ring theory – louis h rowen a course in ring theory advances in ring theory – j l s chen n q ding advances in ring theory – r v kadison j r ringrose commutative ring t

After a brief treatment of those techniques of homology and cohomology needed for rings, the book turns specifically to certain types of rings that have been objects of intense inv[r]

Ramat Gan Israel

ACADEMIC PRESS, INC

Harcourt Brace Jovanovich, Publishers

Boston San Diego New York

Berkeley London Sydney

Tokyo Toronto

Copyright 0 1988 by Academic Press, Inc

All rights reserved

No part of this publication may be reproduced or transmitted in any form or by any means, electronic

or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher

ACADEMIC PRESS, INC

1250 Sixth Avenue, San Diego, CA 92101

United Kingdom Edition published by

ACADEMIC PRESS, INC (LONDON) LTD 24-28 Oval Road, London NW1 7DX

Library of Congress Cataloging-in-Publication Data Rowen, Louis Halle

Ring theory

(Pure and applied mathematics; v 127-128)

Includes bibliographies and indexes

1 Rings (Algebra) I Title 11 Series: Pure and

applied mathematics (Academic Press);

CQA2471

127- 128 QA3.P8 VOI 127-128 510 s [512’.4] 87-14536 ISBN 0-12-599841-4 (v 1)

ISBN 0-12-599842-2 (v 2)

8 8 8 9 9 0 9 1 9 8 7 6 5 4 3 2 1

Printed in the United States of America

rings with polynomial identity (Chapter 6), division rings, and, more generally,

(finite dimensional) central simple algebras (Chapter 7), group rings (8.1 and

8.2), and enveloping algebras (8.3) Since the group ring contains all possible

information about group representations and the enveloping algebra contains all possible information about Lie representations, these two types of rings have been unified in Chapter 8 under the heading, “Rings of Representation

Theory.” These rings often are Noetherian, and the last part of Chapter 8 deals

with general aspects of Noetherian theory that have been used in studying these rings (and also with Hopf algebras, another generalization)

Although the general setup is the same as in Volume I, the pace of the text is

somewhat faster than in the first volume Indeed, the subjects of each chapter have had books devoted exclusively to them in the past, and to deal adequately with current research in any of these subjects would require several volumes The goal in this volume has been to present a unified treatment to the reader and to give the reader some idea, at least, of research in ring theory of the 1980s In certain fast-moving areas, such as polycyclic group algebras and primitive ideals of enveloping algebras, only a brief sketch has been given of recent advances; fortunately, excellent surveys in both subjects have been published within the last five years by Farkas and Passman (for the former); and by Borho, Joseph, and Rentschler (for the latter)

xi

Table of Principal Notation

xiv Table of Principal Notation for Volume I1

5 Homology and

Cohomology

One of the major tools of algebra is homology theory and its dual, cohomology theory Although rooted in algebraic topology, it has applica-

tions in virtually every aspect of algebra, as explained in the classic book of

Cartan and Eilenberg Our goal here is to develop enough of the general theory to obtain its main applications to rings, i.e., projective resolutions, the

homological dimensions, the functors F u ~ and d, and the cohomology groups (which are needed to study division rings)

Duality plays an important role in category theory, so we often wish to replace R - A W by a self-dual category, i.e., we use only those properties of

to consider abelian categories with “enough” projectives and injectives, i.e., for each object M there is an epic P -, M for P projective and a monic

M -+ E for E injective

85.0 Preliminaries About Diagrams

Certain easy technical lemmas concerning diagrams are used repeatedly in homology and cohomology For the sake of further reference we collect them here We shall assume all of the diagrams are commutative, with all

1

coker y yielding the commutative diagram

B

0 + ker p _c* M - N - coker p-0

0 L_* ker y M *N - coker y - 0

If f and g are isomorphisms thenf”and 9 are also isomorphisms

Proofi If Bx = 0 then yfx = gpx = 0 implying f(ker /3) G ker y On the other hand, f(PM) = yfM E yM”, so g induces a map S:N//3M -+ N”/yM”

The commutativity of the ensuing diagram is obvious The last assertion is a special case of the five lemma To wit, adding 0 at the right so that S is in the middle of the appropriate diagram (with f at the left side) shows is an isomorphism; adding, instead, 0 at the left so that 7 is in the middle shows

?is an isomorphism Q.E.D

g5.0 Preliminaries About Diagrams

( 3 ) The snake lemma Given

3

(i) ker a 4 ker p -+ ker y is exact, where the maps are obtained as in (2)

(ii) Iff is epic then there is an exact sequence

ker a + ker p + ker y 4 coker a 4 coker /? -+ coker y

where a is dejned in the proof, and the other maps are in (2)

Proofi

(i) The composition is certainly 0 On the other hand, if z E ker p and fz = 0 then z = f ‘ x ’ for some x’ in M’; then 0 = pz = g’ax‘, so ax’ = 0 and x’ E ker a

as desired

(ii) Define 8 as follows: Given x” in kery we take x in M with f x = x ” ;

then gpx = 0 so px E kerg = g’N’, yielding y’ in N ‘ with g’y’ = px, and we take ax” to be y‘ + aM’ To check a is well-defined, suppose we took instead

x1 with fxl = x“ and we took y’, with g’y; = px, Then f(xl - x) = 0 so

x , - x E f‘M’, and thus

g‘(y‘, - y,) = p(x, - x) E Pf’M’ = g’aM’

Hence y; - y, E aM’ since g’ is monic

Having defined a, which is surely a map, we note that the hypothesis is self-dual, so it is enough to show ker a 4 ker p -+ ker y % coker a is exact;

in view of (i) we need only show ker p-+ ker y 3 coker a is exact We shall use

throughout the notation of the previous paragraph First note that if

x” E j(ker p) then in the definition of ax‘’ above we could take x in ker p, so

g’y’ = fix = 0 and thus y’ = 0, proving df’ = 0 On the other hand, if dx” = 0

then y’ E aM’ so writing y’ = ax‘ we see x - f’x’ E ker and T ( x - f x ) =

4 Homology and Cohomology

keru -P kero + kery -+ cokcl= 0

is exact and f is monic, yielding 0 -+ A’ t A + A” 4 0 as desired Q.E.D

The horseshoe lemma If P’ and P” are projective then the following diagram can be ‘tfilled in” where p: P‘ + P‘ @ P“ is the canonical monic and n: P’ 0 P” P P” is the projection:

$5.0 Preliminaries About Diagrams 5

(You get the horseshoe lying on its side by erasing the dotted lines and the maps to and from 0.)

Pro03 There is a map h: P” -+ M lifting f , i.e., P” = f h Defining B by

P(X’,X’’) = f ’P’x’ + hx” one sees at once that the top right square is commu-

tative Moreover, fP(x’, x”) = fs‘b’x‘ + P”x” = 0 + b”x(x‘, x”) so the bottom right square is commutative Taking K = kerP we can draw the upper row

by means of the nine lemma (turned on its side) Q.E.D

( 6 ) Turning the corner Given

Proofi kerf = iK” = ihK since h is epic; moreover, since i is monic we have ker ih = ker h = gK’ Q.E.D

(7) The pullback and pushout have been introduced, respectively, as inverse

ing some obvious properties First, given

6 Homology pnd Cobontow

n,: lp; -+ Ai be the restrictions of the projection maps from A, 0 A , to Ai Note

ker n, = { (0, a2): a2 E ker g } x ker g, yielding a commutative diagram

we define the pushout P’ = (A‘, 63 A ; ) / N where N = { f‘b, -g’b): b E B’}, and

let p1,p2 be the compositions of the canonical injections into A’, Q A; with

is exact with V T = 0 then z = ILa for some ts: P + A”: (Indeed, replace A by

$A“ Since ZP c ker tp E $A” we see $ lifts.)

Remark: Although for concreteness we have framed the proofs in the cate- gory R-Aod, the proofs could be given in any abelian category Indeed, the five lemma was already proven in general (exercise 1.4.21), and it is quite straightforward to modify the other diagram-chasing lemmas accordingly

The general pullback for abelian categories was described in exercise 1.4.22, and the general pushout is its dual The facts concerning projectives were proved directly from the lifting property, which is categorical Since “abelian”

is self-dual we shall also claim the dual results of any theorems based on these facts

55.1 Resolutions and Projective and Injective Dimension 7 85.1 Resolutions and Projective and Injective Dimension

In this section we introduce two new important dimensions, the projective

dimension and its dual, the injective dimension It turns out these dimensions

have immediate application to ring theory, and we shall use these to develop the motivation of the subject

Projective Resolutions

Definition 5.1.1:

sequence of the form

A projective resolution 9 of a module M is an exact

-+ P, + P,-, -+ ' + PI -+ Po -$ M -+O

where each 8 is projective We say 9 is f.g (usually called Jinite in the

literature) if each & is f.g.; 9' is free if each 8 is free The smallest n for

which P, = 0 is called the length of the resolution 9

Several immediate observations are available

Remark 5.1.2:

(i) Every module M has a free resolution Indeed, take an epic fo: Fo + M

with F, free and, inductively, given fi- c-, -+ &-, take an epic fi: Fi +

kerf;- We could view fi: & + 6- 1; then JF, = ker A- so the sequence

f2 fl

+ F, + Fl + F, + M -+ 0 is exact

(ii) If M is f.g then we could take Fo f.g as well as free If R is left

Noetherian then kerf, I F, is f.g., and, continuing by induction, we may assume each 4 is f.g Thus every f.g module over a left Noetherian ring has an

f.g free resolution

In order to study a resolution we shall repeatedly use the observation that

we can ''cut'' an exact sequence -, M" M -% M’ -+ at g to produce exact sequences

K = kerf, is called the n-th syzygy of M If 9 has length n + 1 then K % P,+

is projective Conversely, if K is projective then we have produced a new

8 Homology and Cobolnology

projective resolution of length n + 1, which we shall compare to 9 by means

of the next result

Proposition 5.1.3: (Generalized Schanuel’s lemma) Given exact sequences

Proofi We cut (1) at f, to get

0 + K + -+ Pz + PI + kers + 0 (3)

and cut (2) at g1 to get

Let M‘ = Pb 0 kers z Po 0 kerd by Schanuel’s lemma (2.8.26) applied to sequences (4) and (6) Then we can modify (3), ( 5 ) to get

0 -+ K + * + P2 + PI 0 Pb

0 + K ’ + - +

M ’ + 0

+ p ; @ Po ’IQ1, M ’ + O

and by induction on the length we get the desired conclusion Q.E.D

Definition 5.1.4: Two modules N, N ’ are projectively equivalent if there are

projectives P, P’ such that N @ P z N‘ 0 P’; if, furthermore, P, P’ are f.g free then N, N‘ are stably equivalent N is stably free if N is stably equivalent

to a free module, i.e., N 0 R(”) is free for suitable n

Note that any module M projectively equivalent to a projective module

is itself projective since M is a summand of a projective module In partic-

ular, all stably free modules are projective (Compare to remark 2.8.4’ and

exercise 10.)

$5.1 Resolutions and Projective and Injective Dimension 9

Corollary 5.1.5: For any n the n-th syzygies of any two resofutions of M are

are stably equivalent

We are ready to open an important new dimension in module theory

Definition 5.2.6: M has projective dimension n (written pd(M) = n) if M

has an f.g projective resolution of length n, with n minimal such In the

literature the projective dimension is also called the homological dimension

Remark 5.2.7: pd(M) = 0 iff M is projective

Proposition 5.1.8: The following are equivalent:

(i) pd(M) I n + 1

(ii) The n-th syzygy of any projective resolution of M is projective

(iii) Any projective resolution d of M can be cut at f to form a projective

Proof: (i) 3 (ii) By corollary 5.1.5 the n-th projective syzygy is projectively equivalent to a projective module and thus is itself projective; (ii) 3 (iii) clear; (iii) e- (i) by definition Q.E.D

Proposition 5.1.9: The following are equivalent:

(i) M has a f.g free resolution of length I n + 1

(ii) The n-th syzygy of any f.g free resolution is stably free

(iii) Any f.g free resolution can be cut and modijied to form a f.g free reso- lution of length n + 1

Proof: (iii)

and 0 -+ P -+ F -+ F' is exact with F, F' free then the sequence

(i) 3 (ii) is clear To see (ii) - (iii) note that if P @ R'"' is free

0 -+ P @ R(")-+ F @ R("'-+ F'

is exact Q.E.D

Elementary Properties of Projective Dimension

Let us now compare projective dimensions of modules in an exact sequence This can be done elegantly using the functor Ext of 55.2 or in a simple but

ad hoc fashion by means of the following result:

Homology and Cohomology

10

Proposition 5.1.10: Suppose 0 M' + M + M" -+ 0 is exact Given projec-

tive resolutions Y, 9" of M', M" respectively, we can build a projective resolution 9 of M where each P, = Pk 0 P i and, furthermore, letting

K b , K l denote the n-th syzyyies we have the following commutative diagram

for each n, obtained b y cutting 9',% and 9" at f:,f., and f:,

$5.1 Resolutions and Projective and Injective Dimension

the resolutions 9', 9" at f b, f :: yields the diagram

Corollary 5.1.12: Suppose 0 + M‘ + M + M” + 0 is exact Suppose 9’,P

an exact sequence

O + K L - , + K,- -+ P i - -+ Pi-2 -+ + P i + M” + 0

K , - , is projective

n” 5 max{n’, n} + 1 Furthermore, n“ = n if n‘ < n, and n” = n’ + 1 if n’ > n

Proofi The first assertion is obtained by “turning the corner” in (7) Every-

thing else follows by choosing m properly Taking m = max{n’,n} we see

KL-l and K,,-l are projective so n 5 rn + 1 = max{n’,n} + 1

have n” < n and thus n” = n by corollary 5.1.1 1

I f n‘ > n then taking rn = n’ we see K,- is projective but K L - is not,

so n“ > m by proposition 5.1.8, i.e., n” 2 n‘ + 1; hence n” = n’ + 1 since

n” 5 max{n’, n} + 1 Q.E.D

Summary 5.1.13: Suppose 0 + M’ + M + M“ + 0 is exact Recapitulating

following possibilities:

Cuse I n < n’ Then n” = n‘ + 1

Case 11 n = n Then n“ < n + 1

Case I l l n > n’ Then n = n

85.1 Resolutions and Projective and Injective Dimension 13

These formulas could be proved directly using induction, cf., Kaplansky

[72B, p 1691; we shall give a less ad hoc proof in $5.2

Example 5.1.14: Suppose f: F + M" is epic with F free and M" not pro-

jective Then pd(F) = 0 so pd(M") = pd(kerf) + 1 In particular, if Rr is

not projective then pd(Rr) = pd(Ann r) + 1

The (Left) Global Dimension

Definition 5.1.15: The global dimension

gl dim(R) = sup{ pd(M): M E R-Aud}

Example 5.1.16:

(i) R is semisimple Artinian iff every module is projective, iff gl dim(R) = 0 (ii) If R is hereditary then gl.dim(R) I 1 (Indeed, every submodule of a projective is projective, so the first syzygy is always projective, implying pd(M) I 1 for any module M.)

There is another way of viewing the global dimension, by dualizing everything

Definition 5.1.17: An injective resolution iE of a module M is an exact sequence 0 + M 5 E, k El -+ with each Ei injective; the n-th cosyzygy is

coker f, IE has length n if Ei = 0 for all i 2 n, with n minimal such The injective

dimension id(M) (if it exists) is the smallest n for which M has an injective resolution of length n

Dually to proposition 5.1.8 we see id(M) = n iff every injective resolution can be cut to a resolution of length n

In the next section we shall see gl dim(R) = sup{id(M): M E R - A d } ,

which is easy to believe since it holds for R semisimple Artinian (gl dim 0)

or hereditary (gl dim 1) Kirkland-Kuzmanovich [87] have constructed

examples of Noetherian R with gl dim R < gl dim R/Nil(R)

Our present interest is to see the connection between R and R[I] for a commuting indeterminate 1 If R is a field then R[A] is a PID and thus

hereditary, so gl dim R[I] = gl dim(R) + 1 Actually, this relation holds much more generally, and we shall verify it for the skew polynomial ring R[;I;o] where R is arbitrary and o is an automorphism of R (In particular,

14 Homology and Cohomology

one could take a = 1 and obtain the ordinary polynomial ring.) To see this

we need to introduce some new module structure

If M is an R-module then we define M [ I ; a ] to be R[I] 8, M as an

additive group, i.e., the elements of M[I;a] have the unique form c I i x i

where xi E M M [ I ; a] is made into RCI; a]-module via scalar multiplication

Remark 5.2.28: Since R [ I ; a] is free over itself we see that F [ I ; a] is a free

R [ I ; a]-module for any free R-module F; consequently, any projective (resp

free) resolution 0 + P, + * -+ Po -+ M -+ 0 of M in R-Aod yields a projec-

tive (resp free) resolution 0 + P,[J.; a] -+ * * -+ Po[A; a] -+ MIA; 03 + 0 in

R [ I ; a]-Aod (This is really a fact about graded modules.) In particular,

pdRI1;,,J M [ I ; a] 5 pd, M; in fact, equality holds by exercise 2

On the other hand, any R[n; a]-module M is viewed naturally as R-module

by forgetting 1, and then we can form M [ I ; a ] as above To distinguish

between the two R [ I ; a]-module structures we write Ax to denote the original

product in M, and 1 - x to denote the new product in M[A; a] We need one more module Define a M to be the set of formal elements { a x x E M }

made into a module under the operations a x 1 + a x 2 = a ( x l + x 2 ) (so that

a M z M as abelian groups) and rax = ax' where x' is the product ( a - l r ) x

$5.1 Resolutiols and Projeetive and Injective Dimension 15

proving g is indeed a map Moreover, we can rewrite g ( 1 I ' axi) as

1 - Axo + xi, Ii (Ax, - xi- l) Hence 1 A' - ox, E ker g iff xo = 0 and Ax, =

We are ready for the first inequality

Proposition 5.1.19: gl.dim R [ I ; a ] I gl dim R + 1 In fact, pdRIA;al M I

pd, o M + 1 for any RIA; o]-module M

Proof: Let n = gl dim R For any R [ I ; u]-module M we have

P~R[,;,] M[n; D l I pd, M 5 n,

and pd,,,,,,(aM)[I;o] I pd,oM I n Applying corollary 5.1.12 to lemma 5.1.18'thus shows pd,[,;,,M I n + 1 Q.E.D

Our next goal is to see, in fact, that equality holds

Proposition 5.1.20: pd(@,,, Mi) = sup{pd(MJ: i E I}

Proof: Write n, = pd(Mi) and n = pd(M); take projective resolutions 3 of

Mi of respective lengths n,, and let 9 = @?, i.e., 9 is formed by taking the direct sum of the respective terms Then, clearly, length (9) = sup(n,), proving n I sup(n,) On the other hand, the (n - 1)-st syzygy of 9 is projec- tive, so the (n - 1)-st syzygy of each is a summand of a projective and thus projective, proving each n, 5 n Q.E.D

Corollary 5.1.21: pd(M")) = pd(M) for any index set I

Proposition 5.1.22: Suppose 9: R + T is a ring homomorphism Viewing any

pdR M I pd, M + pdR T

16 Homology and Cohomology

Proofi Induction on m = pd, M Write n = pd, M and t = pd, T; we

want to show n I m + t For m = 0 we have M projective as T-module

so M 0 M ’ = F for some free T-module F This is also a direct sum as

R-modules, so applying proposition 5.1.20 twice yields n I pd, F = t, as desired

For m > 0 take 0 + M ‘ + F + M -+ 0 exact where F is free Then pd, M‘ =

m - 1 by summary 5.1.13, so pd, M’ I (m - 1) + t by induction Thus

n I 1 + maxtpd, F, pd, M ’ } I 1 + max{t,(rn - 1) + t } = 1 + m - 1 + t = m + t

proving the first assertion To prove the second assertion we note the same

induction argument works if we have the case m = 1 Q.E.D

Remark 5.1.23: Suppose Ra = aR Any M in R - A u d has the submodule

projective) R-module then F/aF is a free (resp projective) RIRa-module

Theorem 5.1.24: (“Change of rings”) Suppose a E R is a regular noninvert- ible element, and Ra = aR If M is an RIRa-module with pd,,,, M = n < 00

then pd, M = n + 1 In particular, gl.dim R 2 gl.dim(R/Ra) + 1 provided

gl.dim R/Ra is jinite

Proof: Write R = R / R a First we show pd, R = 1 Right multiplication by

a gives an isomorphism R + Ra, so applying summary 5.1.13 to the exact sequence 0 + Ra -+ R -+ R + 0 we see pd, R I 1 But R is not projective (for otherwise Ra would be a summand of R, implying Ra = Re for an

idempotent e # 0, and then a(l - e) = 0, contrary to a regular.) Thus

By proposition 5.1.22 we have pd, M I n + 1, and we need to show equality for the case n 5 1 Assume M is a counterexample; we may assume

pd, M I 1 By definition a M = 0 Hence M cannot be a submodule of a free

R-module In particular, M is not projective so pd, M = 1 Hence n = 1

(since M is a counterexample) Any exact sequence of R-modules

55.1 Resolutions and Projective and Injective Dimension 17

Applyingexample5.1.14 to(lO),(l Ifwehavepd, M‘=Oandpd, M’/aF=O But then M‘/aM‘ and M’laF are projective R-modules In particular, (11) splits, implying aF/aM’ is a projective R-module Thus M % F / M ’ z aF/aM‘

is projective Q.E.D

Corollary 5.1.25: gl dim R[A; a ] = gl dim R + 1

Proof: Apply the theorem, with R[A;a] and A replacing, respectively,

R and a Q.E.D

(For gl dim R = co a separate argument is needed, which is an instant application of exercise 4 In fact, infinite gl dim has led to serious errors in the literature, cf., exercise 16 and McConnell[77].) It is also useful to have a localization result

Proposition 5.1.26: If S is a le) denominator set for R and T = S-’R then

pd, S-’M I pd, M for M in R - A o d In particular, gl dim T I gl dim R

Proofi Since the localization functor is exact (theorem 3.1.20), any projec- tive resolution of M can be localized to a projective resolution of S-’M as T-module Q.E.D

There is a right-handed version of projective dimension and thus of

gl dim, which also is 0 if gl dim R = 0 since “semisimple Artinian” is left- right symmetric However, for gl dim R = 1 the symmetry fails because left hereditary rings need not be right hereditary, and Jategaonkar’s example

in $2.1 can be used to find rings of arbitrary left and right gl.dim 2 1

A thorough discussion of global dimension and its peculiar connection to the continuum hypothesis can be found in Osofsky [73B]

Stably Free and FFR

Definition 5.2.27: A module M has FFR (of length n) if M has an f.g

free resolution of length n (In particular M is f.g.)

Our interest in FFR is derived from the next result

Lemma 5.1.28: If P is projective with FFR then P is stably free

18 Homology and Cohomology

f" fi

Proof: Take an f.g free resolution 0 -+ F, -+ F,- -+ -+ F, -+ Fo 5 P - + 0

Since E splits we can write Fo = P 0 P' with flFl = ker E = P', so P' has

a f.g free resolution 0 -, F, + F,- -+ F, -+ P’ +O of length n - 1 By

induction P' is stably free; write P' 0 R(") w F for F free Then P 0 F

P @ P ' @ R(") w Fo 0 R(") is f.g free, proving P is stably free Q.E.D Conversely, we have

Lemma 5.1.29:

pi stably free then Zf M M has has an f.g projective resolution FFR of length I n + 1 of length n, with each

Proofi Take f.g free F, F' such that Po Q3 F' w F If n = 0 then M x Po so

0 + F' -+ F -+ M -+ 0 is the desired sequence In general, we break the resolu-

tion O-+P,-+ -+Po-+M-+O at E to get O-+Pn+- -+P,-+Pl+kerf -+O

and 0 -+ kerf + Po + M -+ 0 By induction kerf has an f.g free resolution

0 -+ F, -+ * * + F, 4 kerf -+ 0; piecing back together yields

Since z[pi] = [@pi] we see every element of K,(R) can be written in the form [PI - [Q] for P, Q f.g projective

Remark 5.1.30': P is stably free iff [PI = [Rlk for some k (Proof (3)

If P 0 R("') w R(") then [PI = [R]"-" (t) If [PI = [RIk then P 0 Q % Rtk) 0 Q for a suitable fag projective module Q; writing Q 0 Q' z R(m) we see P 0 R(m) x R("+lr).)

In view of remark 5.1.30' we see every projective is stably free iff K o ( R ) = ([R]), and we shall use this characterization from now on

Proposition 5.1.31: Suppose R is left Noetherian and gl.dimR < 00

K o ( R ) = <[R]) iff every f.g projective R-module has FFR

Proofi (a) by lemma 5.1.29 (e) by lemma 5.1.28 Q.E.D

$5.1 Resolutiom and Projective and Injective Dimension 19

Our principal interest here is when K,(R) = ([R]) One way of proving this is to show that every f.g module has FFR Thus we might hope for an analog of corollary 5.1.25, to tell us that K,(R[I]) = ([R[A]]) if K,(R) = ([R]) Unfortunately, we lack an f.g analog of proposition 5.1.22, so we start over again

Proposition 5.1.32: Suppose R is left Noetherian and gl.dimR < co If

K,(R) = ([R]) then Ko(S-'R) = ([S-'R]) for any left denominator set S

sf R

Proofi By lemma 5.1.28 it suffices to show every f.g projective S-'R-module

P has FFR P is naturally an R-module; take an f.g R-submodule M with

S-'M maximal If S-'M < P then taking x in P - S-'M we would have

S-'(M + Rx) > K I M , contrary to choice of M; thus S-'M = P But now take an FFR for M Tensoring each term by S-'R yields an FFR for S-'M = P Q.E.D

This argument actually yields an epic K , ( R ) -+ Ko(S-'R), cf., exercise 12

Proposition 5.1.33: Suppose R is left Noetherian and 0 + M' + M -+ M" -+ 0

is exact If two of the modules have FFR then so does the third

Proof: In view of remark 5.1.2(ii) there are f.g free resolutions P , P " (of possibly infinite length) of M', M", and we form the f.g free resolution 9 of

M as in proposition 5.1.10 By lemma 5.1.29 it suffices to find some n for which the n-th syzygies Kb, K,,, and K i are stably free Recall 0 -+ K ; -+

K,, -+ K i -+ 0 is exact If M', M" have FFR then we may assume 9',9'' are f.g., so for large enough n we get KL = K i = 0 and thus K,, = 0 Therefore,

we may assume one of the two modules with FFR is M Now the generalized

Schanuel lemma implies K , is stably free for large enough n But M' or M" has FFR so for n large enough we may assume K ; = 0 or K i = 0; thus K,, is isomorphic to the other syzygy which is thus stably free Q.E.D

Theorem 5.1.34: (Serre's theorem) Suppose R is a ldt Noetherian N-graded ring, with gl dim R < 00 If K,(R,) = ([R,]) then K , ( R ) = ([R])

Before presenting the proof of this important result, we should note the

same proof shows K , ( R ) % K,(R,) This result will be generalized further in

appendix A We follow Bass [68B] and start by noting various properties

20 Homology and Cohomology

of graded modules, of independent interest Let R+ =

R Then R ,

R,, an ideal of R/R+ We consider the category R-9r-Jtibd of definition 1.9.1

Remark 5.1.34‘: If M E R-Y#-Mod and R+ M = M then M = 0 (Indeed,

otherwise, take n minimal such that M, # 0 Then M n n R + M = 0, contradiction.)

Remark 5.1.34”: If a projective module P happens to be in R-Yh-MoA

then there is an epic F + P in R - 9 r - M ~ d for F a suitable graded free

G(R)-module (cf., remark 1.9.7), and the epic splits in R-Yt-Aod by remark 1.9.6 In view of remark 1.9.5 we see if p.d M = t < cc and M is

graded then M has a projective resolution of length t in R-%-Aod

In view of remark 5.1.34” there is no ambiguity in talking about the graded f.g projective R-modules, which we call Y r - f i r ~ j ( R ) , viewed as a full sub-

category of R-%-Aud

Claim 1:

Proof of Claim 1: Take the functor G = R, OR-: R-Yt-Aud -+ R,-Ya-

(graded) for every P in Yr-fiwj(R) Let Q = GP x P/R+P by example 1.7.21’; since Q is projective the epic P 4 Q splits in R,-Aod, and thus in R0-Ya-&~d

by remark 1.9.6 Hence there is a graded monic f: Q 4 P Now FGP =

R ORo Q so we can define a graded map cp: FGP + P such that cp(r 0 x) = r f x

for r in R and x in Q Since G right exact and Gcp is an isomorphism we see G(coker cp) = 0, i.e., coker cp = R, coker cp, implying coker cp = 0 by remark 5.1.34’ But then cp is epic and thus split, so G(ker cp) = 0, likewise implying ker cp = 0 Thus cp is an isomorphism, proving claim 1

The next part of the proof of the theorem involves two clever tricks, the first of which is very cheap Grade R[L] by putting R[IL], = RiL“-i Then R [ I ] , = R,, so R is replaced by the larger graded ring R[I] The second trick is a “homogenization map” @d, given by tjdr = rid-" for any r

in Rn This defines the map tjd: OnsdRn -+ R[I],, so t+bd homogenizes all elements of degree I d Now define the functor H : R[I]-Yt-9imod .+

R - 9 i m o d given by HM = M/(1 - I ) M , i.e., we specialize I to 1 Then H is

given by R where we view R as the R[I]-module R[I]/R[I](l - A)

as in example 1.7.21’ In particular, H is right exact

85.1 Resolutions and Projective and Injective Dimension 21 Claim 2:

For every f.g R-module N there is an f.g graded R[I]-module M

Proof of Claim 2: Write N z R'"'/K where K = If=, Rxi Writing x i =

( x i l , , xi,,) in R'"), we take d greater than the maximum of the degrees

of all the xij, and define x f = ($&lr ,$&in) E (R[IId)(') Letting K' =

H(R[I]("))/HK' z R(")/K z N since H is right exact

Claim 3: H is exact

Proof of Claim 3: We need to show H is left exact, i.e., given a monic

ical map N -, H M = M/(1 - I ) M has kernel N n (1 - I ) M , so it suffices to show N n(1 - 1)M I ( 1 - 2)N Suppose x = E x , E M with (1 - A)x E N ,

where the x, are homogeneous xo = ((1 - A)x), E N , and xi - Axi- = (( 1 - 1 ) ~ ) ~ E N for each i 2 1, so by induction we see each x i E N Thus x E N ,

so (1 - A)x E (1 - I ) N as desired

Proof of the Theorem: Suppose N is an f.g R-module Write N = H M for

M E R[2]-%-A%d, by claim 2 R[A] is Noetherian of finite gl.dim, by proposition 5.1.19 Thus M has an f.g projective resolution

0 -+ P, -, 3 Po + M + 0,

which by remark 5.1.34" can be taken to be graded Then 0 + HP, -+ * -+

HP, -, N +O is a projective resolution of N by claim 3 On the other hand, R[i], = R, so claim 1 shows each P z R[I] &,Qi for suitable f.g

projective R,-modules Qi, which by hypothesis are all stably free Then

HP, z R OR0 Qi are stably free Thus K o ( R ) = ([R]) by proposition 5.1.31

Q.E.D

We now have quite a wide assortment of rings R with K o ( R ) = ( [ R ] )

Corollary 5.1.35: Suppose R is left Noetherian, gl dim R < 00 The hypothe-

(i) R[A; 01 for any monic CT: R -, R

(ii) R[I, ,At] for commuting indeterminates I,, , A,

In particular, if R is a PLID then every f.g projective R I I l , , I,]-module

is stably free

22 Homology and Cohomology

(proposition 3.5.2) of finite gl dim (corollary 5.1.25)

The hypothesis R left Noetherian of finite gl.dim could be weakened throughout to “Every f.g, R-module has an f.g projective resolution of

finite length,” without change in proof Such rings are called (homologically)

regular

0 Supplement: Quillen’s Theorem

One good theorem deserves another, and Quillen [73] proved a startling extension of Serre’s theorem Before stating Quillen’s theorem let us make another definition

Definition 5.1.36: A ringR is filtered if R has a chain of additive sub- groups R, 5 R, E - * * such that RiRj G Ri+j for all i, j and UieN Ri = R Note that this concept is not new to us; R is filtered iff R has a filtration over (Z, +) for which R(l) = 0 (Just take the new Ri to be the old R - i ) Thus example 1.8.14(i) and the results applicable to that example can be used here, and, in particular, we can form the associated graded ring G(R); also note that R, is a subring of R The terminology introduced here has become standard when applied to enveloping algebras, and the results to be given here are fundamental in the theory of enveloping algebras, cf., $8.4 Quillen’s Theorem: Suppose R is filtered, such that the graded ring G(R) is Noetherian and of finite gl dim If G(R) is pat as R,-module then the natural injection R , -+ R induces an isomorphism q: K,(R,) -+ K,(R)

Quillen actually proved a much stronger result, that the corresponding isomorphism also holds in the “higher” K-theories Our exposition (also, cf.,

the appendix and exercise 13) follows the very readable account of McConnell

[SS], which in turn relies on Roy [SS]

The main object is to develop the homological module theory of filtered rings to the same stage that we developed the module theory of graded rings

To this end we assume throughout this discussion that R is a filtered ring; we

85.1 Resolutions and Projective and Injective Dimension 23

say an R-module M is filtered if M has a chain of subgroups M , C M I E such that RiMj I Mi+j and UitN Mi = M R is itself filtered as R-module in the natural way; on the other hand, any R-module M can be trivially Jiltered

If M is a filtered R-module we define the associated graded module

G ( M ) = @(Mi+JMi), viewed naturally as G(R)-module by the rule

(Ti + Ri+l)(xj + Mj+1) = rixj + Mi+j+l

for ri in R, and xi in Mi Given x in Mi we shall write X for the corresponding

element of G ( M )

Any filtered map f: M -+ N gives rise naturally to a map C ( f ) : G(M) -+

G ( N ) , so G is a functor The point of strictly filtered maps lies in the following

“exactness” feature of this functor:

If M“ M 3 M’ is exact with f , g strictly filtered then

We say M is Jiltered-free if G ( M ) is graded free over G(R), cf., remark 1.9.7

Remark 5.1.37: Any filtered-free R-module is free, for if {Fi: i E I > is a base

of homogeneous elements of G(M) then {ei: i E I } is a base of M; in fact, writing ei E M,,,i, we have

Conversely, any graded free G(R)-module T gives rise to a filtered-free R-module F such that G ( F ) = (Proof (a) Clearly the above holds for

k = 0, and thus is easily seen to hold by induction for all k, by passing to

G(M) at each step; likewise, any dependence would be reflected in some Mk

and thus pass to G(M), (t) Let {Pi: i E I ) denote a base of F over G ( R ) and

define F = R(” Labeling a base of F as {ei: i E I ) use the above equation to

filter F ; then G(F) = F by inspection.)

Now that we want to see that our object is really “free”

Lemma 5.1.38: Suppose M is a Jiltered R-module

24 Homology and Cohomology

(i) l f F is a Jiltered-free module then for every graded map 3 G ( F ) -+ G ( M )

(ii) There exists a filtered-free module F and a strictly Jiltered epic F -+ M

(iii) lf P is Jiltered and G ( P ) is G(R)-projective then P is projective

there is a jiltered map f : F -+ M such that G ( f ) = f

Proof.-

(i) Let { e i : i E I} be a base of F taken as in remark 5.1.37, and define

is filtered

(ii) Let F be a graded-free G(R)-module, together with a graded epic

f F + G ( M ) By remark 5.1.37 we can write F in the form G(F), so by (i) we have a filtered epic fi F -+ M It remains to show f is strictly filtered, i.e.,

$6 -+ Mi is epic for each i; we do this by induction on i, noting it is clear

for i = 0 and thus follows for general i by applying proposition 2.1 1.15 to the

We need to show g is split, i.e gp = 1, for some p: P + F Note PJe- is

a summand of G ( P ) and thus is projective; hence the epic fi -+ Pi/e- splits But then Pi z pi - I 0 (pi/pl, - I), so we can build p inductively on i (i.e., piecing

together elpi- I 4 &IF,- and pi-, 4 to get pi c, fi) Q.E.D

Proposition 5.1.39: For any filtered module M we have pd, M I pd,(,, G(M)

Proof.- Let n = pdGo, G(M) There is nothing to prove unless n is finite, so

we proceed by induction on n Lemma 5.1.38(iii) gives the result for n = 0

For n > 0 take an exact sequence 0 -+ N -+ F -+ M -+ 0 where F is filtered- free and the maps are strictly filtered, cf., lemma 5.1.38(ii) Then we get an exact sequence 0 -+ G ( N ) -+ G ( F ) -+ G ( M ) 4 0 But pd,,,, G ( N ) 4 n - 1 by summary 5.1.13, so pd, N I n - 1 by induction, implying pd, M I n by summary 5.1.13 Q.E.D

55.1 Resolutions and Projective and lnjective Dimension 25

Corollary 5.1.40: ff R is a jiltered ring then gl dim R gl dim G(R)

Proofi Any R-module M can be filtered trivially, and then by the propo-

sition pd, M I pd,,,, G ( M ) I gl dim G(R) Thus gl dim R I gl dim G(R)

Q.E.D

Digression: These results actually yield more We could define the graded

pd of a module and graded gl.dim of a ring, in terms of graded projective resolutions Then we see at once the graded gl.dim of a graded ring equals its usual gl dim

We are ready for our watered-down version of Quillen’s theorem

Theorem 5.1.41:

@.dim ff Ko(R,) = ([Ro]) then K , ( R ) = ([RI)

Proofi R is Noetherian by corollary 3.5.32(i), and has finitely gl.dim by

corollary 5.1.40 We want to show that every f.g projective R-module P is stably free Viewing P as filtered trivially, we see G ( P ) is f.g over G(R) and

thus we can build a graded resolution of f.g graded free modules, as in

remark 5.1.34” By hypothesis the n-th syzygy K , is projective and is the

kernel of a graded map, so is graded projective By claim 1 (following remark 5.1.34”) there is a projective R,-module Qo with K , z G(R) OR,, Qo But Qo is stably free by hypothesis, so K , is stably free as G(R)-module Thus

K , is “graded” stably free by remark 5.1.34”, so just as in lemma 5.1.29 we have a graded FFR of G ( P ) having finite length Using lemma 5.1.38 we have

the resolution in the form 0 + G(F,) + G(F,- I ) + -+ G ( P ) + 0, and this

“pulls back” to an FFR 0 + F, + F,-, -+ * -+ P 4 0 in R-Aud, as needed

Euler Characteristic

The FFR property has an important tie to topology

Definition 5.1.42: Suppose R has IBN The Euler characteristic of a module

M with an f.g free resolution 0 f F, + + F, + M + 0 is defined as

x ( M ) = CY=,( - l)irank(&)

Remark 5.1.43: x ( M ) is independent of the FFR because of the generalized Schanuel lemma In particular, if 0 + F, + + F, + 0 is exact then

c( - l)i rank(4) = 0

26 Homology and Cobomology

Remark 5.1.44: Suppose S is a left denominator set of R and S-‘R also has IBN If an R-module has FFR then S-’M has FFR in S-’R-&O~, and

x(S-’M) = x ( M ) (Just localize the resolution, since the localization functor

is exact and preserves “free” and the rank.)

The name “Euler characteristic” comes from Euler’s observation that

#faces - #edges + #vertices of a simplicia1 complex is a topological invariant The standard reference on properties of the Euler characteristic

is Bass [76]

Theorem 5.1.45:

gl dim with K,(R) = ([R]) then R is a domain

(Walker [72]) If R is a prime l@t Noetherian ring of Jinite

Proofi By Goldie’s theorem R is a left order in a simple Artinian ring

S ’ R x M,,(D) We claim n = 1, which would prove R is a domain Indeed,

L = R n Mn(D)el, is a left ideal of R which is nonzero since M,,(D) is

an essential extension of R; thus S-’L 5 M,,(D)e,, so 0 < [S-’L:D] I n

On the other hand, take an FFR 0 + F, -+ -+ F,, -+ L -+ 0 Then each

S-’F, is a free S-’R-module so n2 divides [S-’F,:D] Hence remark 5.1.43

shows n2 divides [S-’L: D], a contradiction unless n = 1 Q.E.D

Walker’s theorem is more general, with “semiprime” replacing “prime”,

cf., exercise 14 However, the proof given here will be relevant when we

consider group rings later

0 Supplement: Serre’s Conjecture

One of the outstanding recent problems in algebra has been to verify

Serre’s conjecture: Every f.g projective module over the polynomial ring

R = F [ I , , , At] is free, for any field F This is trivial for n = 0 and is easy

for n = 1 since R is a PID (and so every submodule of a free module is free,

by corollary 2.8.16) Attempts to verify Serre’s conjecture for larger n spurred the development of algebraic K theory, a deep and useful subject which also ties in to results about division algebras; Serre’s conjecture was finally

established by Quillen [76] and Suslin [76], two masters of K-theory, although the solutions were (surprisingly) not so difficult Following Rotman [79B] we present Vaserstein’s simplification of Suslin’s proof

The solution is comprised of two parts:

(i) f.g projective modules are stably free (by corollary 5.1.35);

(ii) Stably free modules are free, which we consider now

55.1 Resolutions and Projective and Injective Dimension 27

The techniques are not homological, but instead involve linear algebra over

PIDs Thus this part of the proof lies in the realm of noncommutative

algebra, although the theorem itself is over a commutative ring We say a vector u = ( I l , ,rn) E R'") is unimodulur if cf= r,ri = 1 for suitable ri in R ;

Any strongly unimodular vector certainly is unimodular, and we are inter-

ested in the converse In the following discussion let e, , , en be a standard

base of R'")

Remark 5.1.46:

(i) v = (rl, , r n ) is unimodular iff there is an epic f : R ( " ) + R with

fu = 1 (Proof (-) Given r l , ,rk with x r i r i = 1 define f:R("' t R by

f ( x , , , x,) = E x i ( (e) 1 = f v = f ( C r i e i ) = C r i f e i , so take r: = fei.)

(ii) u is strongly unimodular iff there is an invertible transformation

T: R(") -, R'") for which Te, = u (seen at once by viewing T as a matrix)

Digression 5.1.47: Remark 5.1.47(i) enables us more generally to define an

element x of an arbitrary R-module M to be mimodular iff R x is a summand

of M isomorphic to R

We shall describe "stably free implies free" in terms of the unimodular

are strongly unimodular

Remark 5.1.48: Suppose P = R(")/Ru where v is a strongly unimodular

vector of R(") Then P z R("- " (Indeed, take an invertible transformation

T: R@) 4 R(") with Te, = u Then P = R(")/Ru = TR'"'/T(Re,) % T(R(")/Re,) z

-1

R(n-1)

Proposition 5.1.49: Suppose R has IBN (invariant base number) Every f.g

stably free R-module is free iff R has the unimodular vector property (Note:

'Yg." actually is supeJfluous by exercise 10.)

Proof: (a) Suppose v , is a unimodular vector Taking f:R("' + R epic with ful = 1 we note R'") z R 0 kerf since R is projective; by hypothesis

kerf is free with some base { u2, , v,} Thus we have an invertible transfor-

mation T: R'"' + R'") given by Te, = ui, corresponding to an invertible matrix whose first row is u l

(e) First assume P 0 R z R("), i.e., the canonical map f: R(") + P has kernel Ru z R for suitable v in R(") Since f splits we see by remark 5.1.46 that

28 Homology and Cohomology

v is unimodular and thus strongly modular, so P is free by remark 5.1.48 By

induction on m we now see that if P 0 R(m) z R(") then P is free Q.E.D

We see from proposition 5.1.49 (*) that local commutative rings have the unimodular vector property, since all projectives are free

Proposition 5.1.50: (Horrocks) Suppose C is a local commutative ring, and

u = ( fi, , f,) is a unimodular vector in C[A]'"' If some 1; is monic then u is strongly unimodular

Proof: Reordering the 1; we may assume f l is monic Suppose m = deg( f i )

We can use fl to cancel higher degree terms in the other entries; since this involves elementary transformations and thus affects neither unimodularity nor strong unimodularity, we may assume deg(f,) < m for all i > 1 For

m = 0 we have u E C(") is strongly unimodular Thus we may induct on

m > 0

For n = 1 the assertion is vacuous For n = 2 we have ( g 1 , g 2 ) in C[%]"'

such that figl + f z g 2 = 1, so is invertible For n 2 3, letting- denote the canonical image in the field c = (C/Jac(C)), we see V is

unimodular in c[A]("' But f, = 1" + is not invertible in c[A] since

m > 0, so # 0 for some i 2 2 Assume f, # 0 Then some coefficient of f2

is not in Jac(C) and thus is invertible If CIA]fl + C [ A ] f i contains a monic

polynomial of degree m - 1 then using elementary transformations we can change f3 to a monic of degree m - 1, and are done by induction Thus it remains to prove the next result Q.E.D

Lemma 5.1.51: (Suslin) Suppose f = A"' + xy= and g = bil"-j

in C[A] Then for any j -= rn the ideal C[i.]f + C [ 1 ] g has a polynomial h of degree s m - 1 and leading coeficient bj

Proof: Induction on j For j = 1 we take h = g In general, let A = C [ A ] f +

C[A]g and define g' = 1 g - b,f = xy=l b:Am-' E A, where b: = bi+ - blai

By induction b J - , is the leading coefficient of a polynomial h' in A having

degree rn - 1, But now h' + a j P 1 g has leading coefficient bj - bluj- +

a j - ,bl = bj, as desired Q.E.D

Next we need the following passage from C[1]'") to C [ 1 ]

g5.1 Resolutions and Projective and Injective Dimension 29

Lemma 5.1.52: (Suslin) Suppose u = (f,(A), , f,(A)) is a unimodular vector

with some f; monic Then, uiewiny u as 1 x n matrix, we have u = wA for some

A E GL(n, C [ l ] ) and w E Cfn)

Proof: Write G = GL(n,C[A]) Given any h in C[A] let u(h) denote the

vector (f;.(h)) E C[A](") (Thus u = u(I).) Clearly u(0) E C'"), and we shall prove the result by showing u E u(0)G Let I = (c E C : u E u(g + cAh)G for all g, h in

C [ A ] ) , an ideal of C If I = C we are done by taking g = A, c = 1, and

h = - 1 Thus we assume I # C and aim for a contradiction

Taking a maximal ideal P 2 I of C we work in Cp[A, p ] where p is another

indeterminate commuting with A and C By proposition 5.1.50 u(A + p) is the

first row of some matrix A = A ( I + p) in GL(n, Cp(A + p)), i.e., u(A + p) = e l A

Write A , for the image of A under p + 0, also an invertible matrix Then

B = A-'A, E GL(n, C,[A, p]) satisfies

u(A + p ) B = e , , A , = u (12)

View B = B(A,p) as a matrix whose entries are functions in A,p; and

likewise for B-' = B-'(A,p) Then B(I,O) = A i ' A , is the identity matrix,

so the coefficients for Ai for i > 0 must all be 0 There is some c 4 P such

that cB and cB-' are in C[A,p] It follows that B(A,cp)€ M,(C[A,p]) with

inverse B-'(A,cp), and (12) implies u = u(A + cp)B(A,cp) Specializing A -, g

Q.E.D

Theorem 5.1.53:

field F

Euery f.g projective FIAl, , A,]-module is free for any

Proof: Induction on t; E = 0 is obvious We need to verify the unimodular vector property Let u = (f;.(ll, , A,)) be a unimodular vector with fi # 0

By the Noether Normalization Theorem (cf., Lang [64B, p 2601) we have

some ct in F such that afi is monic over pi, , p, - where the pi are algebra- ically independent polynomials By Suslin's lemma au E GL(n, KIA1, ,A,])w

for some w in F[pl, , p, - , I Clearly, w is a unimodular vector and thus by

induction is strongly unimodular Hence uu is strongly unimodular, and so u

also is strongly unimodular, as desired Q.E.D

Of course, theorem 5.1.53 is the solution to Serre's conjecture This proof illustrates one of the features of the Russian school-translate a problem

to matrices and tackle it there Quillen's solution is more structural and also

30 Homology and Cohomology

has several very interesting ideas, cf., Quillen [%I An excellent treatment of Serre's conjecture is to be found in Lam [78B]

M Supplement: Cancellation and Its Consequences

Unfortunately, the Quillen-Suslin theorem does not generalize directly to noncommutative rings, in view of the fact that D[A,, I, J has nonfree projec- tives for any noncommutative division ring D (cf., exercise 9) Nevertheless, there are noncommutative results to be had, when the projective module is required to have sufficiently large rank Suslin [79] lists a host of interesting theorems, including the following:

Suppose R is f.g as a module over a central Noetherian ring C of classical Krull dimension n Let T be any ring obtained by localizing RCA,, .,,I,]

by a submonoid generated by various Ai Then any projective T-module

of rank > max( 1, n) has the form T B R M for a suitable R-module M

In particular, if D is a division algebra of dimension 22 = 4 over its center then every projective D[,ll, , I,]-module of rank 2 5 is free Actually,

by using theorem 3.5.72 we can obtain some rather sweeping results which imply, for example, that "big enough" f.g projective modules over any Noetherian ring R with K,(R) = ([R]) are free Recall s(R) from definition 3.5.71

Lemma 5.1.54: Suppose P 0 R z R(") and s(R) < n Then P is free

Proofi More precisely, P 0 Ru = R(") where u is unimodular, cf., the proof

of proposition 5.1.49 (F) Writing u = (a,, ,an) in R(") we thus have

1 Ra, = R, so the stable range condition yields ri in R for which

Factoring out Rv yields P = R'"'/Rv % R("-') Q.E.D

(This is sharp even for R commutative.)

Proposition 5.1.55:

which cannot be generated by k elements then P is free

Suppose s(R) = k If P is an f.g stably free module

gS.1 Resolutions and Projective and Injective Dimension 31

Proofi Write P Q R('") FZ R(") and induct on m, the case m = 1 given by

the lemma (since, clearly, n > k) Q.E.D

Corollary 5.1.56: If R is lefr Noetherian with K,(R) = ([R]), and if

K-dimR = k, then every f g projective which cannot be generated by k elements is free

Proof: Apply theorem 3.5.72' to the proposition Q.E.D

Projective modules of low rank need not be free, as we mentioned earlier, also, cf., $8.4 The following two properties obviously are related to the freeness of projectives:

Property I (Cancellation) I f M 0 R(") z N 0 R(") then does M z N? Property 2 Does a given module have R as a summand?

These properties hold under much more general conditions than the Quillen-Suslin theorem The exposition here follows Stafford [82] We recall the notation of the normalized reduced rank

&M, P ) = p ~ , ~ ( M / P ~ ~ / p ~ ~ ~ ( R / P ) where p is the reduced rank

Lemma 5.1.57: Suppose M , N are modules over a left Noetherian ring R, and

K is an f.g projective R-module such that @(K, 5) 2 @(M, 5) for given prime

ideals PI, , P, of R Given maps f: I< -+ M and f’: N + M we can find

g(M/(f + f ' h ) K , 5 ) = @ ( M / ( f K + f ' N ) , q ) f o r e a c h l s j s m

Proofi First assume m = 1 and let P = PI Since K is projective any

K

32 Homology and Cohomology

Thus we can assume that P = 0, i.e., R is prime left Noetherian and so an

order in a simple Artinian ringQ Thus we are given p ( M ) I p ( K ) and,

letting M ‘ = f K + f ‘ N we want to find h: K + N for which p ( ( f + f ’ h ) K ) =

p ( M ’ ) Since (f + f ’ h ) K I M‘ we can find h: K + N such that p ( ( f + f ’ h ) K )

is maximal Replacing f by f + f ’ h we may assume p ( f K ) 2 p ( ( f + f ’ h ) K )

for all h: K + N ; we want to show p ( f K ) = p ( M ’ ) , or by proposition 3.5.7

map h,: K -+ L such that hotkerf) # 0 Writing x = f ’ a for suitable a in N

we have a map h: K + N given by hy = (h,y)a Note f ’ h K = f‘((h,K)a) =

(h0K)f’a = h0Kx; thus p ( f ’ h K ) = p ( h 0 K x ) = p ( h 0 K ) > 0

We claim the contradiction p((f + f ’ h ) K ) > p ( f K ) or, equivalently,

p(ker(f + f ‘ h ) ) < p(ker f ) Indeed, if y E ker(f + f h) then fy = f hy E

fK n Lx = 0 so y E kerf Thus ker(f + f’h) I kerf On the other hand,

0 f p(ho(kerf)) = p(h,(kerf)x) = p ( ( f + f ” k e r f ) ,

proving p(ker(f + f’h)) = p(ker f ) - p((f + f’h)ker f ) < p(ker f ) , as desired

This contradiction arose from the assumption p(kerf) > 0, so we have p(ker f ) = 0

But now by hypothesis p ( M ) I p ( K ) = p ( f K ) I p ( M ’ ) I p ( M ) , so equal- ity holds and we are done for m = 1

In general we apply the same argument inductively Namely, we arrange

the primes Pl, , P,,, such that if i < j then pi $ 8 (by starting with the largest ones and working down) and assume by induction the result has been proved for m - 1 Let A = PI n n P,,, - 1 Now A $.Z P,,, by arrange- ment of the Pi, so the canonical image A of A in RIP, is nonzero But now we reduce to the case m = 1 by passing to RIP,,,, K/P,,,K, MfP,,,M, and AN/P,,,N since this has the same reduced rank as N/P,,,N Q.E.D

Recall b ( M , P ) and b ( M , S ) from definition 3.5.67, and define b ( M ) to

be b(M, J-Spec(R)) where J-Spec(R) is the set of prime, semiprimitive ideals

of R Remark 3.5.68(iii) shows b ( M ) I max {K-dim RIP + g ( M , P): P E J-

Spec(R))

Proposition 5.1.58: Suppose R is left Noetherian with n = K-dim R < CC,

and K is an f.g projective R-module for which g^(K,P) 2 K-dim R I P + 1

55.1 Resolutions and Projective and Injective Dimension 33 for all P in J-Spec(R) If b(R/f,K) I n for some fl: K -+ R then there are f2, ,fn+ , in Hom(K, R) such that b(R"+"/(f;)K) I n (viewing (h) as

(f,? ? f n + 1): K -+ R("+ I ) )

froofi We shall prove by induction on t that for all t I n + 1 there are

f i , ,f, for which b(R")/(f,)K) I n, where ( A ) = (fl, ,f,); the assertion is

given for t = 1 Let N = R(f)/(A)K By induction we may assume that we have f2, ,f, with b ( N ) I n; our first task is to find f,+l Let Y = { P E J-SpecfR): b(N, P ) = n or K-dim R I P = n) Y is finite by proposition 3.5.70

and corollary 3.5.45'

Let R,+, denote thesubmodule(0, , 0,R)of R"+')andview R ( ' k R('+')

via the first t coordinates Letting f be the composition ( J ) : K 4 R(') < R('+')

and letting f’: R,+ , -+ R('+" be the canonical inclusion, we have by lem-

ma 5.1.57 some map h: K -+ R, + for which

g^(R"+"/(f + f ' h ) K , P ) = ij(R"+"/(fK + f'R,+,,P)

= cj(R'"/fK, P ) = g^(N,P)

for all P in 9 We take f,+ , to be the composition K -+ R,+ -+ R and

It remains to show b(R('+')/fl<, P ) I n for all P in J-Spec(R) If P E Y we have

b(R"+"/TK, P ) I g(R('+l)/fK, P ) + K-dim(R/P) = g ( N , P ) + K-dim R I P I n

7=(L ,f,+J = f +f'h

(The last inequality is clear if g ( N , P ) = 0, and if g ( N , P ) > 1 then

y(N, P ) + K-dim R I P = b(N, P ) I n.) O n the other hand, if P $9' we use the exact sequence

0 - + Rf+l/(Rl+ I n ( f ~ + PR"+ I ) ) ) -+ ~ ( ' + l ) / ( f l Y + PR('+'))

-+ R(')/((L)K + PR'") -+ 0;

since ij(M, P ) I 1 for any cyclic module, the additivity of reduced rank yields

g^(R"+ "/TK, P ) I 1 + g^(N, P )

Thus b(R"+"/fK,P) 2 K-dim(R/P) + g ( N , P ) + 1 I b ( N , P ) + 1 5 n since

P $ X so we have proved b(R"+ "/fK) I n as desired Q.E.D

We are finally ready for our main result, followed by the promised applications

34 Homology and Cohomology

Theorem 5.1.59: Suppose R is a left Noetherian ring of K-dimn < a, and K

is an f.g projective R-module with g^(K,P) 2 K-dimR/P + 1 for all P in Spec(R) Given r E R and f: K + R for which R = f K + Rr, we can find some h': K + R such that R = (f + Fh')K, where ? denotes the right multiplication map by r

Proofi Let Pl, , P, be the primes of R for which K-dim R / e = n, cf.,

corollary 3.5.45' By lemma 5.1.57 there is h in Hom(X,R) for which g(R/( f + Fh)K, 5) = g(R/( fK + Rr), 6) = 0 by hypothesis Let fl = f + Fh

Note g(R/flK,P) I 1 for each P in Spec(R), since R/flK is cyclic Hence

proposition 5.1.58 gives us f 2 , , f n + in Hom(K, R) such that b(R("+ ')/Z?)I

n, where Z?=(fi)K ={(fix , , f + l ~ ) : ~ ~ K } Let el , , en+l be the standard base of R("+l), and write 7 = (A) Clearly Re, = flKe, + Rre, so

In particular, el = (X)x + 112: ai(ei + rirel) for suitable x in K and a, in R

Matching coefficients of e, shows

Corollary 51.60: (Serre's theorem generalized) Suppose R is a lejl Noetherian

ring, and K is an f.g projective R-module satisfying G(K, P) 2 K-dim RIP + 1

for each P in J-Spec(R) Then R is a summand of K

Proofi Taking f = 0 and r = 1 in the theorem, we have an epic K .+ R which splits since R is projective Q.E.D

$5.1 Resolutions and Projective and Injective Dimension 35 Corollary 5.1.61: (“Cancellation”) Hypotheses as in corollary 5.1.60, any iso- morphism a: K 0 R + M 0 R yields an isomorphism K -+ M

Proofi Let x,: M 0 R + R be the projection; then there are f l in Hom(K, R )

and f2 in Hom(R, R ) for which x 2 a ( k , r ) = f , k + f 2 r for all k in K and r

in R But f2 = F2 for some r2 in R and R = flK + Rr, since no is onto, so

theorem 5.1.59 provides h’: K -+ R for which R = (fl + F2h’)K In particular,

there is x , in K such that 1 - r2 = ( f l + Fh’)x, Define h: K 0 R + K 0 R by

0 0 R, so we have a natural chain of isomorphisms

This result should be compared to exercise 2.9.16 In certain instances the bounds can be lowered, e.g., polynomial rings over simple Noetherian rings

and Weyl algebras The interested reader should consult Stafford [77, Sec-

tions 5,6,7] for such results

As mentioned earlier, more sweeping results are to be found in Stafford [81] and in Coutinho [86] One of Coutinho’s results involves a sweeping gener-

alization of the Eisenbud-Evans “basic element” theorem Write M * for

Hom(M, R) An element x of M is called pre-basic if the left ideal { f x : f E

M * ) of R contains a nonzero ideal; x is basic if its image in the RIP-module

for any given x,, , x - ~ in M and P in Spec(R) there are x , E M and

f E M * such that 0 = f x , = = f x , - but f x , is regular modulo P

Coutinho [86, theorem 4.6.11 proves a “descent” for basic elements which

yields the conclusions of corollaries 5.1.60 and 5.1.61 under the weaker hypothesis that r-rk(K) 2 K-dim R + 1 His result is very strong in view

of the following:

(i) K is not required to be f.g

(ii) K is not required to be projective!