group; center Q = N = Q', the commutator subgroup of Q ; Q has a balanced presentation (exhibited above). It follows from Corollary 10.16 that H , ( Q , Z ) = 0. The five-term sequence with A = Z is H 2 ( Q , Z) + HAG, Z

) 5 Hi(N, Z ) G

+

H ,(Q, Z ) 5 Hi(G, Z )

+

0.

Since H , ( Q , Z ) = 0, the map ci is monic. By Corollary 10.3, H , ( Q , Z ) E Q/Q' = Q / N z G and H , ( G , Z ) 2 G/G' = G ; since both of these groups have order 4 and is epic, 0 must be monic. It follows from exactness that c1 is epic, i.e., H,(G,Z) 2 H,",Z)G. Now Q acts on n = {n - l : n E N J by conjugation: if x E Q, then x . n = x n x - ' ; since N is the center of Q, however, xnx-' = n. It follows that G = Q / N acts trivially on it and hence on n/n2 = H , ( N , Z ) . Therefore,

3 10

11 Spectral Sequences

H I ( N , Z ) ,= H , ( N , Z ) Y N / N ‘ = N , and the multiplier of G has order 2 (as we saw in Chapter 10). We shall return to composite functors, so let us give a simple necessary condition guaranteeing acyclicity. Theorem 11.8: Assume (F, G) is an adjoint pair of functors, where F : % -+23 and G: 23 -+ ‘u. (i) If G preserves epimorphisms, then F preserves projectives; (ii) If F preserves monomorphisms, then G preserves injectives. Proof: Let P be projective in ‘u, and consider the diagram in 23 FP

1

B-B‘-0

If t:Hom,(F , ) diagram in %

-+

Horn,( ,G ) is the natural bijection, then there is a P

GB

GB’-0

Gt:

The row is exact, since G preserves epimorphisms, and so projectivity of P provides a map +’ E Hom(P, GB) with (Gs) $’ = ~(9). 0

We claim z-’($’):FP -+ B completes the original diagram. Naturality of gives commutativity of

Hum (FP,8)

HomfFP, B’)

T

bHom(P, GB)

Hom(P, GB’)

t

311

Derived Couples and Spectral Sequences

Evaluating on T-'$', we have T E * ( T - ' $ ' ) = ( G E ) . + d - ' $ ' ) . But the left side is T ( E T-'$'), while the right side is (GE)*$'= ( G E ) $' = ~ ( c p ) . Since T is a bijection, 1

i

=

E ' T-'$'

cp,

as desired. The dual proof gives the second statement. Exercises: 11.8. Show that the functor Hom,(Z, ) of Theorem 11.5 satisfies the hypothesis of Theorem 11.8. 11.9. Show that the functor ZON of Theorem 11.6 satisfies the hypothesis of Theorem 11.8.

DERIVED COUPLES AND SPECTRAL SEQUENCES The main observation in this section is that an exact couple is more than a clever organization of long exact sequences; it determines other exact couples. Consider the exact couple, with maps of the indicated bidegrees: (-1,I)

D

x

E Define d ' : E -+ E by d' = B y (first y and then P!). Since y P = 0, it follows that d'd' = 0 and E has homology groups H ( E , d ' ) = kerd'/imd'. One usually denotes H ( E , d ' ) by E 2 and regards it as a bigraded module. In more detail, d;,q:Ep,q5 D,-

1.4

B +

E,

- 1.4'

so that d' has bidegree ( - 1,O). By definition of bigraded quotient module,

E%,q= kerdb,,/imdi+ Define a second bigraded module D 2 = im u. Since the definition of bigraded image gives

D;,¶= u p - 1 , q + i ( D p - 1 , q +

I

=

im M p -

c1

has bidegree (1, - l),

1, q + 1 C

Dp,q.

312

11 Spectral Sequences

We next define a2, j2,y2 as pictured below:

Set a2 to be the restriction of a (for D 2= ima c D).Since the inclusion map i: D2 = im a 4 D has bidegree (0, 0), the map a2 = a 0 i has the same bidegree as a, namely, (1, - 1). Define 8' :D2-+ E by B2Y = CBa-'yI,

where bracket denotes homology class (more details in a moment), and define y2 :E2 + D2 by Y:,qczP,,I

= Yp.4Zp.q E DP- 1 . 4 .

It is a brief exercise that y z is well defined, so that y2 has the same bidegree as y, namely, (- 1,O). Let us return to p'. If y E Di,q, then y = @ p - l . q f l ( X p - l , q + l ) ; define

Cap-1,

+ l(xp- 1 . 4 + ,,I E E p - 1 , q + 1 . Again, we let the reader check a map is well defined. Visibly, j2has bidegree

p;.q = Pp- 1 , q + 1Mp-- I 1 , q + 1 :Y +t

q

2

(-191).

Observe that the bidegrees of a2,j2,y z are the same bidegrees occurring in the exact couple of Theorem 11.1. Note also that all definitions make sense if we begin with an exact couple whose maps have any bidegrees.

Theorem 11.9:

With the definitions aboue,

E2

is an exact couple; az has bidegree (1, - l), j2has bidegree ( - 1, l), and y2 has bidegree ( - 1,O).

Proof: This is a simple diagram chase. The reader may verify that successive composites are 0, i.e., im c ker, and we shall prove the reverse inclusions. We drop subscripts.

313

Derived Couples and Spectral Sequences

Assume x E D2 = im a and a 2 x = a x = 0. Then x E ker a = im y , SO that x = yy for some y E E. But x E im o! = ker P implies 0 = Px = fiyy = d ' y . Therefore, y is a cycle and x = y 2 [ y ] . Assume x E D2 = imu and fi2x = 0. Thus, x = ay and Py = d'w = pyw. Hence y - yw E ker P = im a = D2,and r 2 (y - yw) = a( y - yw) = ay = x (for ay = 0). Assume x E E 2 and y2x = 0. Now x = [y] E E 2 , and y 2x = yy = 0; hence y E ker y = im fi, so y = Pw. One checks that P2(aw)= [y ] = x . Definition: The exact couple ( D 2 ,E 2 ,az,P2, y 2 ) is called the derived couple of (4E, a, P, Y).

Of course, we may (and do) iterate this construction to get a sequence of exact couples ' 2

D'

D'

E' where, by definition, the ( r + 1)st exact couple is the derived couple of the rth one. It is useful to have more explicit descriptions of the data of the rth derived couple. Theorem 11.10: Let (D, E,a,/?,y) be an exact couple, where a, 8, y have bidegrees (1, - l), (O,O), and ( - l,O), respectively. If the rth derived couple is (D',E', a', /l'y'), , then

(i) a' has bidegree (1, - l), has bidegree (1 - r, r - l), and y' has bidegree (- 1,O); (ii) d' = py' has bidegree ( - r , r - 1 ) and is induced by pa-'- ' y ; (iii) E'p,q = kerd'p,q/imd;+r,q-r+l. Proof: An easy induction on r, using the fact that bidegrees add.

I

Definition: A spectral sequence is a sequence (E*,d':r 2 l } of bigraded modules and maps with d'd' = 0 such that E" as bigraded modules.

= H(E',

d')

314

11 Spectral Sequences

Of course, one may denote the original E by E' (as we have just done). Thus, every exact couple determines a spectral sequence. An elementary definition is useful here. Definition: A subquotient of a module M is a module of the form h4'/h4'', where M" c M ' are submodules of M . Obviously, one may also speak of subquotients of graded modules or of bigraded modules. In our context, the most obvious example ofa subquotient is homology: if C is a complex, then for each q, H,(C) is a subquotient of C,. Exercises: 11.10. Subquotient is transitive: if B is a subquotient of A and C is a subquotient of B, then C is a subquotient of A . 11.11. If an abelian group G has any of the following properties, then so does every subquotient of G :

(i) G = 0; (ii) G finite; (iii) G f.g.; (iv) G torsion; (v) G p-primary; (vi)mG = 0 for some m > 0; (vii) G cyclic. Returning to spectral sequences, we see that each E' is a subquotient of E = E l , or of E 2 , indeed, of any earlier term. This remark can be pictured. Write E2 = Z2/B2 (actually, Ep',, = Z,Z,,/B,Z,,).Since E 3 = Z3/B3 is a subquotient of E2, the third isomorphism theorem allows us to assume 0 c B2 c B3 c Z 3 c Z 2 c E 2 .

Iterating, 0 c B2 c . . . c B' c B'+l c . . . c Zr+' c Z'

Definition: 22, = (')rZ*p,q; B;.q

=

urB;.q;

E;;4

c

. . . c Z 2 c E2.

= Z;,/B&.

The bigraded module E" is called the limit term of the spectral sequence { E r ) , and it is clear that, as r gets large, the terms E' do "approximate" the limit term. There are other descriptions of the modules ZL,,, BL,q, EL,,, for r I co; indeed, many accounts of spectral sequences use these descriptions as their starting point. The appropriate context is that of "dfg-modules" (differential, filtered, graded). It is instructive to see this approach, and the reader is advised to look since other viewpoints are valuable [MacLane, 1963, XI.1 and XI.31. Exercise: 11.12. If drp,4 = 0 for all p , q, then E' = E'". If dL,q = 0 for all for all p , q. p , 4 and all r 2 s, then E;, =

315

Filtrations and Convergence

FILTRATIONS AND CONVERGENCE Definition: Let 2l be a category and let A be an object in 2l. A filtration of A is a family of subobjects of A, (FPA: p E Z ] , such that

. . . c FP-LA

c FPA

FP+'A c . . .,

We are especially interested in two special cases: 2l = complexes; '? =I graded modules. Thus, a filtration of a complex C is a family of subcomplexes { F P C : pE Z ) with FP-'Cc FPC for all p . A filtration of a graded module H = { H , : n E Z} is a family of graded submodules { F P H : p E Z ) with F P - ' H c FPH for all p . Note that this last condition merely says that there are (ordinary) submodules F P H , of H , , for each n, such that FP- ' H , c F P H , , all p . Theorem 11.11: Every filtration { F P C : pE Z} of a complex C determines an exact couple

in which CI has bidegree (1, - l), /3 has bidegree (0,0), and y has bidegree ( - 1,O). Proof: Let us abbreviate F P C to FP. For each fixed p , there is a short exact sequence of complexes o+FP-'+FP+FP/FP~'

+O.

The corresponding long exact homology sequence is . . . + H P + , ( F p - ' )5 H p + , ( F p )5 H p + q ( F P / F P - l )

HP+4l(FP-')+ . .. .

Observe that there are only two types of terms: homology of some FP or homology of some quotient F P / F P -'. Accordingly, define

DP,,= H p + , ( F P )

and

EP,, = H p + , ( F P / F P -').

The long exact sequence now reads

. . . + D,-

',,+

0

5 DP,,+ EP,¶A Dp-

..

4..

The reader may now equip the maps or, /j, y with bidegrees, namely, (1, - l), (0,0), and ( - 1,0), respectively. All has been verified. I

316

11 Spectral Sequences

It is at this point the reader may begin to feel that he could have invented exact couples; he also sees why the bidegrees before Theorem 11.9 were so chosen. Corollary 11.12: Every Jiltration { F P C } of a complex C determines a spectral sequence. Proof: The filtration determines an exact couple with maps of the proper bidegrees; Theorem 11.9 gives a sequence of derived exact couples whose terms EZ,E 3 , . . . ,E', . . . form a spectral sequence. I It would be pleasant to know the E 2 term of this spectral sequence. Let us be content for the moment with exhibiting the differential d ~ , , : E P , , - + E p - 1, ,. By definition, d 1 = f l y ; with subscripts, d;,, = B,- 1,,Yp,,:Hp+,(FP/FP- ')

-,H,+,-

IFP-

1/Fp-2);

moreover, yP,, is the connecting homomorphism Hp+,(FP/FP- ') -+ Hp+q-l(Fp-l),while flp-l,q is the map induced by the natural map FP- 1 -,FP- 1/FP-2.

Exercise: 11.13. Consider the short exact sequence of complexes

o-,

F P - ~ I F P - ~-+

F P / F P - -+ ~ F P / F P - ~ +0.

Show that the map d;,, is the connecting homomorphism Hp+,(FP/FP-')-+

H,+,-l(FP-1/Fp-2). Hint: Use naturality of the connecting homomorphism

in the situation

A filtration { F P C } of a complex C defines a filtration of the (graded) homology module H J C ) in the following obvious way. If i,: FPC -+ C is the inclusion map, then there is the induced map H,,(i,):H,(FPC) -+ H,,(C) for each n. Clearly im H,,(ip-1) c im Hn(ip)for all p , i.e., the (graded) submodules {imH(ip)} form a filtration of H , ( C ) . Wouldn't it be nice if the spectral sequence of ( F T } , whose E' term is H,(FP/FP-'), determined the factor modules of this filtration of H , ( C ) ? If so, the spectral sequence would allow us to commute homology and filtration, better, homology and factor modules.

317

Filtrations and Convergence

Definition: Let H be a graded module. A spectral sequence { E ' ) roughly of H converges to H , denoted Ei,q 7H , if there is some filtration such that Epqq 2 @ p H p + q / @ p - l H p + q for all p , q. There is a universal notational agreement in this subject that

+ q.

n =p

Hence, the isomorphisms in the definition of rough convergence are written E& z (DPH,/OP-'H,. The index p written under the arrow E;,¶ 7H is called the filtration degree; it tells which of the two subscripts index the filtration of H , and this is necessary because EZq and EZP may not be isomorphic. Let us state some difficulties implicit in the notion of rough convergence by considering a filtration {FPCf of a complex C and its corresponding spectral sequence {Er}. First, we need some criterion to guarantee rough convergence. Next, even if we have rough convergence, it may be useless. For example, define FPC = 0, the zero subcomplex, for all p . Clearly 0 = E = E 2 = E 3 = . . . , so that E;q = 0 for all p , q. Moreover, the "obvious" filtration {imH(i,)} of H , ( C ) consists of zero submodules, hence has zero factor modules. Thus, we do have rough convergence, and the information it conveys is 0 = 0. Plainly, some hypothesis is needed on the filtration {FPC).In most applications, the following condition does hold. Definition: A filtration {FPH}of a graded module H is bounded if, for each n, there exist integers s = s(n) and t = t ( n ) such that F"H,=O

and

F'H,= H,.

Since a complex C is a graded module if we forget its differentiation, it makes sense to say that a filtration { FPC) of C is bounded. In any filtration, FP-'H c FPH for all p . In particular, if { F P H } is bounded, then for each n, FPH, = 0 for all p Is, FPH, = H, for all p 2 t , and there is a finite chain 0 = F'H, c F'+'H"

c ...c

PH,

= H,.

Definition: Let H be a graded module. A spectral sequence ( E ' ] converges to H, denoted Ef,¶3 H,, ifthere is some hounded filtration {OpHJof H such that P

Epqq 2 @,PH,/@p- H ,

for all p , q (remember, n

=p

+ 4).

318

11 Spectral Sequences

The previous notion of rough convergence may be forgotten; convergence is the important notion. Theorem 11.13: Assume (FPC)is a boundedjltrution of a complex C, and let {E'} be the spectral sequence it determines. Then (i) for each p , q, we have E2q = E;,q for large r (depending on p , q ) ; (ii) E;,, =. H,(C). P

Proof: (i) How can we say anything at all about E;,q? The one thing we do know is that its differentiation d' has bidegree ( - r, r - l), Choose p large, i.e., p t(n). For any such p , F P - ' = FP and FP/FP-' = 0. Hence Ep,q= H p + , ( F P / F P - ' )= 0, and so E;:q = 0 for all r, all such p (for Erp,q is a subquotient of IY,,~).Similarly, choosing p small, i.e., p < s(n), gives Ek,q = 0 for all r. Now choose any p , q. Since d' has bidegree ( - r , r - l), dr(Ek,q)c E ; - r , q + r - l .For large r, the index p - r is small and EL-',,+'-' = 0. Hence E;,q = ker for large r. Let us now compute EL:' = ker CP,,$n G+,+,, q-r+ When r is large, the index p r is large, and so im G+r,q-r+l = b ( E ; + , , q - r + l )= 0. Therefore, EL:' = kerd;,q = E;,q for large r (depending on p and 4).The spectral sequence is thus ultimately constant, whence E:, = for large r. (ii) Equip H J C ) with the obvious filtration

=-

G,,

'.

+

OPH,(C)= im(H,(FPC) -+ H,(C)), where the map is induced by the inclusion FPC-C. That F is bounded implies 0 is bounded :

0 = (DSHH,(C) c @"+'H,(C) c . . . c @IH,(C) = H,(C). Consider the exact sequence arising from the rth derived couple (the indices are determined by the bidegrees of the maps, which were calculated in Theorem 11.10): (*)

*

..

-+

D;+r-

2, q - r

+2

ar +

D',+r - I , q --I + 1 5 E ; , ~5 D;-

', -+. . . .

Recall that D' = a 0 . . . o ctD ( r - 1 factors equal to a) and that ct has bidegree (1, - 1). It follows that ( r - 1 factors of ct) D ; + r - l , q - r + =l c t o . . . ctD,,, = irn(H,+,(FP) + Hp+q(FP+'-I)), 0

for

is just the homology map induced by inclusion FP-+ FP+ For fixed

p,q, boundedness of the filtration gives Fp+'-'C = C for large r. We con-

319

Bicomplexes

clude that for large r D',+r-

l.q-r+ 1

= @"'p+q(Ck

A similar calculation shows that, for large r, the first term in the exact sequence (*) is @ p - l H p + , ( C )Substituting, . exact sequence (*) becomes

l H p + q ( C ) - ' @ P H p + q ( C ) j E ' , . q ' D' , - l , q ' . . . .

...-*@p-

,= 0 for large r, then we have

If we can show that DL-

@ p H , + q ( C ) / ~ p - l H p + ~EL,¶ ( C ) = E&, and we are done. But this is essentially the same calculation:

D ' , - l , q = a0

...o

= a o . . .

0

aDp-,,q+r-l aHp+,- *(FP-'C)

= 0,

( r - 1 factors of a) ( r - 1 factors) if p - r < s . I

In practice, the vast majority of spectral sequences arise from a bounded filtration ( F P C )of a complex C; moreover, the typical application involves a certain complex (the "total complex") associated to a "bicomplex". For this reason, we proceed to bicomplexes. (The reader finding a more general spectral sequence in his hands is referred to any other textbook treating spectral sequences; we maintain that an understanding of the special case suffices for an understanding of the general.)

BICOMPLEXES Definition: A bicomplex (or double complex) is a bigraded module M = ( M P , ¶ with } maps d ' , d " : M + M of bidegree ( - 1,O) and (0, - I), respectively, such that d'd' = 0,

d"d"

= 0,

and

d'd"

+ d"d'

= 0.

As usual, picture a bigraded module as a family of modules in the p - q plane. The maps db,¶:MP,¶ M,- 1, go one step left; the maps d;,,: MP,¶-+ M p ,¶ - go one step down. The first two conditions d'd' = 0 and d"d" = 0 say that each row and each column is a complex. Before giving examples of bicomplexes, let us associate a complex to M in order to explain the anticommutativity equation d'd" + d"d' = 0. --$

Definition: If M is a bicomplex, its total complex Tot(M) is the complex defined by Tot(M), = M,,,,

LI

p+q=n

320

11 Spectral Sequences

and

is given by

d,

db,q

= p+q=n

+ di34.

Lemma 11.14: If M is a bicomplex, then Tot(M) is a complex. Proof: Begin by drawing a picture.

Tot(M), is the sum of all Mp,qlying on the 45" line p + 4 = n. Note that if p + q = n, then imdb,q c M p - l , q and imd;,, c Mp,q-1; in either case, the sum of the indices is p + 4 - 1 = n - 1, so that irnd, t Tot(M),- 1. Finally, we must show that dd = 0. But dd = C ( d ' + d")(d'

+ d") = Cd'd' + C(d'd'' + d"d') + Cd"d",

and each of the summands is zero, by definition of bicomplex.

I

The first example of a bicomplex is very general, and we state it as a very trivial lemma.

Lemma 11.15 (Sign Lemma): Let M be a bigraded module with maps d', d": M -,M of bidegree (- 1,0) and (0, - l), respectively; assume d'd' = 0, d"d" = 0, and the diagram commutes. I f di-, is replaced by = ( - l)Pd:,,, then ( M ,d', A") is a bicomplex. Proof: Changing sign does not alter kernel or image, so that A ' A ' For the remaining identity, = ( - 1)'d" P- 1 . 4 d L d;, 4 - lA;,q = (- l)pdb,q = ( - 1 ) P ( - 1 ) ' - 1 A ~ - l , 4 d ~ ,= 4 -A'' P - 1 , 4 d'P.4'

= 0.

I

In short, a big commutative diagram whose rows and columns are complexes can be modified by a simple sign change to be a bicomplex.

321

Bicomplexes

Examples: 1.

Let A be a right R-module and B a left R-module; let

be a deleted projective resolution of A, and let QB

= . . . + Q,

2Q,-

+ . . . + Qo + 0

be a deleted projective resolution of B. Define a bigraded module M by M,,, = P , 0 Q,; the reader may check that M becomes a bicomplex if one defines and

db,, = A; 0 1,

d;,,

l)pl, @ A:,

= (-

where 1, and 1, are identity maps on Q , and P,, respectively. Observe that MP,, = 0 if p < 0 or q < 0; such a bicomplex is an example of a first quadrant bicomplex. 2. Nowhere in the construction just given did we use the hypothesis that the complexes were deleted projective resolutions. Thus, if A and B are complexes, we may form a bicomplex M as above. Tot(M) is usually called the tensor product of the complexes A and B, and one defines

A 6 B = Tot(M). Thus if A

= (A, A') and

B = (B, A"), then

(A 0 B),

=

LI

p+q=n

Ap@B,,

and

d,:(A 0 B),

+

(A 0 B)n-

I

is given by

up 0 b,

H

A'a, Q b,

+ ( - l ) P a p0 A"bq,

where up E A , and b, E B,. 3. The Eilenberg-Zilber theorem [Maclane, 1963, VIIl.81 states that if X and Yare topological spaces, then H , ( X x Y ) z H,(S(X)@S(Y)), where S(X) is the singular complex of X. 4. Let A and B be left R-modules, let pA -- . . . + P p * PA' p - ] + ' . '

+PO+O

be a deleted projective resolution of A, and let EB -- O + E

O+

...+

be a deleted injective resolution of B.

E 4 2 E9+1+.

11 Spectral Sequences

Define a bigraded module M by

M - , , - ¶ = Hom(Pp,E4); define d'-

~ , = 4

:Hom(P,, E4)-,Hom(P,+

(A;+

1,

E4)

and = (- 1)p+4+

ti:,.,

'(A;)* : Hom(Pp,E 4 )-, Horn(!',,,

E 4 + 1.

First of all, the signs of the indices have been chosen to ensure that d' has bidegree ( - I, 0) and d" has bidegree (0, - 1). Second, the sign does give anticommutativity so that M is, indeed, a bicomplex (the reason for the more complicated exponent will be seen soon). = 0 for p > 0 or q > 0, M is an example of a third quadrant Since Mpvq bicomplex. As usual, we eliminate negative indices by raising them. Thus, write Mpq4instead of M-p,-q.In visualizing M, replace points ( - p , - q ) by points ( p , q ) , so the original picture (third quadrant) is replaced by a new picture (first quadrant) by reflection through the origin. In particular, all arrows retain their slope but reverse their direction.

Example: 5. If (A,A') and (C,A") are arbitrary complexes, define a bicomplex M having modules

M,,,

=

Hom(A,,C,),

and differentiations

d;.4 = (A-,+,)*:Hom(A-,,C,)-+ Hom(A-,+,,C,) and d"P.4

= (-

I),+,+

1

(A&:Hom(A-,, C,)

+

Hom(A

-,,,C,-

Because interesting maps between complexes usually involve infinitely many indices (see Exercise 11.2), Tot(M) is not the reasonable complex to associate

323

Bicomplexes

to M (but see Exercise 11.14). Define

Hom(A,C), =

n

p+q=n

Hom(A-,,C,).

wheref,,,:A-, Thus,f E Hom(A, C), is a family o f m a p s j = (f,,,), p + q = n. Observe that a harmless notational change yields

Hom(A,C), =

n

Hom(A,, C,) =

,-p=n

which is precisely the maps A

+C

n P

+ C,

and

Hom(Ap, C p + n ) ,

of degree 11 as usually defined.

Definition: If (A,A') and (C,A") are complexes, then Hom(A,C) is the complex defined by Hom(A,C), =

n

p+q=n

Hom(A-,,C,)

and

D, : Hom(A,C), + Hom(A, C),is given by

f = {fp,,)

where g p , , : L P+ C, (with p

+b

Dn.f = k7p.q;,

+ q = n - 1 ) is defined by

9p.4=(-1)p+q.fp+l,qA'~p + Ab'+If,.,+l. It is a mechanical check that Hom(A, C) is a complex. Moreover, the definition of D, is just

D, = =

n n

[(

p+,=n-

1

p+q=n-

1

= (-

1y- 1

-

1 )"'(A'-

,I*

[( - l ) P + q d ; +

n

p+q=n

+ (A;+ 1)*] + ( - l)-p-(q+L)-ldb).q+11

M;,, + (I;.,)

Exercises: 11.14. If A is a positive complex (e.g., a deleted projective resolution) and C is a negative complex (e.g., a deleted injective resolution), then Hom(A, C), = (Tot(M)), (where M is the bicomplex with Mp,, = Hom(A-,,C,)) and the differentials agree up to sign. Conclude that, in this case, Hn(Hom(A,C ) ) -1 H,(Tot(M)). (Hint: For each n, there are only finitely many ( p , q ) with p

+ q = n.)

324

11 Spectral Sequences

11.15. If A, B, and

plexes

C are complexes, there is an isomorphism of com-

Hom(A @ B, C) z Hom(A,Hom(B,C)). Here is the strategy. Given a bicomplex M, consider the complex Tot(M). There are two rather natural filtrations of Tot(M), each of which gives rise to a spectral sequence. If M is either first or third quadrant, then we shall see both spectral sequences converge to H,(Tot(M)). Moreover, the EZ terms of these spectral sequences can be computed explicitly, and, in each case, E z q is for large r. We now give the filtrations IF and "F of Tot(M), where M is a bicomplex. Mp,q. Recall that Tot(M), = Definition: The first filtration IF is defined by

('FPTOt(M)), =

LI

i s p

A picture will make this clear.

The degree n term of 'FPTot(M) consists of the sum of all Mp,qwith ( p , q ) on the 45" line p + y = n and to the left of the vertical line. To check that 'FP is a subcomplex of TotfM),recall that d = x ( d ' + d"). Thus, d(Mi,,-i) = (d' for i - 1

-= i 5 p .

+ d")Mi,,-i = d'Mi, " - i + d"Mi, ,-i c M i - n - i 0 M i ,n - i -

Definition: The second filtration "F is defined by

A picture will make this clear.

c ('FPTot(M)),-

325

Bicomplexes

The degree n term of "FPTot(M) consists of the sum of all M,,¶ with (p,q) on the 45" line p + q = n and below the horizontal line. One checks, as above, that I I F p is a subcomplex of Tot(M). (Be alert to a possible danger that will be overcome: p signals the first index, but we are using it to restrict the second index.)

Lemma 11.16: (i) Zf M is a bicomplex, then both filtrations of Tot(M) ure bounded ifand only ifeach 45" line p + q = n meets M in ajinite number of points (i.e., there are only finitely many such ( p , q ) with M,,¶ # 0). (ii) I f M is either afirst or third quadrant bicomplex, then bothfiltrations of Tot(M) are bounded. Proof: (i) Fix a 45" line p + q = n. The first filtration is defined by moving a vertical line to the right, and collecting all the terms in back of it. If the 45" line meets M in a finite number of points, a vertical line can be drawn to left of all of them (giving s ( n ) )and a vertical line can be drawn to the right of all of them (giving t(n)). Similarly, the second filtration is defined by moving a horizontal line upwards, and collecting all the terms below it. (ii) This is obvious from the first part. However, we can give explicit formulas for s(n) and t(n).If M is first quadrant, then s(n) = - 1 and t(n)= n works for both filtrations; if M is third quadrant, then s( - n) = -n - 1 and t( - n) = 0 works for both filtrations. I

Theorem 11.17: Let M be u j r s t or third quadrant bicomplex, and let {'Er} and {"E') be the spectral sequences determined by thejrst and secondfiltrations of Tot(M), respectively. (i) For each p , q, we have IE?, = 'EL,¶for large r ; similarly for "EFq. (ii) IE;,, =-H,(Tot(M)) and * H,(Tot(M)). P

P

Proof: Lemma 11.16 and Theorem 11.13.

I

We have created notational monsters, 'EL,¶and *'EL,, (though they could be worse; one corner is still undecorated), but the next two theorems will replace them by something less ugly.

326

11 Spectral Sequences

Having two spectral sequences converging to H,(Tot(M) ) is, of course, an advantage. With Theorem 11.17, we may contemplate calculating factor modules of H,(Tot(M)) as certain subquotients of E 2 ; obviously, we should at least be able to compute EZ.Let us consider the first filtration (we omit Mi, the differential the prescript I). Fix nand recall (FPTot(M)), = d is, of course, the restriction of the differential d = x ( d ' + d") of Tot(M). Now (Fp/Fp-'), is just M,,,, where p + 4 = n, for F:-' consists of all

His,

+

terms on p q = n to the left of the vertical line at p - 1, while F: consists of all such points to the left of the vertical line at p. Also, dp,,(MP,,)c Mp- 0 M,, (since only db,, d;,, are relevant). However, Mp-l,qc FP-', so that only d' survives in FP/FP-'. We conclude that FP/FP-' is the pth column with differential d". Therefore

,-

+

EP,, = H,(FP/FP-',d) = H,(M,,,,d") = ker(db',,:MP,,

-+

Mp,4 - ,)/im(db',4 + 1 Mp,q + 1

+

MP.,).

A better notation for EP,¶is Hi,,(M). For each fixed p, H;,,(M) is a complex: the differential d' induces maps d':H;JM)-+ Hi-,,,(M)by

a;,,: [ZP,,I

[db.¶ZP,,l. where [ ] denotes homology class. It is a routine exercise (left to the reader) that 2 is well defined. Taking homology of H;,,(M) with respect to J' gives a bigraded module {HbH;,,(M)}, the first iterated homology of M with respect to the first filtration. In short, first take homology of the pth column, then take homology of its rows. Theorem 11.18: if M is a first or third quudrunt bicomplex, then 'E;,¶ = H;Hi,JM)* H,(Tot(M)), P

where { ' E r } is the spectral sequence arising from the $first filtration. Proof: In light of Theorem 11.17, it suffices to compute E 2 . We have already verified that Ep,q = Hi,,(M). Now invoke Exercise 11.13 which

327

Bicomplexes

identifies d’ = By with the connecting homomorphism 8 arising from the short exact sequence of complexes O ~ F P - ~ I F ~ - ~ ~ F P ~-,oF ~ - ~ ~ F P ~ F ~ - ~

(where F

= IF). Consider

this sequence in degrees n = p

I

I/

0 + ~ , - 1 , q

+

+ 4 and n - 1:

i/”

A M,-l.q@M,-l,,

+ +

(actually, there should be a third term, M p - Z . q + lin, im(d’ d”), but it becomes 0 upon dividing out F P p 2 ) Let . z E M p , , be such that x = [ z ] E H ; , J M ) ; thus d”z = 0. Choose C ‘ z = (z,O). The formula d = i-’(d’ d”)n-l now gives

’

ax = [ i - d ’ z ] , since d“z = 0. This says d’ = (7 is induced by d’, for i is merely an inclusion map whose presence guarantees the element lies in the correct place. Thus EZ = H ( E , d ’ ) = H’H”, the first iterated homology. 1 Let us evaluate the E 2 term arising from the second filtration of Tot(M). It is at this point that a little trick enters, necessitated by the “danger alert” mentioned when the second filtration was defined. Definition: If M = {Mp,q,dk,,,dy,q)is a bicomplex, its transpose ‘ M is the A;.,, A;,,}, where bicomplex {rMp,q,

‘M,,,

= M 4.p,

A;.,

= d;,,.

and

= di,p.

It is a simple matter to check that ‘ M is. indeed, a bicomplex; obviously, ‘M is just M reflected about the line p = q. There are two elementary, but important, observations. First, Tot(‘M) = Tot(M) as complexes, i.e., they + A’). have the same nth terms and the same differential d = x(d’+ d ” ) = c(A’ Also, the second filtration of M is the first filtration of ‘ M . Theorem 11.19: I f M is u frrst or thiril rpudrunt bicomplex, then

‘‘E;,,

= H;H;,,(M)

* H,(Tot(M)), r

where {“E’} is the spectrul sequence uri,\ing from the second filtration.

328

11 Spectral Sequences

Proof: Since the second filtration of M is the first filtration of ' M , Theorem 11.18 gives H;H;,,('M) H,(Tot('M)) = H,(Tot(M)) (our first observation P above). Now HE,,('M) = ker A;,,/im A;, + = ker d;,,/im d; + = H;,,(M). Finally, A = d" yields

,

HlpHiJ'M) = HiHbJM).

j

The second iterated homology H;H,,,(M) first takes homology of the pth row, and then takes homology of its columns. It is important that EZ terms can be computed, so let us review the procedure. First filtration: fix p, so that Hi,, = H,(Mp,*); now regard this as a complex and take homology H;H;,,(M). Second filtration: relabel the modules as Mq,p;fix p, so that H;,,(M) = H,(M*,J; now compute H;H;,,(M). Although both spectral sequences converge to H,(Tot(M)), the corresponding filtrations and factor modules need not be the same, i.e., it is possible that 'E& $ "E&. Remark: If M is a third quadrant bicomplex, we have already mentioned that one gets rid of negative indices by raising them: M - p , - q = MPs4.The same convention applies to the bigraded modules E', except that r does not change sign: thus

ELp,-,, = E,P.,. The differentiald,: E, -P E, now has bidegree (r, 1 - r)and filtrations decrease : F, 3 F p + l for all p . The reader is requested to restate the definition of convergence E,P., H"(Tot(M))for the spectral sequences arising from either P

filtration of Tot(M) when M is a third quadrant bicomplex. The proof of the next lemma will illustrate this sign change and its notational consequences. Definition: Let M be a bicomplex; {E'} is a usual spectral sequence of M if it is the spectral sequence arising from either the first or second filtration of Tot(M). The following simple situation arises often enough to deserve a name. Definition: A usual spectral sequence {E') of a first or third quadrant bicomplex collapses if E;,, = 0 for q # 0. Thus, collapsing says the bigraded module E 2 can have nonzero terms only on the p-axis. Lemma 11.20: Let {E'} be a usual spectral sequence of a bicomplex M ,

329

Bicomplexes

where M i s jirst or third quadrant. I f { E'

EE4 = E:,¶,

all

p,q,

collapses, then

H,(Tot(M)) Z E;,,.

und

Proof: The differential d' on E' has bidegree ( - r, r - I), so that the arrow in the p-q plane representing d;,¶ (connecting ( p , q ) to ( p - r , q r - 1)) has slope (r - l)/- r. Since the only nonzero terms lie on the p-axis, d' = 0 for r 2 2, for every d;,, either starts or ends at a 0 module. Therefore, E' = kerd'= kerd'limd' = H(E') = E"' for all r 2 2, and so E2 = E". If M is first quadrant, there is a filtration 0 of H,(Tot(M)) with

+

0 = W ' H , c . . . c W"',

+

= H,

and

@pH,/OP-'H,

z EE4 = E;,¶,

where n = p q. As there is only one such nonzero term, namely, E;,-,, the filtration 0 has only one nonzero factor module and H,(Tot( M) ) = H, E E;,o. If M is third quadrant, there is a filtration Y of H-,(Tot(M)) (Lemma 11.16(ii)givess(-n)= - n - 1 andt(-n)=O)with 0 = Yfl-'H-,,

c

Y-"H-, c . . .c Y ' H - ,

c

Y'H-,

= H-,

and Raise indices and rename the filtration @, so that

H" = @OH" 3 @'H"

3

' '

.

= Y PWe .now have

CyH" 3 @+, ' H " = 0

= I

and

z EEp,-, OPH"/@P+'H"

=

E?,.-,

=

E';.¶.

Again, there is only one nonzero factor, namely, E3' z WH"; that all previous factors are 0 says H" = @OH" = Q'H" = . . . = W H " , and so H"(Tot(M)) z E?'. It is necessary to see the index raising in boring detail one time; we shall not be so tiresome again!

Remark: There is a similar result to Lemma 1 1.20 if a spectral sequence collapses on the q-axis. Let us use spectral sequences to give another proof of Theorem 7.9: the definition of Torf(A, B) is independent of the variable resolved (we shall also obtain a bit more information). What is more natural than to resolve both variables simultaneously?

Theorem 11.21:

Let PA and Qe he deleted projective resolutions of a right

330

11 Spectral Sequences

R-module A and a left R-module B, respectively. Then

O B) HAPA 0 Q B )

2 H A A 0 QB).

Hfl(PA

Proof: Consider the first quadrant bicomplex M of Example 1: {Mp,qj

=

(Pp

@ Qq}?

so that Tot(M) = PA@J Q B . The first filtration has E2 term the iterated homology H,H;,,(M). Now H;,,(M) = kerd;,,/imd;,,+,, and this is just the qth homology of the pth column P, 0 Qe. Since P, is projective, hence flat,

for H;,o

= coker(P, @ Q, -+

P, @ Qo)

= P, @ B.

Hy;,,(M) = H,(P,

It follows that

OB)

and

thus, this spectral sequence collapses, and

H,(P,

@J Q B ) = H,(Tot(M)) 2

'E;,, z H,(PA 0 B).

A similar argument works for the second filtration, now using the fact that each Q, is projective, hence flat. Thus

this spectral sequence also collapses, and

HAPA 0 Q B )

HAA O QB).

I

We have actually proved more (namely, Corollary 7.10), for we only used projectivity of the modules P , and Q, to guarantee they are flat. Corollary 11.22: If PA and respectively, then

HAPA 0 B)

Q B

are deleted flat resolutions of A , and RB,

HAPA 0 Q B ) 2 Hn(AO QB).

Exercises: 11.16. Let PA be a deleted projective resolution of A , and EB

331

Bicomplexes

be a deleted injective resolution of B. Prove that H"(Hom(P,, B))

H"(Hom(P,4, ER))2 H"(Hom(A, Es))

using the third quadrant bicomplex of Example 4 and Exercise 11.14. 11.17. Let M be a first or third quadrant bicomplex all of whose rows (or columns) are exact. Prove that Tot(M) is acyclic. (One can prove this without the spectral sequence machine.) 11.18. Let G = G' x G" be a direct product of groups. If P' is a deleted G'-projective resolution of Z, and P" is a deleted G"-projective resolution of Z, then P' @ P" is a deleted G-projective resolution of Z @ Z = Z. (Hint: Use Exercise 5.5: there is a ring isomorphism ZG 2 ZG' OZZG".) 11.19. If G is a free abelian group of finite rankr, prove that cd(G) 5 r. (Hint: Use induction on r and Exercise 11.18 to construct a G-projective r.) resolution of Z of length I 11.20. Let cp: R --t R* be a ring map and let A be an R*-module. Prove pd,(A) I pdR*(A)+ pdR(R*). (Hint: Take an R*-resolution of A, then take an R-resolution of each term, and construct a bicomplex.) The added information obtained in Theorem 11.21 allows us to complete the discussion of anticommutativity of i terated connecting homomorphisms; we had reached the point, in Theorem 7.26, that a 3 x 3 commutative diagram of complexes having exact rows and columns yields a certain anticommutative diagram involving the homology of its four outside corners. Theorem 11.23: Let 0 + A' -+ A + A" + 0 he un exact sequence of right R-modules, und let 0 - B'+ B + B" - 0 he an exact sequence of left Rmodules. For each n 2 0, there is un unticvimmutatioe diagram Tor,(A", B")

Tor,- l(A', B")

P

Tor,, - A", B')

+ Tor,,- z(A',B')

where the maps are connecting homomorphisn1.s

Proof: By Lemma 6.20, there are deleted projective resolutions P,P, P" of A', A , A", respectively, such that 0 + P' + P + P" + 0 is an exact sequence

332

11 Spectral Sequences

of complexes. Similarly, there are deleted projective resolutions Q’, Q, Q” of B’ B, B ‘ , respectively, such that 0 + Q’ + Q + Q” + 0 is an exact sequence of complexes. There results a commutative diagram of complexes having exact rows and columns 0

I

o-P@Q’-

0

I-

P@Q

0

0

0

P @ Q ” -0

0

(one has such a diagram in each degree n, for tensor is a bifunctor and each P i , P,, P t , Qh,etc., is projective, hence flat). This is exactly the situation to which Theorem 7.26 applies: we obtain the anticommutative diagram with connecting homomorphisms

Theorem 11.21 allows us to identify H,(P‘ 0 Q”)with Tor,(A”, B”), etc., and the result follows. I The dual result holds for Ext, using Exercise 11.16 in place of Theorem 11.21.

Theorem 11.24: Let 0 + A’ -+ A + A” + 0 and 0 -+ B’ + B + B” + 0 be exact sequences of left R-modules. For each n 2 0, there is an anticommutative

333

Bicomplexes

diagram

’Ext”’ ‘(C’,A’)

Ext”(C’,A”)

Ext”+’(C”,A”)

Ext”+*(C”,A’)

where the maps are connecting homomorphisms. There is a nice application of this, but it is better if we first present a general lemma.

Lemma 11.25: (i) A commutative diagram of modules c4

A

f

D

v

B

lg

.I

gives rise to a complex X, where

x =O+D-CCOBA-+ U-9)

(ii) If i is monic, then

H,(X) 2 ker(7:coker g

-+

coker i).

(iii) If i and j are monic, then

H,(X) 2 (im i

n imj)/im if.

Proof: (i) If we replace g by - 9 , we obtain an anticommutative diagram which may be regarded as a first quadrant bicomplex M (the displayed terms are at positions (O,O), (0, l), (l,O), and (1,l); all other terms are 0), and Tot(M) is precisely X. (ii) Let us compute the iterated homology HbHi,,(M), which may be displayed as a 2 x 2 matrix. Since i is monic,

Hb’,,(W = coker i coker kerg g

1

334

11 Spectral Sequences

and = HbHi,,(M) =

E P.4 2

["*

*

1-

ker(2:coker g + coker i)

= 0 for all r 2 2, since EL,l is a subquotient Now E& = 0 implies of E i , l ; therefore, E& = 0. We know that E:.o = kerj; what is E?,? The differential d': E' + E' has bidegree ( - r, r - I), so that

E;:,'

= kerd;,,/imd:+,,

l-r.

If r 2 1, both maps d;,o and d:, 1 , are 0 (the first ends at 0, the second begins at 0, for it is only a question of being outside the little 2 x 2 square). Therefore E:,, = E:,, = . . . = E y 0 (Exercise 11.12). Further, E:,4 = H,(X). In particular, there is a two-step filtration of Hl(X) with factor modules E& = 0 and ET,o = kerj. Therefore, H , ( X ) z kerJ. (iii) Recall that

x = o + D -(1. --c~ ~

9)

~ i + jA-+ O ,

so that

H,(X) = ker(i +j)/im(f, -9).

Define cp: ker(i

+ j) .+ im i n im j by

cp:(c, b) H ic.

Since (c, b) E ker(i + j), ic + j b = 0 and ic = -jb E im i n imj. It is clear that cp is epic and im(f, -g) c ker cp. Therefore, cp induces an epimorphism tp:ker(i

+ j)/im(f, -9)

-+

(im i n im j)/im if

(just compose the epimorphism onto imi n im j with the natural map mod if). A simple calculation, using i and j monic, shows that tp is monic, hence is an isomorphism. I (Cartan-Eilenberg): Let A , and RB be modules, and let O - + K $ P + A + O and O + L $ Q + B + O be exact, where P and Q are pat. Then Theorem 11.26

(i) Tor,(A, B) z Tor,- 2 ( K ,L ) for n > 2; (ii) Tor,(A,B) 2 ker(a 0 p); (iii) Tor,(A,B) [im(a @ lQ) n im(l,@ /3)]/im(a 0 B). Corollary 11.27:

Let R be a ring with right ideal Z and left ideal J . Then

(i) Tor,(R/Z,R / J ) 2 Torn-2(Z,J ) f o r n > 2 ;

335

Bicomplexes

(ii) Tor,(R/Z, R / J ) 2 k e r ( l 0 J -, I J ) ; (iii) Tor,(R/Z, R / J ) ( I n J ) / Z J . Proof: This follows immediately from the Theorem applied to the exact sequences 0 + Z + R + R / I + 0 and 0 -+J R + R / J + 0. I --t

As an instance of this, let R be a commutative local ring with maximal ideal m and residue field k = R/m. Then Tor?(k, k ) 2 m/m2. Of course, one may prove this more simply.

Proof of Theorem 11.26: (if Theorern 11.23 gives an anticommutative diagram of connecting homomorphisms:

Tor,,- , ( AL)

Tor,, - ( K ,B )

-

Tor, - , ( K , L )

If n > 2, these maps are isomorphisms, for all the bordering terms are “honest” Tor’s (i.e., not Tor,) which vanish because a variable is flat. Therefore, for n > 2 Tor,,(A,B ) z Tor,- J K , L). (ii) There is a diagram with exact rows and columns that is almost commutative (the upper left square is only anticommutative): 0

0

It follows that Tor,(A,B) z Tor,(A,L) 2 ker(cr 0 I L ) = ker(y ( S I @ 1t)), = ker(a 0p).

since y is monk,

336

11 Spectral Sequences

(iii) Apply Lemma 11.25 t o the commutative diagram 10 1,

p@Qt

Z@lQ

KQQ

Since P and Q are flat, the maps i = l p @ fl and j = a @ 1, are monic. But the map J:coker(l, @ B) -,coker(l,@ B) is just a @ lB, whose kernel is Tor,(A, B).The final part of the Lemma gives the desired isomorphism. Exercise: 11.21. Let 0 -+ K 5 P -+ A -,0 and 0 -+ B -,E 5 L -+ 0 be exact sequences with P projective and E injective. Then

(i) Ext"(A,B) z Ext"-'(K, L ) for n > 2; (ii) Ext2(A,B ) z coker Hom(a,B); (iii) Ext'(A,B) z kerHom(a,P)/(kera* ker&).

+

KUNNETH THEOREMS

The Eilenberg-Zilber Theorem (Example 3) poses the problem : given' complexes A and C whose homology is known, find H(A @ C). Even in the special case that C consists of a module C concentrated in degree 0 (i.e., C , = C if q = 0 and C, = 0 otherwise), the problem is not simple. If one further assumes R is a hereditary ring and each term A , of A is a projective R-module, then the solution is the Universal Coefficient Theorem 8.22. To indicate why the general problem involving two complexes is difficult, even with the extra assumptions just mentioned, let us see what goes wrong with one's first idea of considering the bicomplex M of Example 2 having M,., = A , 0 C,; after all, Tot(M) = A @ C. We I compute 'E& = HlpH;,,(M). The pth column of M is A , @ C;since A , is projective, hence flat, we know that H,(A, 0 C) z A , Q H,(C) (Exercise 6.4: homology commutes with exact functors). Hence H;,,(M) = A , @ H,(C). The Universal Coefficient Theorem now gives Ei,q = H,(A 0 H,(C)1 = Hp(A) 0 H,(C) 0 Tor?(H,- i(Ah H,(C) ). Though it is true that E 2 * H(A @ C), we are essentially helpless because we cannot compute E" (the other spectral sequence, arising from the second

337

Kunneth Theorems

filtration of A @ C, is worse: the Universal Coefficient Theorem may not apply, and one may not even compute E’). Indeed, there seems to be just one meager result that can be salvaged: if A is projective, i.e., each A , is projective, and if either A or C is acyclic, then A @ C is acyclic (either observe that Ef,, = 0, or use the much more elementary Exercise 11.17). We give two solutions to the problem. The first, strong enough for use of the Eilenberg-Zilber Theorem, does not use spectral sequences (the proof has been deferred to this chapter since it involves bicomplexes and graded modules). The second, more general, solution uses spectral sequences, as well as a special case of the first (Corollary 11.29).It also illustrates a second technique in using spectral sequences. Our first illustration treated a case in which both spectral sequences collapse; this proof is an instance in which only one of them collapses. Here is a collection of easy exercises which will be used in the forthcoming proof. Exercises: 11.22. Consider the diagram of modules

H

in which the row and the triangle are exact. Prove that there exist unique maps cr:Z + H and P : H + W making the outer triangles commute, and that there is an exact sequence 0-Z:

H 5 W +O.

The result remains true if one replaces “modules” by “graded modules”. 11.23. For fixed k 2 0 and complexes A and C, show there is a bicomplex M with M,,q = Tor,(A,, CJ. (Use the facts that Tor, is a bifunctor that is additive in each variable, together with the Sign Lemma 11.15.) Denote the complex Tot(M) by Tor,(A, C). 11.24. If 0 + A‘ + A + A“ -+ 0 is an exact sequence of complexes, and if C is a complex, then there is an exact sequence of complexes

. . . +Tor,(A,C) + Tor,(A”,C)+ A ’ @ C + A @ C

-+

A”@C --t 0.

(Use naturality of connecting homomorphisms and the fact that each Tor,, k 2 0, is a bifunctor.)

338

11 Spectral Sequences

11.25. If A is a complex, then its cycles Z = [ Z , ) and boundaries B =

{ E n )are subcomplexes having zero differentiation (i.e., each d, = 0).

1 1.26. Let (A, A) and (C, d ) be complexes. If A has zero differentiation, then the differentiation in A 0 C is { f 1 d,,). 11.27. Let A be a flat complex (i.e., A , is flat for each n) having zero differentiation. Then, for any complex C,

H,(A 0 C) = (A 0 HJC)),,

where H J C ) is construed as a complex with zero differentiation. Theorem 11.28 (Kiinneth): Let A be a complex of right R-modules whose subcomplexes of cycles Z and boundaries B are Jut. Given a complex C, there is an exact sequence for each n

0

-+

u

p+q=n

H,(A) ORH,(C) 5 H,(A OR C) 5

u

p+q=n- 1

Tor:(H,,(A), H,(C) )

where a : [ a ] 0 [ c ] I-+ [ u 0 c ] , a E Zp(A), c E Z,(C). Moreover, natural in A and C.

c(

and

0,

-+

B are

Proof: (A. Heller): The usual exact sequences

+O

O-+Z,+A,+E,,-,

give rise to an exact sequence of complexes 0 + Z 5 A 5 B+ + 0,

where (B'),, = B,Obviously, B flat implies B+ flat, so Exercise 11.24 gives an exact sequence of complexes

0 +Z@ C

- i@ 1

A0C

1

B+ 0 C -+ 0.

?I@

There results an exact triangle of graded modules in which a has degree - 1 and the other maps have degree 0:

HJB+ c i ~C)

H,(Z Q C)

H*(A Q C )

Note that H,(B+ 0 C) = H,- l(B 0 C). Also, Exercise 11.27 permits us to write HJB' 0 C) = B+ 0 HJC)

and

H,(Z 0 C) = Z 0 H,(C),

339

Kiinneth Theorems

where H J C ) is construed as a complex having zero differentiation. Rewriting with these two observations in mind gives an exact triangle of graded modules i

H*(A @ C) where we now regard (71 Q l), as having degree - 1 and 13as having degree 0. The next claim is that the connecting homomorphism i' is the map induced by the inclusion B 4 Z. Let A be the differentiation in A, let d be the differentiation in C , and consider the defining diagram for the connecting homomorphism (having replaced B + by B) (A 0 C)n

(Z 0 C h - 1

I

1\81

'(BBQ-1

D

(A 0 CL-1

Ignoring needed summation, a typical cycle in (B @ C ) ,- I has the form b Q c , where b E B , - , , c E C,, p y = 1 7 . I f A u = h, then

+

d(b @ c ) = j - ' D ( u @ c) = j - ' ( A u @ c + (-l)PaQdc) = j - ' ( b @ c ( - - 1)P(1Q d ) ( a Qc)).

+

But j - ' tells us to regard b Q c + ( - I),( 1 Q d ) ( u @ c) in (Z @ C ) n - ,, and Exercise 11.26 tells us this element is homologous to b @ c. The claim is established. Finally, consider the exact sequence of complexes

0 4 B -+ Z + H,(A) + O , where H J A ) is regarded as a complex with zero differentiation. Since Z is flat, by hypothesis, there is an exa,ct sequence of graded modules to which we affix our exact triangle (the maps B @ H , ( C ) -+ Z @ H J C ) being the same):

340

11 Spectral Sequences

Exercise 11.22 applies at once to give an exact sequence of graded modules

0

-+

HJA) 0 H,(C) 5 H,(A O C) 5 To 0.

The first spectral sequence thus collapses, and Lemma 11.20 gives H,(Tot(K)) = 'E,Z,o = H,(A @ C). Since the second spectral sequence also converges to H,(Tot(K)), it is only necessary to show its E2 term H;Hb,,(K) is as stated in the Theorem. For fixed p (noting the transposing of p and q), we must first consider the A, 0 M,,t; this complex homology of the complex with gth term is just Tot(A @ M,). Since M arose from a proper projective resolution of C, the cycles and homology of M , are projective; Corollary 11.29 applies to give Hb,,(K)

= H,(Tot(A

0 M,) ) =

U

3+t=q

H,M) 0 H,(M,).

We also know that

.

'

+

H1(M2) H,(M,)+ H,(Mo) + H,(C)+ 0 -+

is a projective resolution. The homology of this complex tensored by H,(A)

347

Kunneth Theorems

is precisely the definition of Tor:

We merely state the dual result for cohomology. Theorem 11.34 (Kiinneth Spectral Sequence): Let A be a positive complex und C u negative complex. If either A is projective or C is injective, there is a third quadrant spectral sequence

Eq3q=

fl

s+t=q

ExtP(H,(A),H,(C)) H"(Hom(A,C)). j

P

We give a second proof of the Kiinneth Theorem 11.28 using spectral sequences. As the proof will yield an E 2 consisting of only two columns, we first dispose of this situation. Lemma 11.35: Let M be a jirst quadrunt hicomplex for which E 2 ( = ' E 2 ) consists of two columns: there is t > 0 with E i . g = 0 for p # 0 and p # t. Then

(i) E 2 = E 3 = . . . = E'; (ii) E'+' = E"; (iii) there are exact sequences 0 -+ EbT:

+ H,(Tot(M)) + Ei.+n\t + 0 ;

(iv) there is an exact sequence

d' dt . . . +H,+ ,(Tot(M))+ E; " + ,-,+ E i . n + H,,(TOt(M))+E?.-t+E;, "- 1 +.".

Proof: The differential d': Er + E' has bidegree ( - r, r - 1 ) : d' goes r steps left and r - 1 steps up. It follows that d' = 0 for r # t, since it either begins or ends at 0. Exercise 11.12 now applies t o prove (i) and (ii). Consider the picture of the E2 plane below.

348

11 Spectral Sequences

Since M is first quadrant, we know E&

3 P

H,(Tot(M)), i.e.,

0 = @ - ' H , c @OH, c . . . c @"H, = H ,

@pH,/@p-'Hn z E g q .

and

Thus @ O H , = E?,

c H,.

Also, @'H,/@'H, = E c n- = 0, so that @OH, = Q'H,. Indeed, @OH, = @ ' H , = . . * = 0'-H , . When we reach t, and then all remains constant again: W H , = 0" 'H,= * . . = W H , = H,. We conclude there is an exact sequence 0 -+E Z ,

-+

H,

+ E r n - ' + 0.

In light of (ii), this may be rewritten as

(*I

0 + EbT:

+H ,

-+

Ei,+,Lt

0,

which is item (iii). There is an exact sequence obtained from d = 2,,+ 0 + ker d'

+ Ei, ,+

-1:

d'

+ Eb,, -+ coker d' + 0.

By definition,

Ei.+,\

1

= ker

4,

n+

I

-Jim

4,.,+ 2

- z1

= ker

4,,+

1 --I

and

E'+1 O,n

= kerd~,,/imdi,,+l-t= Eb,n/imdi,,+l-t= c ~ k e r d i , , + ~ - , .

The sequence may thus be rewritten

(**I

0 -+ Ei,:\

-+

Ei, ,+

df

--f

EL*, + EbT,'

-+

0.

Splicing (*) and (**) together and remembering that E' desired exact sequence 2

d'

=

- . . + H , + l -*Er,,+ l - r - + E ~ , n + H , + E : , _ I - r . . . .

E 2 yields the

I

Observe that it is possible that the terms E' of a spectral sequence may remain constant for a while before moving. Thus, it is not true in general that E' = E'+l implies E' = E m . Exercise: 11.31. Let M be a first quadrant bicomplex for which E 2 consists of two rows: E:,¶ = 0 for q # 0 and q # t. Show there is an exact

349

Kunneth Theorems

sequence

..

* -+

H,+,(Tot(M))-+ E,&

,j'+ I

--+

E,Z-,- l , t

+

+ 1.)

(Hint: d' vanishes unless r = t

H,(Tot(M)) +. . . .

We really need only a special case of this lemma.

Lemma 11.36: Let M be a first quadrant hicomplex for which E 2 ( = ' E 2 ) consists of two adjacent columns: EE., = 0 for p # 0, 1. Then there are exact sequences 0

-+

E&,

+

H,(Tot(M))+ E:. ,- -+ 0.

Remark: If M is third quadrant, arrows are reversed, and there is an exact sequence under the adjacent column hypothesis: 0 -+ E i * n - l+ H"(Tot(A4))-+ E:*" -+ 0.

Here is the fancy proof of the Kunneth theorem when A and C are positive complexes. Theorem 11.37 (=Theorem 11.28): Let A be a positioe complex whose subcomplexes Z and B are j u t . Given a positice complex C, there is an exact sequence for each n 0

-+

1l

p+q=n

H,(A) 0 H,(C)

+

H,(A 0 C) --+

p+q=n-

1

Torl(H,(A), H,(C))

+

0.

Proof: First, let us show A is flat. For each p , exactness of 0 -+ Zp -+ A, -+ B,- -+ 0 gives exactness, for any module X, of

,

Tor,(Zp,X)+Torl(A,,X) +Torl(Bp-l,X). The outer terms vanish because Z, and B,- are flat; hence Tor,&,, X) = 0 for every X, and so A, is flat. Since A and C are positive, Theorem 1 I .34 applies to give

Now exactness of 0 -+ B, -+ Z,

+ H,(A) --+

0 gives exactness of

Tor,(Z,, X) -+ Tor,(H,(A), X ) If p

-

-+

Tor,- l(B,, X).

1 2 1, flatness of Z, and B, force the middle term to vanish:

Tor,(H,(A),X)

=0

for p # 0, 1.

350

11 Spectral Sequences

Thus Eiqq= 0 for p # 0, 1, which is precisely the adjacent column hypothesis of Lemma 11.36. There are thus exact sequences O-+ E;,"

-+

H,(A @ C)

-+

E:."-,

-+

0,

and this is what is sought, once one replaces the E 2 terms by their values.

I

GROTHENDIECK SPECTRAL SEQUENCES The main theorem of this section shows that a composite of functors often leads to a spectral sequence. For a change, we will prove the third quadrant (cohomology) version and merely state the dual first quadrant (homology) version. As usual, all functors are between module categories. Theorem 11.38 (Grothendieck): Let G :'u -+ 8 and F : 23 3 6 be functors with F left exact such that E injective in 2I implies GE is right F-acyclic. For each module A in 2I, there exists a third quadrant spectral sequence with

-

E t q = R P F ( R 4 G ( A ) ) R"(FG)(A). P

Proof: Choose an injective resolution 0 -+ A G to its deletion to obtain a complex

GE,

= O-+

GEo -+ GE'

-+

--*

Eo + E'

. . , and apply

3.

GE2 +. . . .

By the dual of Lemma 11.33, there is a proper injective resolution of GEA. The corresponding bicomplex M may be pictured, after indices are raised, as a diagram of injective modules M P a q :

T T

T

T

0

T T T T

0

T

0

351

Grothendieck Spectral Sequences

Consider the bicomplex FM and its total complex Tot(FM). It is very easy to compute the spectral sequence arising from the first filtration. For fixed p , the pth column MP** is a deleted injective resolution of GEP.Therefore, FMPq*is a complex whose homology is R q F : we have

H ” P * q ( F M= ) HY(FMP.*)= ( R 4 F ) ( G E P ) . But EP injective implies GEP is right F-acyclic. so (RqF)(GEP)= 0 for 4 > 0. Also, F left exact implies R’F = F . We conclude that

All that survives is on the p-axis: 0 + FG(E’)

+ FG(E’) + F G ( E Z )+ .

..

Applying HIP, we see that if q = 0, if 4 > 0.

EP,.q = {ORP(FG)(A)

Therefore, the first spectral sequence collapses, and we have H”(Tot(FM))z R “ ( F G ) ( A ) . Now consider the spectral sequence arising from the second filtration. Let us recall that our choice of a proper injective resolution of GEA means the following. If the qth row is M4, and Z’q, B’q, and H’qdenote the complexes with zero differentiation of cocycles, coboundaries, and cohomology, respectively, then for each p there are injective resolutions Z’P.’ . . . , 0 + BP(GEA)+ B’p.’ + B‘P,‘ + . . . , 0 + ZP(GEA)+ Z’P.’

and

4

0 + HP( GE,) + H ’ p . @ + H‘P- + . . . .

For every p , q, we transpose indices and consider the usual exact sequence 0 -, Z’4.P

+

M4.P

!!+B‘q+ 1 . P + 0;

it splits because all terms are injective. The sequence thus remains split after applying F , and

ker(Fd’4.P)= FZ’4.P

and

im(Fd’q-’,P)

= FB’4.P.

Therefore, H’qJ’(FM) = Hq(FMP) = ker(Fd’q*P)/im(Fd’q-‘ s p )

= FZ’q,P/FB’4*P.

352

11 Spectral Sequences

Since 0 +B q s P

+ Z'4.P + H'Q.P+

0

is also split exact, FZ'4*P/FB'q*P = FH'q.P.Weconclude that H'q*P(FM)= FH'4.P. Now "Ep4 = H"PH'qJ'(FM) = HP(FH'4*P). But the modules H'4.p form an injective resolution of H4(GEA)= (RqG)(A);by definition of right derived = RPF(R4G(A)). Hence functor, Hp(FHrqsP) ( R p F ) ( R q G ( A= ).) R"(FG)(A). P

I

Here are three more versions; the proofs are the same.

Theorem 11.39: Let G:2I + 23 and F:b + 6 be functors with F right exact and P projective in 'ill implies G P left F-acyclic. For each module A in a, there is a first quadrant spectral sequence Ei.4

= LpF(LqG(A) => Ln(FG)(A).

Theorem 11.40: Let F : B + 6 be a contravariant left exact functor, and let G:2I + 23 be a functor such that P projective in 2I implies GP right F-acyclic. For each module A in 'ill, there is a third quadrant spectral sequence ET4 = RPF(L,G(A)) R"(FG)(A). P

Theorem 11.41: Let F : B + 6 be a contravariant left exact functor, and let G : X + 23 be a contravariant functor such that P projective in % implies GP right F-acyclic. For each module A in 'ill, there is ajirst quadrant spectral sequence Ep',4= RPF(R4G(A))* L,(FG)A. P

There are other Grothendieck spectral sequences for other combinations of variances whose statements are left to the reader. Before giving applications of the Grothendieck spectral sequence, let us give the standard proof of the five-term exact sequence. We could have given the proof some time ago, but the advantage of waiting is that we may now give another proof of Theorem 11.2.

Theorem 11.42: Assume Ei.q * H,(M), where {E'} is afirst quadrant specP tral sequence.

353

Grothendieck Spectral Sequences

(i) (ii) (iii) (iv) (v)

For each n, there is an epimorphism E;,, + EC,; for each n, there is a monomorphism E:o Ei,o; for each n, there is a monomorphism E& -+ H , ( M ) ; for each n, there is an epimorphism H , ( M ) -+ EZo; there is an exact sequence -+

H , ( M ) -+ E;,o 2 E;,,

-+

H,(M)

-+

E:,o

-+

0.

Remark: The maps in (i) and (ii) are called edge homomorphisms.

Proof: (i) By definition, E;,, = ker di,,/im d ; , ,-1 . But kerd;,,: E;,fl E Z , , , + , is the zero map (since E;,q is first quadrant). Therefore kerd;,, = E;,, and there is an epimorphism E;,, E;,,. The argument can be repeated for each d':E' -+E', using only the fact that d' has bidegree ( - r , r - 1). There is thus a chain of epimorphisms E;,, -+ E;,, -+ E&, -+ . . . -+ E',,,. This completes the argument, for we know E;,, = EL,, for large r. (ii) This argument is the dual of that just given: E:,o 4 Elf,obecause E.",o = kerdi,o/imdlf+2,- = ker d,',o c Elf.o. Now iterate, noting that all imd' = 0. (iii) The definition of convergence yields -+

-+

EZ,

= @ o H , / @ - l H , = @OH,, c

H,

(iv) The definition of convergence yields E20

= @"H,/W-'H, =

H,,/W"'lH,,.

(v) There is an exact sequence

0

-+

ker d;,o

-+

E:,o

5 E& ,

-+

coker d k. We proceed by induction on k. If k = 1, then n r 2. Now N " ( n , Z )= Z" = Z and H'(ll,Z)= Hom(l7, Z ) z Z, so the induction begins. For the inductive step, choose a subgroup N with n / N z Z ( N is necessarily free abelian of rank k - 1). The Lyndon-Hochschild-Serre spectral sequence satisfies E54 = H P ( I I / N H4( , N , Z ) ) * H"(ll, Z). P

358

11 Spectral Sequences

Since l l / N 2 Z, we know Epiq = 0 for p # 0, 1. Therefore, the "two adjacent column" hypothesis of Lemma 11.36 (rather, the third quadrant version of that Lemma) holds, so there are exact sequences

(*I

0 + E:* "-

' + H"(ll,Z) + Ei*"

--*

0.

Now E i V n - ' = H ' ( l l / N , H " - ' ( N , Z ) ) . By the remark before Theorem 11.5, H"- ' ( N ,Z) is ll/N-trivial, and so

'

H ' ( l l / N ,H" - ( N ,Z) ) z Hom(Z,H" - ' ( N ,Z) ) 2 H"- ' ( N ,Z);

by induction, this group is free abelian of rank (:I:). The term E;." = H o ( l l / N ,H"(N, Z)) E H"(N,Z), using the remark once again. By induction, this group is free abelian of rank (,k '). It follows that the exact sequence (*) of abelian groups splits, whence H"(ll, Z) is free abelian of rank (:I i) =

+

(9. I

Corollary 11.50: Zf

n is free abelian of finite rank k, then cd(ll) = k.

Proof: Exercise 11.19 shows cd(ll) I k, while Theorem 11.49 shows this inequality cannot be strict. I

Lyndon [19483 computes H"(ll, Z) for any f.g. abelian group ll.

MORE MODULES We now examine composites of suitable pairs of functors and the corresponding Grothendieck spectral sequences. Let R and S be rings, and consider the situation ( A R R, B S , &); the associative law for tensor product gives a natural isomorphism A@R(B@sC)~(AORB)@SC. Thus, if G = B @ s : s % R ~ R . ( n nand F = A@R:R.(nn+Ab, then FG = (A BRB)B S Clearly . F is right exact, so Theorem 11.39 applies when there is acyclicity.

Theorem 11.51: Assume A , and R B S satisfy Tof(A, B BSP ) = 0, all i > 0, whenever ,P is projective. Then there is a first quadrant spectral sequence for every ,C: To 1 t, then either p > 1 or q > t. If p > I, then EP,.q = 0 because we have exceeded pd,(L); if q > t, then EP,.q = 0 because we have exceeded pdR(T). It follows that E g g = 0 for all p, q with p + 4 = n, whence Ext;h(L,B)= 0. As B is arbitrary, pd,(L) I1 t. I

+

The reader should now be convinced that virtually every purely homological result may be proved with spectral sequences. Even though “elementary” proofs may exist for many of these results, spectral sequences offers a systematic approach in place of sporadic success.

References

Albrecht, F., On projective modules over a semihereditary ring, Proc. Amer. Math. SOC.12 (1961), 638-639. Anderson, F., and Fuller, K., Rings and Categories oJ' Modules. Springer-Verlag. Berlin and New York, 1974. Auslander, M., and Buchsbaum, D. A., Unique factorization in regular local rings, Proc. Nut. Acad. Sci. U.S.A. 45 (1959). 733-734. Bass, H . , Big projective modules are free, I//inois J. Marh. 7 (1963). 24-31. Brewer, J. W., and Costa, D. L., Projective modules over some non-Noetherian polynomial rings, J . Pure Appl. Algebra 13 (1978), 157-163. Cartan, H., and Eilenberg, S., Homological Algebra. Princeton, Univ. Press, Princeton, New Jersey, 1956. Chase, S. U., Direct products of modules, Trans. Arner. Math. SOC.97 (1960), 457-473. Chase, S. U., A generalization of the ring of triangular matrices, Nagoya Math. J. 18 (1961), 13-25. Cohen, D. E., Groups of Cohomological Dimemrrui One. Lecture Notes 245, Springer-Verlag, Berlin and New York, 1972. Cohn, P. M., Some remarks on the invariant basis property, Topology 5 (1966), 215-228. Cohn, P. M., Free Rings and Their Relations. Academic Press, New York, 1971. Curtis, C., and Reiner, I., Representation Theory OJ Finite Groups and Associative Algebras. Wiley (Interscience), New York, 1962. DeMeyer, F., and Ingraham, E., Separable Algebru.) over Commutatice Rings. Lecture Notes 181, Springer-Verlag, Berlin and New York, 1971. Endo, S., On semihereditary rings, J . Math. So(..Japan 13 (1961), 109-119. Endo, S., On flat modules over commutative rings, J. Math . SOC.Japan 14 (1962), 284-291. Faith, C., Algebra: Rings, Modules, and Categories, Vol. 1. Springer-Verlag, Berlin and New York, 1973. Fossum, R., Griffith, P., and Reiten, I., Trivial E.trension.7 of Abelian Categories. Lecture Notes 456, Springer-Verlag. Berlin and New York, 1975.

References Fuchs, L., Infinite Abelian Groups, Vol. I. Academic Press, New York, 1970. Fuchs, L., Infinite A b e h Groups, Vol. XI. Academic Press, New York, 1973. Goodearl, K. R., Global dimension of differential operator rings, Proc. Amer. Math. SOC.45 (1974), 3 15-322. Griffith, P., On the problem of intrinsically characterizing the functor Ext, Symp. Math. X l l l , p. 207-232. Academic Press, New York, 1974. Gruenberg, K., Cohomological Topics in Group Theory. Lecture Notes 143, Springer-Verlag, Berlin and New York, 1970. Helmer, O., Divisibility properties of integral functions, Duke Math. J . 6 (1940), 345-356. Herstein, I. N., Noncommutative Rings. Carus Mathematical Monograph, Math. Assoc. Amer., Providence, Rhode Island, 1968. Isaacs, I. M., Character Theory of Finite Groups. Academic Press, New York, 1976. Jans, J., Rings and Homology. Holt, New York, 1964. Jategaonkar, A. V., A counter-example in ring theory and homological algebra, J . Algebra 12 (1969), 418-440. Jensen, C. U., Homological dimensions of rings with countably generated ideals, Math. Scand. 18 (1966), 97-105. Kaplansky, I., Projective modules, Ann. Math. 68 (1958a), 372-377. Kaplansky, I., On the dimension of rings and modules, X , Nagoya Math. J . 13 (1958b),85-88. Kaplansky, I., Commutative Rings. Allyn and Bacon, Boston, Massachusetts, 1970. Kong. M., Euler classes of inner product modules, J . Algebra 49 (1977). 276-303. Lam, T. Y., Serre's Conjecture. Lecture Notes 635, Springer-Verlag, Berlin and New York, 1978. Lambek, J., Lectures on Rings and Modules. Ginn (Blaisdell), Boston, Massachusetts, 1966. Lang, S., Algebra. Addison-Wesley, Reading, Massachusetts, 1965. Lyndon, R. C., The cohomology theory of group extensions, Duke Math. J . 15 (1948). 271-292. MacLane, S., Homology. Springer-Verlag. Berlin and New York. 1963. MacLane, S., Category Theory ,fiw the Working Mathematician. Springer-Verlag, Berlin and New York, 1971. Matsumura, H., Commutative Algebra. Benjamin (Addison-Wesley), Reading, Massachusetts, 1970. Milnor, J., Introduction to Algebraic K-Theory. Princeton, Univ. Press, Princeton, New Jersey, 1971. Nagata. M., A general theory of algebraic geometry over Dedekind rings, 11, Amer. J . Math. 80 (1958), 382-420. Northcott, D. G., An Introduction to Homological Algebra. Cambridge, Univ. Press, London and New York, 1960. Nunke, R. J., and Rotman, J. J., Singular cohomology groups, J . London Math. Soc. 37 (1962), 301- 306. Ojanguren, M., and Sridharan, R., Cancellation of Azumaya algebras, J. A/gebra 18 (1971), 501-505. Osofsky, B. L., Upper bounds of homological dimension, Nagoya Maih. J . 32 (1968), 315-322. Osofsky, B. L., Homological Dimensions of' Modules, CBMS Regiunal Conf. Ser. Math. 12. Amer. Math. SOC..Providence, Rhode Island, 1973. Rotman, J., The Theory of Groups: An Iniroduction. 2nd ed. Allyn and Bacon, Boston, Massachusetts, 1973. Shelah, S., Infinite abelian groups, Whitehead problem, and some constructions, Israel Math. J . 18 (1974), 243-256. Small, L., Hereditary rings, Proc. Nut. Acad. Sci. U.S.A. 55 (1966), 25-27. Stallings, J., On torsion-free groups with infinitely many ends, Ann. Math. 88 (1968). 312-334. Stammbach, U., Homology in Group Theory. Lecture Notes 359, Springer-Verlag, Berlin and New York, 1973.

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Index

A Ab (category of abelian groups), 4 ACC (ascending chain condition), I10 Acyclic complex, 170 Acyclic module, 302 Acyclic space, 285 Additive functor, 7 Adjoint isomorphism, 37 Adjoint pair of functors, 37 Albrecht, 128 Algebra, 13 Annihilator, 254 Anticommutative diagram, 2 I5 Artinian ring, 1 13 Associative law for 0,16 Augmentation ideal, 146, 163 Augmented algebra, 146 Auslander, 237,241 Auslander-Buchsbaum, 263 Auslander-Buchsbaum-Nagata, 264 AxiomsforExt.212.214 Axioms for Tor, 226

B

B2(C.A ) , 155 B,(A) (n-boundaries), 169 Baer, 122 Baercriterion, 67

Baer sum, 209 Balanced presentation, 277 Bar resolution, 287 Ban-Rinehart. 297 Basis of free abelian group, 7 Basis of free group, 268 Basis of free module, 57 Bass. 112, 114, 138 Bezout ring, 127 Biadditive function, 9 Bicomplex (double complex), 3 19 Bidegree, 301 Bifunotor, I% Bigraded module, 300 Bigraded quotient, 301 Bigraded submodule, 301 Bilinear function, 14 Bitnodule, 12 Boolean ring, 119 Boundaries, 20, 169 Boundary operator, 19 Bounded filtration, 317 Brewer-Costa, 149 Bumside basis theorem, 268 Burnside problem, 278 C C = complex numbers

Cartan-Eilenberg, 124,334

371

372

Index

Category, 3 cd (cohomological dimension), 295 Central "xtension, 279 Center of ring, 7 Chain map, 168 Chain map overf, 18I Change of rings functor, 245 Change of rings spectral sequences, 364, 365 Character module, 87 Chase, 113, 114, 127 Coboundary, 155 Codimension (cod), 263 Coefficients, 21 Cofinal subset, 49 Cogenerator for category, 79 Coherent ring, 113 Cohn, 60.95, 126 Cohomological dimension (cd), 295 Cohomology groups of group, 265 Cohomology groups of space. 230 coker (cokernel), 25 Collapses, 328 Commutative diagram, 9 Comp (category of complexes), 169 Comparison theorem, 179 Complement of projective, 64 Complement of subgroup, 152 Completion, 54 Complex, 166 Composition, 3 Concentrated in degreek. 167 Connected sequence of functors, 212 Connecting homomorphism, 171 Constant direct system, 40 Constant functor, 6 Constant inverse system, 50 Contracting homotopy, 178 Contravariant functor, 6 Convergence of spectral sequence, 317 Correspondence theorem, 26 Cosyzygy, 235 Cotorsion group, 343 Covariant functor, 6 Cycles, 20, 169 Cyclic module, 25

D D(R) = global dimension of R , 236 DCC = descending chain condition, I13 Dedekind ring, 120

Degree, 300 Deleted complex, 178 Der(G, A), 157 Derivation, 157 Derived couple, 313 Differentiation, 167 Dim(R) (Krull dimension ofR), 261 Dimension shifting, 222 Dir(1) (category of direct systems overl), 44 Directed index set, 44 Direct l im i t ,l h , 41 Direct system, 39 Divisible module, 69 Domain = commutative integral domain Duality theorem, 294 Dual statement. 32

E f&,E!*'', 313, 328

d G , A ) , 155

Eckmann-Schopf, 72 Edge homomorphisms, 353 Eilenberg, 64 Eilenberg-MacLane space, 286 Eilenberg-Zilber, 321 End,(A) = Hom,(A. A) (endomorphism ring ofA), 59 Endo, 135 Enlargement of coset map, 271 Epic, epimorphism, 25 Equivalence (morphism), 6 Equivalent categories, 82 Equivalent extensions, 155,202 Equivalent functors, 75 Essential epimorphism, 136 Essential extension, 72 Euler characteristic, 255 Exact couple, 301 Exact functor, 35 Exact sequence of bigraded modules, 30 1 Exact sequence of complexes, 171 Exact sequence of modules, 26 Exact triangle, 173 Exponent ofgroup, 292 Ext",, 185, 186 Extended module, 145 Extending by linearity, 8 Extension ofgroups, 150 Extension of modules, 199

373

Index F Factor set, 154 Faithfully flat module, 249 Family, 252 fd (flat dimension), 238 FFR (finite free resolution)), 251 f.g. (finitely generated), 25 Filtration, 315 Filtration degree, 317 Finitely generated (f.g.), 25 Finitely related group, 276 Finitely related module, 90 First filtration, 324 First iterated homology, 326 First quadrant bicomplex, 32 1 Five lemma, 76 Five term sequence, cohomology, 304 Five term sequence, homology. 306 Fixed points, 162 Flat complex, 338 Flat dimension (fd), 238 Flat module, 84 Flat resolution, 85 Fossum-Griffith-Reiten, 236 Frattini subgroup, 268 Free abelian group. 7 Free group, 268 Free module, 57 Free resolution, 60 Frobenius algebra, 132 Functor. 5

G G-module (left W-module), 152 Gabel, 140 Generator for category, 82 Generators of module, 4 Generators and relations, groups, 275 Generators and relations, modules, 58 GL(n,A ) , 138 Global dimension, 235,236 Golod-Safarevif ,278 Goodearl, 250 Graded module, 300 Green, 355 Griffith, 214 Grothendieck spectral sequences, 350,352 Groupring, 117, 151 Gruenberg resolution, 271

H H".2I. 184,265 H,. 20. 169,265 Half-exact functor, 171 Helmer, 127 Hereditary ring, 120 Hilbert basis theorem, 108 Hilbert syzygy theorem, 247 Hom. 3 , 6 Horn, 323 Homology functor, 169 Homology groups of group, 265 Homology groups of space, 20 Homotopy, 177 Hopf s formula, 274 Hopkins-Levitzki, 114 Horrocks. 142 Horseshoe lemma, 187 Hurewicz, 287 I IBN (invariant basis number), 58 id (injective dimension), 235 Idempotent, I19 Identity functor, 5 Identity morphism, 3 im (image),24 Indecomposable module, 13I Induced map, 169 Injection. 29 Injective dimension (id), 235 Injective envelope, 73 Injective equivalence, 236 Injective module, 65 Injective resolution, 71 Inner automorphism, 158 Inv(1) (category of inverse systems over I ) , 5.5 Inverse limit, 50 Inverse system, 50 Invertible ideal, 125 Isomorphism theorem. first, 25 Isomorphism theorem, second, 25 Isomorphism theorem, third, 26,27,28 Iterated connecting homomorphism, 215

5,

J Jategaonkar, 236 Jensen. 236

374

Index

K Kaplansky, 121, 135, 139,236 ker (kernel), 24 Kong, 139 Kulikov, % Kunneth formula, 341,342 Kunneth spectral sequences, 345,347 Kunneth theorem. 338

L Laurent polynomials, 295 ID(R)(left global dimension), 235,236 Left derived functor, 181 Left exact functor, 34 Lifting. 153 lim, 41 4 lim, 50 c Limit term of spectral sequence, 314 Lindel-Liitkebohmert, 149 Local ring, 133,259 Localization, 97 Long exact sequences, 172 Lyndon, 358 Lyndon-Hochschild-Serre, 355

M RW (category of left R-modules), 4 !UIR(category of right R-modules), 4 MacLane, 227 Map of bigraded modules, 301 Map of graded modules, 300 Map of modules, 4 Mapping cylinder, 175 Maschke, 117,293 Mayer-Vietoris, 176, 177 McConnell-Roos, 250 Minimal left ideal, I15 Module, 4 Mohan-Kumar, 149 Monk polynomial, 142 Monic. monomorphism, 25 Monoid, 5 Morphism. 3 Multiplication, 14,200 Multiplicatively closed, 98 Multiplier (Schur). 268

N Nakayama, 137 Natural equivalence, 75 Natural map, 25 Natural transformation, 43 Naturality of connecting homomorphism, 173 Nielsen-Schreier, 269 Nobeling, 123 Noether normalization lemma, 144 Noetherian ring, 108 Normalized coboundary, 280 Normalized factor set, 280 Nullhomotopic, 177 Nunke-Rotman, 230 0

Objects in category, 3 Obstruction, 195 Ojanguren-Sridharan, I49 Operates without fixed points, 285 Opposite orientation, 18 Opposite ringROP, 16 Orientation, 17 Osofsky. 237,242

P pd (projective dimension), 233 PDer(G. A ) , 159 Perfect group, 279 Perfect ring, I14 Presentation. 275 Preadditive category, 6 Principal derivation. 159 Principal indecomposable module, I3 1 Product, 28 Projection. 29 Projective basis, 6-4 Projective cover, 75 Proiective dimension (pd), 233 Projective equivalence. 234 Projective module, 62 Projective representation, 281 Projective resolution of complex, 343 Projective resolution of module, 71 Proper essential extension, 72 Proper projective resolution, 345 Proper submodule, 24

375

Index Priifer ring, 127 Pullback. 51 Pure exact sequence, 94 Pure submodule, 94 Pushout. 41

Q Q = rationals Quasi-Frobenius ring, 130 Quasi-ordered set, 5 Quillen, 146, 147, 149 QuiUen-Suslin, 145, 148, 149 Quotient complex, 170 Quotient module, 25

R R = real numbers rD(R)(right global dimension), 236 R-sequence, 261,263 Radical (Jacobson), 136 Rank of free module, 58 Realizes operators, 152 Rees, 248 Regular local ring, 261 Regular ring, 262 Representation of ring, 23 Residue field, 259 Right derived functor, 184, 185 Right exact functor, 34,35 Right global dimension, 236

Shelah, 238 Short exact sequence, 27 Sign lemma, 320 Simple module, I IS Simplex, 19 Singular complex, 167 Small, 109 Small module, 82 Snake lemma, 174 Specker, 123 Spectral sequence, 313 Split exact sequence, 33 Split group extension, 152 stab((;. A), 159 Stabilizing automorphism, 158 Stably free. 139 Stallings, 297 Standard basis, 139 Standard resolution, 284 Standard simplex A,,. 17 Steinitz, 122 Stem cover, 279 Stem extension, 279 Strongly connected sequence of functors, 212 Subcomplex, 170 Submodule. 24 Subquotient, 314 Sum (direct sum), 28 sttpemuous. 75 Suslin. 141 Swan. 298 Syzygy. 233

S

S ( X ) = singular complex, 167 Schanuel, 92 Schreier transversal, 289 Schur, 281 Schur multiplier, 268 Schur-Zassenhaus, 293 Second filtration, 324 Second iterated homology, 328 Semidirect product, 152 Semihereditary ring, 127 Semiperfect ring, 135 Semisimple module, 115 Semisimple ring, I16 Serre, 141,253,262,298 Serre's problem, 138 Shapiro, 2%

T 3 x 3 lemma (9 lemma), 175 Tensor product of algebras, 16 Tensor product of complexes, 32 I Tensor product of modules, 9, 12 Third quadrant bicomplex, 322 Top (category of topological spaces), 4 Top element, 49 Tori, I 8 I Torsion-free module, 70, 128,224 Torsion module, 224 Torsion submodule, 128,224 Total complex, Tot(M), 319 Transversal, 152,289 Trivial module, 152 Trivial quasi-order, 40

376

Index

U UFD (unique factorization domain), 127 Unimoduiar column, 138 Unimodular column property, 138 Universal central extension, 279 Universal coefficient theorems, 228,229, 230,286 Universal mapping problem, 10

W

wD(R) (weak dimension),239 Watts, 77,78,79 Wedderburn, 116 Whitehead’s problem, 238

Y Yoke, 238 2

V Valuation ring, 127 Villamayor, 90 Virtual degree, 141 von Neumann regular ring, I19

Z = integers Z2(G,A ) , 154 Z,(A) (n-cycles), 169 Z ( R )(center of ring), 7 Zero differentiation, 170,300 Zero morphism, 7

Pure and Applied Mathematics A Series of Monographs and Textbooks Editors

Samuel Eilenberg and Hymen Bass Columbia University, N e w York

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