Commutative algebra

COMMUTATIVE

ALGE~RA

-, N~.GOPALAKMSHNAN OefMrtment of M.rhemetIP Urilver6/ty of

"oon.

"une 411 007. Imile

m. LTD.

OXONIAN PlUlSS New DeIJ,If,"

,."

TO MY GRANDPARENTS

© 1984.tcs. Gopalakrl,hllQn This book hal b.M ,ab8ld/ud by the Go.....nm.nt olllfdlD, thro",h the Nat/oltal Book Trwt, IndID.lor the b.lfejit ol,tad.II1,.

(4'-37/1983)

R,. 22.50

Pabli,hed by Oxonian Prel' Pvt. Ltd•• N.S6Co_",ht C/r.w. N.w D.lh/ 110001 ~ printed tJt Prabhat Pru'. M.erat.

II

r PREFACE

This book is intended as a textbook in Commutative Algebra. In a vast field like Commutative Algebra there is a variety of choice of topics for a textbook. However, the choice in the present book has been made with the following two aims. The first aim is to present basic results on Commutative Algebra necessary for elementary Algebraic Geometry. The second aim is to introduce homological methods in .Commutative Algebra. Some efforts have been made to make the book self-contained. Chapter I introduces projective modules and their elementary properties. It also deals with flat and faithfully flat modules. The second chapter introduces localisation and gives some applications of localisation to projective modules. The third chapter contains basic properties of Noetherian and Artinian rings. The fourth chapter introduces integral extensions and includes results such as the going up and' going down theorems, normalisation theorem and the finiteness of integral closure. The fifth chapter deals with valuation rings and Dedekind domains and ends with a proof of ramification formula in the finite separable case. The sixth chapter contains basic results on completion such as Artin Rees Lemma, Krull's intersection theorem and Hensel's lemma. The seventh chapter deals with Ext and Tor functors and the projective dimension. The eighth and the ninth chapters deal with dimension theory at some length. Homological properties of Cohen Macaulay rings and homological characterisation of regular local rings are given. We also give a proof of Cohen's structure theorem on complete local rings. The last chapter includes a proof of the Big C.M. module conjecture in the chp case (p> 0) and its consequences to other homological conjectures. The style of the book is maintained in the form oflecture notes. This will facilitate readers with some mathematical maturity to study the book on their own. Motivations ate given at a number

6

II

T t

vi

PREPACB

of places, examples follow every definition and exercises are given at the end of each section. I take this opportunity to thank Professor R. Sridharan, Dr. Amit Roy and Professor Avinash Sathaye for their advice and help in the preparation of this book. My thanks are also due to Professor P.K. Sharma and Dr. C. Musili for their useful suggestions. My thanks are due to the University Grants Commission for their assistance in the preparation of this book and to the National Book Trust, Government of India, for subsidising this book for the benefit of students. Suggestions for improvement of the book are welcome.

Pune, 1984

N.S.

GoPALAnIllHNAN

CONTENTS

Preface

v

Chapter I MODULES 1.1 Free modules 1.2 Projective modules 1.3 Tensor products 1.4 Fl•• modu1~s Chapter

II 2.1 2.2 2.3 2.4

Chapter HI

3.1 3.2 3.3 3.4 Chapter IV

Chapter

1 1 7 14 21

LOCAUSATION

30 30 38 43 53

Ideals Local rings Localisation Applications NOETHERIAN RINGS

63 63 67 76 81

Noetherian modules Primary decomposition Artinian modules Length of a module INTEGRAL EXTENSIONS

4.1 4.2 4.3 4.4

Integral clements Integral extensions Integrally closed domains Finiteness of integral closure

V

DEDEKIND DOMAINS

88 88 92 95 101 107 107

5.1 Valuation rings

..., (

......

,~~-j( l'~

-~.~...::,

I

I

I \

viii

..,.

CONTIlNTS

Chapter

I, f ,...

Chapter

\

Chapter

I'

1

1 \

-,,-

116 120

VI

COMPLETIONS

6.1 6.2 6.3 6.4

Filtered rings and modules Completion I-adic filtration Associated graded rings

130 130 136 143 149

VII HOMOLOGY

Chapter

VIII DIMENSION

8.1 8.2 8.3 8.4 8.5

Hilbert Samuel polynomial Krull dimension Dimension of algebras Depth Cohen-Macaulay modules

IX 9.1 9.2 9.3 9.4

Regular local rings Homological characterisation Normality conditions Complete local rings

Chapter X

REGULAR LOCAL RINGS

SOME CONJECTURES

10.1 Big Cohea-Macaulay modules conjecture 10.2 Intersection conjecture 10.3 Zero divisor conjecture

10.4 Bass' conjecture Index

"

"'~.",~.i,.r .........

191 191 198 206 212 223 232 232 237 245 249 260 260 266 268 272 287

.!

CHAPTER I

MODULES

156 156 165 173

7.1 Complexes 7.2 Derived functors 7.3 Homological dimension

( I

5.2 Discrete valuation rings 5.3 Dedekind domain

;.)

\.

f'

j 4,

I

This chapter is concerned with a preliminary study of modules. The notion of an ideal which arose from number theory is also important in Algebraic Geometry. It is useful to study ideals from a module theoretic set up, as operations of linear algebra such as formation of quotients, products and tensor products are closed for modules but not for ideals. We then study special types of modules such as projective modules, fiat modules and faithfu.lIy fiat modules. Projective modules play the role of a vector space while studying linear algebra over a general commutative ring. Flat modules are more general than projective modules. Sometimes, geometric objects defined over a field behave differently over a bigger field. Such questions are best studied by scalar extension of rings using tensor products. Some properties are preserved under fiat extension but faithfully fiat extensions are more useful as they have nice descent properties. 1.1. Free modules A ring R will always mean a commutative ring with unit element 1. Definition: An R-module M is an abelian group M together with a map R x M .... M given by (a, x) .... a.x, satisfying the following conditions.

I

J

a.(x+y)=a.x+a.y, (a + b).x = a.x + b,», a.(b.x) = (ab).x,

aeR,x,yeM. a,beR,xeM. a, beR, xeM.

2 COMMUTATIVE ALGEBRA MODULES

Lx=x,

xEM.

Proof: Clearly K satisfies the conditions for a submodule. The map!: MIK-+Ngiven by1(x +K) =/(x). XE M, is well defined and is an isomorphism.

We denote a.x by ax. Examples: (i) Any vector space V over afield K is a K.module. (ii) Any abelian group G is a Z-module.

Corollary: Let M be an R-module. N a submodule of M and . K a submodule of N. Then MI N ~ ~~.

DefInition: A subset N e M is called a submodule, if N is a subgroup of the abelian group M and ax E N for aU a E Rand xEN.

Proof: The composition of the projections M -+ M/K -+ M/K . surjective " N/K an d has k ernel N. Hence by Proposition 1, MIK M/N~ N/K'

Examples: (i) Any subspace W of a vector space V. (ii) All polynomials of degree utmost n is a submodule of the R·module R[X]. . (iii) The modules 0 and Mare submodules of M caned improper submodules, \1

J

I

IS

We now consider .some operations on modules. If Nand K are submodules of M, then NnKis a submodule of M but NUK is not in general. a submodule of M. The smallest submodule of M containing NU K is called the submodule generated by N and K.

Definition: Let M and N be R-modules. A map I: M -+ N is called a homomorphism of R-modules if f (x + y) =f(x) +I(Y), x,yEM. I(ax) = al(x), aER xEM.

Proposition 2: The submodule S generated by Nand K is the submodule N+ K-{x +YlxE N. YEK}.

Examples: (i) For any fixed a E R, the map I: M -+ M, given by I(x) = ax, is a homomorphism. (ii) For any submodule N of M, the inclusion map i : N -+ M is a homomorphism. Let N be a submodule of M. Consider the quotient abelian group M/N with the scalar multiplication given by a.(x+N)=ax+N, aER, xEM. Then M/N acquires the structure of an R-module called the quotient module M/N. The map p :M-+M/N defined by p(x) = x + N, x E M is a homomorphism of modules called the projection. Definition: A homomorphism of modules which is both (I. I) (injective) and onto (surjective) is called an isomorphism. Proposition 1: Letj": M -+ N be a homomorphism of M onto N. Then the kernel of 1= {x E M I/(x) = O} is a submodule K of M and the quotient module MI K is isomorphic to N. .

r

m

3

Proof: Clearly N + K is a submodule of M, N e N + K. and KeN+Kso that SeN + K. Oonversely for any XEN, YE K. we have x, YES so that x + YES. Thus N + K e S. and S=N+K.

t

Corollary: If N 1• N I , '" Nk are submodules of M. thesubmodule generated by N lO N., .... N k is equal to k

p; XI J X,E N/}=N1 + N.+ ..,.+Nk. 1-1 Let A be a subset of M and I an ideal in R. {x = I a/x, I a, E I, x, E A}

Then the set

I

is a submodule of M and is denoted by IA. and A = {x}, IA is denoted by Rx,

In particular if I = R

Definition: An R-module M is called cyclic if M = Rx for some xEM.

l)

4

MODULES

COMMUTATIVS ALGEBRA

Proposition 3: An R-module M is cyclic if and only if M for some ideal [ in R.

=- RI[

=-

Definition: The annihilator of an R-module M is defined as

Anl/(M) = {a E R I aM = O}.

= (0).

DefiDition: M is called a finitely generated R-module if M~= M l + M I + ... + M •• where each M, is cyclic. If M, = Rx" then {Xl> XI••• • x.} is called a generating set forM. .

Example: The module of polynomials over R of degree utmost n is generated by 1. X. Xl: ...• X'. . Clearly this generating set is not unique, as 1. 1 + X. XI• .. . X" il also a generating set for the same module. Definition: M is called a direct sum of submodules Ml> M I , M., if every X E M can be uniquely expressed as Xl

! I \

\

••••

+ XI + .... + X., X, EM" 1 ~ i ~ n.

The direct sum is denoted by M

= Ml

E9 M I E9 ... E9 M •.

Proposition 4: An R-module M = M l E9 M I E9 ... E9 M •• if and only if (i) M =Ml + M I + ... + M n and (ii) M,

for all

n (Ml + M.+MI-J + M i +1"

'+

M n) = 0

t, 1 ~ i .,..; n.

Proof: Suppose M prove (ii), suppose

= Ml X

I

place. we have by uniqueness. X = O. Conversely assume conditions (i) and (Ii). By (i), each x EM can be expressed as

+ XI + + X., X, E Mi' Suppose X = Yl + Ys + +Yn; Y, EM,. Then 0= (Xl - Yl) + + (X, - YI) +... + (x n -Y.), x

so that

Clearly Ann(M) is an ideal of R. If M is cyclic and is generated by X. Al/n(M) is denoted by Ann(x).

x=

+ Y + ... + Y'-l + Yi+l +.. .Yn. Yl E M h Since x = 0 + 0 +...+ 0 + x + 0 +...+ 0, with X in the i-th

that XE M, and x =Yl j oF i.

Proof: If M is cyclic. then M = Rx for some x E M. The natural map 6 : R-+ M. given by 8(a) = ax. is a surjective homomorphism. If [ = Ker 8. then R{[ 0< M. Conversely if M R{I, then M is cyclic. as RI[ is cyclic being generated by 1= 1 + 1.

Definition: M is called a faithful R-module if Ann(M)

t

S

E9 ... EB Mn• Then clearly (i) is true. To is in the intersection on left hand side. so

(Xi -

x, - Y,

= Xl

y,) EM, =0 -

and

[(Xl - Yl)

+ ... + (X/-1 -

YH) +

(XI+! - Y'+l) +... (X. - Y.)]

E(MI +...+MI-J +M'+l + •.. M.). Hence by (ii), Xi - Y, = 0, l.e, x; = Y"l = [p, t.e. the exact sequence o -+ M' -+ M -+ P -+ 0 splits. To prove the converse, express P as the quotient of a free module F with kernel K so that the sequence 0 -+ K -+ F -+ P -+ 0 is exact. By assumption, this sequence splits so that P is a direct summand of F and hence projective.

t

MODULES

13

Proof: Since M is finitely presented, there exists anexact sequence o -+ K -+ F -+ M -+ 0, F free, F, Klg. By Proposition 6, KEBF'=K'EBF

Since K EB F' isf.g. so is K' EB F. Since K' is its quotient, K' is

r.s.

1.2. EXERCISES.....-

.Corollary: P is Ig. projective " P is a direct summand of a free module of finite rank. Definition: An R·module M is said to be finitely presented, if there exists. an exact sequence 0 -+ N -+ F -+ M -+ 0 with F free and F, N finitely generated.

1. Let :M -+ F be a surjective homomorphism of a finitely generated module M onto a free module F. Show that Ker (I/» is finitely generated. 2 . Let n be a positive integer with n = dld l • Consider the sequence given by

" o -+ diZ. _~. _ Z. _

Proposition 6: (Shanuel's Lemma). Let 0 -+ N -+ P -+ M -+ 0, and o -+ N' -+ P' -+ M -+ 0 be two exact sequences of R·modules wit~ P,P' projective. Then P EB N' c« P' EB N. Proof:

3.

Consider the diagram

o -+N ~ o -+

N'

P

r:_ P'. _

~

(i) HomB (R, M) ""- M. M -+ 0

(ii) HomB (EB

"Til-+ 0 _s'_ M

<jJ(x, y) = O(x)-f'(Y), x E P, yEN' and

•

r/>: N

-+

PEBN', by

Then the sequence

0-+ N_PEBN' _

oj.

i

P' -+ 0

is exact. Since P' is projective, by Proposition 5, the sequence splits. Hence P EB N' """ P' EB N. Corollary: LetMbe finitelypresented and 0 -+ K' -+ F' -+ M -+ 0 be any exact sequence with F' free of finite rank. Then K' islg.

N) "'" ED

M

'_1

(iii) HomB (M, ED

Since P is projective, there exists an R·linear map 0: P -+ P' such that g'O = g. Now 0 induces an R-Hnear map 0': N -+ N'. Define <jJ: P ED N' -+ P' by

(y) = (f(Y), O'(y»).

d1Z. -+ O.

where ~ is the inclusion map and IX is the multiplication by dl • Show thai the sequence is exact. Show further that the sequence splits if and only if dt and dl are relatively prime. For any R·module M, show that

"

i N ,)

1-1

"4.

Of

EB

oj

HomB(M" N).

i

HomB (M, N,).

1-1

1-1

Show that if P is a projective module,. there eJists a free moduIeFwithPEDFfree. P:j pIe Pc1.,;P L'f.' , S. An element e E R is called idempotent if el = e. Show that if every projective R':.Illodule is free, then the only idempotents of R are 0 and 1. 6. Let M be an R-module. An element x E Mis called unimodular if the cyclic submodule Rx is a free direct summand of M. Show that x E M is unimodular if and only if there exists an R.homomorphism ;:M -+ R with r/J(x) = 1. In particular if M is free with basis {e,hclc., x = 'Ea,e, is unimodular if and only if (a1 , • •• a.) is a unimodular row. "..,. A projective R-module P is said to be stably free if P EB Rm"", R" for some m, n. Show that every stably frcc

r-s-

14

COMMUTATIVE ALGEBRA'

MODUUlS

IS

R-module is free es, Every unimodular row over R can be completed to a non-singular matrix. [If R=k[X1•... X n) . k field, every J.«. projective R-module is free. See proof of Serre's conjecture by D. Quillen and A.A. Suslin in the Lecture Notes in Mathematics, No. 635, Springer Verlag).

1,3. Tensor products Let M. N, K be R-modules. We recall that a map 6: M X N -. K is R-bilinear, if O(x. y) is R-linear in x, for each fixed YEN and R-linear in y for each fixed x E M. . The study of bilinear maps from MXN to K can be reduced to the study of linear maps on a suitably defined module (tensor product of M and N) with values in K. Definition: Let M and N be R-modules. The tensor product of M with N is a pair (T, 6) where T is an ,R-moduleand 6: M x N -. T an R-bilinear map with the property that for any R·module K and R-bilinear map f: MxN -+.Jr, there exists a' unique R-!inear

m~~T-+K~ilin=~

.

Theorem: The tensor product of M and N exists and it is unique up to)Upmorphism. lA..... ~".,..

Proof: We first prove the uniqueness. Suppose (T. 0) and (T', 6') satiBfy the conditions for the tensor product. Consider the diagrams

(x+ x', y)-(x, y)-(x', y) (x, y

+ y')-(x, y)-(x, y'),

(ax, y)-a(x, y)

x, x' E M, Y, l' EN,

a E R ..

(x, ay)-a(x, y)

Let T = FIG and 0: MxN -+ T given by 6(x, y) = (x, y) + G. Then we show that (T, 6) satisfies the conditions of the tensor product. Clearly 6 is R-bilinear. Consider now any R-module K and any R-bilinear mapf:MxN -+ K. Thenf can be extended to an R-linear map h.: F -. K by defining

it II, a, (x" y,)] = ,:& a, f(x" y,). It induces therefore an R-linear map]: FIG = T -+ K, which sa-

MXN~T

I

~"J

and

T'

,6

,

and y, E N. with formal addition and natural scalar multiplication. Let G be the submodule of F generated by elements of the type

h. vanishes on G, by the definition of G and the bilinearity off.

e

-,

Since (#)6 = <jI ("" xn} and {Yl,' .• , Ym} respectively. Then M®N is free with basis -/ {x,® Yl}' By the definition of a free module, M = Ei3

Proof:

B

Proof: (i) The map (M

maps M, -+ M and N J -+ N. induce R-linear maps M, ® NJ -+ , M®N. for each i and}. These in' turn induce an R-linear map 1/>: Ee }j (M, ® Nt) -+ M ® N. Clearly 1/>8 = 1 and 81/> = I showing

.+

M ®.(N ® P) given by

(x® Y. z) -+ x® (y® z) is a well defined R-bilinear map on (M® N) X P which can be extended to an R-linear map (M ® N) ® P -+ M ® (N ® P) given by (x ® y) ® z -+ x ® (y ® z). This is an isomorphism ~t~ inverse given by the map x ® (y ® z) -+ (x ® y) ® z defined S1lD.1larly.

N = EEl

i, Rx,

and

i

RYl> where Rx, C>!. R "" RYJ for all i and}. By ProposiJ-1 tion 3, M ® N "'" Ee I (Rxi ® RYl)' But Rx, ® RYl "'" R ® R""" R ~I

B

and Rx, ® RYJ "'" R(x, ® YI} under the map ax, ® bYJ-+ ab(XI ® YJ). Hence M®N "" EB I R(xi ® YI), i.e, it is free with basis {(x, ® Yl)}' 1./

Corollary 2: The tensor products of free modules of ranks m and n is free of rank mn. .

18 COMMUTATIVE ALGEBRA

Letf: R -+ S be a homomorphism of rings. -Any S-module M can be considered as an R-module through the mappingfby defining ax = f(a)x, a E R, x EM. Such an R-module is called

the module obtained by restricting the scalars through! In particular the S module S can be considered as an R-module through! Proposition 4: If M is a finitely generated S-module and S is a finitely generated R-module thenM is a finitely generated R-module. Proof: Let {Xl,. .. , x n } be a generating set of Mover Sand {Sl' •• , generating set of S over R. For any X E M, we have x = l& a,x" a, E Sand aj = I rlJSh rl} E R. Hence x = I ( I r/jSj)x, = , j / j iE rlJ(SjX,), Thus the set {SjX,}"j is a generating set of Mover R.

SOl} a

4j

19

MODULES

Definition: A homomorphism f:A -+ B of R·a1gebras is an Rmodule homomorphism which satisfies the condition f(xy) = f(x)f(y), x, YEA andf(l) = 1. If A and Bare R-algebras, the tensor product module A ®'B II

has a well defined multiplication defined by (a ® b)(a' ® b') = aa' ® bb', a, a' E A, b, b' E B. This makes A ® B into an RR

algebra, called the tensor product of the algebras A and B. Definition: An R-algebra A is said to be finitely generated if there exists a finite set {al'" . a.} of elements of A such that the 'natural map R[Xb XI'" . X.] -+ A. given. by [(Xl'" .X.) -+ f(a l , . : • a.) is surjective.

.

Definition: For any R-module' M, the

module S® M is an .R

S-module for the operation defined by s(SI ® x) = SSI ® X, S, SI E S, xE M. This S-moduleis called the module obtained by extension of scalars.

ExampIes: (.) . I A = R[XlI I••• ,X.] ,quotIent of a polynomial algebra.is finitely generated, generated by {aI' a l , •• •an}, where

a, = X, + I (1 (ii)

Proposition 5: If Mis a finitely generated R-module, its scalar " extension S ® M is a finitely generated S~module. R

Proof: If {xa (I S®MoverS.

~ i~

Examples; (i) A = R[XI , X 2, • • • X n], the ring of polynomials in Xl> X., ... X n over R. (ii) Any quotient ring of R[XI .. • • X.]. (iii) Any extension ring S:JR is an R-algebr~. (iv) A = R [[X]], the ring of all formal power series In X over R

{i:, a/X' Ia, E

The Z-algebra Q is not finitely generated.

Proposition 6; Let A and B be finitely generated R-algebras. Then A ~ B is also a finitely generated R-algebra. Proof: LetA be generated by {al,. .. ,a.} and B by {bt, ... ,b..}. Then there exist surjective algebra homomorphisms R[X1 , ••• , X.]

Definition: An R-algebra A is a ring A, which is also an R-module satisfying the condition - a(xy) = (ax)y = x(ay), a'E R, x, yEA.

of 'power series.

i -< n).

n) generate Mover R, {I ® x,} generate

R

i.e. A =

~

R} with the usual addition and multiplication

I·

- - - A andR[Y1 , R[X, •. . . , X.] ® RIYI ,

g

,

Y..] _ _ B. The induced map f®g: , Ym] -+ A ® B is also surjective. But

R

R

~IXl" .. ~ X.] ® R[ Y1 , • • • , Y01] "'" RIXI , • • • , X., Yu •..• Y..] cise I, 1.3). Hence A ® B is finitely generated.

(Exer-

R

~~marks: (i) If an R-algebra A is finitely generated as R-module. It IS clearly a finitely generated R-algebra but the converse is not true as is shown by A = R[X]. . (ii) rff:R -+ S is a ring homomorphism and A is anRalgebra, its scalar extension S ® A is an S-algebra. It will be a R

finitely generated S-algebra if A is a finitely generated R-algebra.

l

MODULES

20 COMMUTATlVB ALGEBRA 1.3. EXERCISES I.

Show that for any commutative ring R R[XJ®RIYJ e: HO";R (M, HomR(N, K» ... HomR(N®M, K) by setting R

vi:

r/>(f)(x® y) =f(Y)(x), x E N, Y E M. Show that '" is an .' isomorphism. ' (Direct product). For any collection {M..} of.R-mod.u.les, the product 11: M. is an R-modulefor componentwise addition and

m~tiplication.

Show by an example that n(MCI) ® N CI R need not be isomorphic to 11: (Mil ® N).

scalar

II

8.

and extended by distributivity , The symmetric algebra S(M) iJ; defined to be the quotient of T(M) by the two-sided ideal generated by elements of the type {x ® Y- Y ® x}, x, Y E M. Prove that if M is a free R-module, with base {e,hel' S(M) is isomorphic to the polynomial algebra in {X'}iel' 9. Show that for R-modules M and N, S(M EB N) =- S(M) ® S(N). Deduce that if P is a projective R-module there exists an Ralgebra T such that S(P) ® T is a polynomial algebra.

R

4.

R

(Symmetric Algebra). Let M be an R.module. The tensor algebra T(M) of M is defined to be the R-aigebra

10. (Bxterior algebra). Let M bean R-module and T(M) the tensor algebra of M. The quotient of T(M) by the two-sided ideal generated by elements of the type {x ® x}, x E M is called the exterior algebra I\.(M). The canonical image.of XI ® XI ® ... ® x, under the natural projection is denoted by xd..x•... I\. x, (called elements of homogeneous of degree r), Let ~ (M) denote the roth homogeneous component of A(M). Prove (i) If M is a free module with basis {el •.• . ,.en }, then A(M) is free with basis {e" I\.e,. 1\ ..• I\.e,r} 1 ~ II < i. < ...< ir ~ n. n

M ®' ® M (i times). To(M) = R. R R R multiplication in T(M) by (x 1® X. ®'. .. ® x r) (YI ® Y.® ... ® y,)

e
(iii) and it is sufficient to show that (iii) => (ii). Let I: N' -+ N. be injective and consider J*:M®N' -+ M®N. Let I(xr®Y/)

x (iJ Y + ImJ* -+ x (iJ .JI + Ker g*

is:an isomorphism.

23

R

=

R

R" is fiat. Clearly R is R-fiat as M

® R "" M. R

By Proposition 3, 1.3, tensor product commutes with direct sums. Hence M = Rn is fiat. (ii) Z. is-not fiat over Zby the Example at the 'end of Theorem 1, 1.4. Proposition 2: Let I: R -+ S be a homomorphism of rings. (i) If M is R-llat, the scalar extension S ® Mis S-llat. R

.

24

MODULES

COMMUTATIVE ALGEBRA

(ii) If Mis S-flat, and S as an R-module throughfis R-flat, then Mis R-flat. Proof: (l) Let 0 -.. N' ~ N be an exact sequence of S-modules. Since N' ® S ® M"" N' ® M and since 0 ~ N' ® M ..... N ® Mis s

R

R

R

R

exact, M being R-flat, it follows that S ® M is S-flat. R

Let 0 -+ N' -+ N be an exact sequence of R-modules. Since S is R-flat, the sequence 0 ..... S® N' -.. S ® N is exact. Since (ii)

R

(i) A sequence 0 ..... N' ~ N ~ N" -+ 0 of R-modules is exact if and only if the tensored sequence

0 ....M® N'

/.

~

11*

M ® N - + M® N" -+ 0 is exact.

Mis R-flat and for any R-module N, M ® N =0 implies R

N=O. Mis R-flat and for any 'Rrhomomorphism f: N' -.. N, the induced mapf*:M®N' -+ M®N is zero implies thatf=O. (iii)

Proof: (i) => (li) Clearly (i) implies that M is flat. If M ® N = 0, consider the sequence 0 --? N -+ O. The tensored sequence 0-+ M ® N .... 0 is exact so that 0 --? N -+ 0 is exact, i.e, N = O. R

(ii) ~ (iii) Let K = lmf, so that f: N' -+ K -+ 0 is exact.

Hence M®N'~M®K-+Oisexact. Sincef*=O,M®K=O and this implies K = 0, i.e. f = O. (iii) ~ (i) Since M is flat. the exactness of

O-+N'~N~N".-+O implies the exactness of o--?

Definition: An R-module M is called faithfully flat if it satisfies any one of the equivalent conditions of the above Theorem.

R

Theorem 2: For an R-module M, the following conditions arc equivalent.

(ii)

I' g. tcnsored sequence M ® N' - - 4 M® N ~ M ® Nil is exact. Since g*f* -= 0, by (iii), gf = O. Hence K = Imf c Ker g = L. Consider the exact sequence 0 -+ K -+ L LjK -+ O. Since Mis flat. the sequence 0 ..... M ® K -+ M ® L M ® LjK -+ 0 is exact. Hence M ® LjK"" M ® L/(M ® K) = 0, since (M ® K) -= Imf* = Ker g* = M ® L. If p is the projection, p: L -+ Lj K, then p*:M®L-+M®LIK is zero and hence p=O. This implies / N _K_ Nil IS " exact. K = L. i.e. t he sequence N r ---+

R

Mis S-flat, the sequence O-..M®S®N'-+M®S®N isexaet, Ii R s R i.e, 0 -+ M ® N' -.. M ® N is exact. Hence Mis R-flat. R

2S

.

/. M ®N' ~ M ®N ~ M ®N" --? O.

f N -g+ N", th e Conversely assume that for any sequence N ' --+:.

Corollary 1: A faithfully flat module M is flat and faithful, t.e. Ann (M)=O. Proof: Let a E R with aM = O. Consider the mappingf.: R-+R given by f. (b) = abo Then f:: M ® R -+ M ® R is zero because x ® b -+ x ® ab = ax ® b = O. This implies that f. = O. Hence f.(I) = a = O. i.e. Ann (M) = O. Corollary 2: M is faithfully flat if and only if M is flat and for· each maximal ideal m of R, mM =1= M. Proof: MjmM"" M ® R/m =1= 0, if M is faithfully flat, since Rjm =1= O. Conversely assume M/mM =1= 0 for alI maximal ideals m and that M is flat. Let N =1= 0 be an R-module. Then N contains a non-zero cyclic submodule K"" RjJ. Since 1 isa proper ideal, 1 C m for some maximal ideal m. Now M =1= mM implies M =1= 1M, i.e, M ® R/l =1= O. Hence M ® K =1= O. Since M is flat, M ® K is a submodule of M ® N and so M ® N #- O. This implies that M is faithfully flat. Example: A module which is fiat and faithful need not be faithfully flat. For instance Q is flat over Z (See Exercise 10) and faithful but Q is not faithfully Z-flat, as Q ® Z» = 0 but Z. =1= O. z We now study faithfully flat algebras. Let A be an R-algebra, M and N, R-modules. Then there exists a natural A-homomorphism

26

COMMUTATIVE ALGEBRA

"'M: A ® HomR (M, N) -+ HomA (A ® M, A ® N) given by R R R "'M(a®f){(b®x)}= ab ®f(x), a, bE A, x E M f E Homs. (M, N) We want to investigate condijions under which "'At is an isomorphism for all R-modules N. Clearly",,,, is an isomorphism fol' M = R. Using the property that tensor product and Hom commutes with finite direct sums (Exercise 3, 1.2) it follows that "'''' is an isomorphism when M is a free module of finite rank. PropositioD 3: Let A be a flat R-algebra. M and N are R-modules. Let "'M be the natural map defined above. Then (i) "'Mis injective if M is finitely generated. (ii) is an isomorphism if M is finitely presented.

"'M

Proof: (i) If M is finitely generated, there exists an exact sequence O-+K -+F-+M-+ 0, where Fis a free module of finite rank. For any fixed R-module N, define T(M) = A ®' Homs. (M, N) and R

T'(M) = HomA (~ ® M, A ® N) II

R

Since A is R-:flat and Hom is left exact (Proposition 2, 1.2) we have a commutative diagram

o -+

T(M) ..... T(F) -+ T(K)

1"'K

1

9M

1F 0-+ T (M) -+ T'(F) ..... T'(K)

with exact rows. Since"', is an isomorphism, it follows from the diagram that is injective. (ii) If M is finitely presented, by Corollary to Proposition 6, 1.2, K is finitely generated. We can replace K by a free module of finite rank Ft so that there exists an exact sequence F t -+ F -+ M -+ 0 with both Fan!! Ft free of' finite rank. Consider the commutative diagram

"'M

o.....

11.1M) ..... T(F) -+ 111Ft )

"'M

"'It!

""1

0-+ T'(M)-+ T (F) -+ T'(FJ

1

MODULES

"'It

"'It,

27

with exact rows. Since and are isomorphisms, it follows from the above commutative diagram that is an isomorphism.

"'M

PropositioD 4: Let A be a faithfully :flat R-algebra and M an R-module. If A ® M is finitely generated (finitely presented) over R

A then M is finitely generated (finitely presented) over R. (ii) If A ® M is finitely generated and projective over A, R

M is finitely generated and projective over R. Proof: (i) Suppose A

® M is finitely generated over A. Choose R

a generating set of the type {I ® x,}, X, E M 1';;; i';;; n. Let N be the submodule of M generated by {x,}, I ,;;; I,;;; n. Then the natural map A ® N -+ A ® M induced by the inclusion t: N -+ M is R

II

surjective. Since A is faithfully :flat, i is surjective, i.e. N = M. Hence M is finitely generated. If A ® M is finitely presented, it is also finitely generated and so M is finit~ly generated. Choose an exact sequence 0 ..... K

..!.-..

F ~ M -+ 0 where F is free of finite rank. Then the . f·· sequence 0 -+ A ® K _ A ® F -.!..c..... A ®M -+ 0 is exact. R

Since A

R

® M is finitely presented,

R

A

R

® K is finitely generated over II

A. Hence K is finitely generated over R showing that M is finitely presented over R. (ii) Let A ® M be finitely generated and projective over A. R

Since it is projective, by Proposition 5, 1.2, it is also finitely presented. Hence, by (i), M is finitely presented over R. To show that Mis R-projective it is sufficient to show that for any surjective R-homomorphism f: N -+ L, the induced homomorphism HomR (M, N) -+ HomR (M, L) is surjective. Since A is faithfully :flat over A, it is sufficient to show that

r:

t, ®r: A ® Homv (M, N) R

is surjective.

-+ A

® HomR(M, L) R.

Consider the commutative diagram

28

COMMUTATIVE ALGl!BRA

MODULES

29

(iii) Deduce that M is fiat. if and only.if M is flat for R, i.e.

where the horizontal maps are induced by f: N -+ L and the verti. cal maps are isomorphisms by Proposition 3. Since A ® M is ,R

A-projective, the horizontal map below is surjective. Hence is surjective.

fA

®f·

1.4. EXERCISES 1. 2. 3.

4. S. 6.

• 7.

Show that M = EB IM" is flat ifand only if each M" is flat. Show that the tensor product of two flat (faithfully flat) modules is flat (faithfully flat). Let 0 -+ M' -+ M -+ M" ~ 0 be an exact sequence of R-modules with M" fiat .. Show that M is fiat if and only if M'.is flat. "Dlustrate by an example that intersection or.two fiat modules need not be flat. Show that M = EB I"M" is faithfully fiat if each M" is fiat and at least one M" is faithfully fiat. Let ¢: R -+ S be a homomorphism of rings. Show that S is faithfully R-fiat if and only if ¢ is injective and S/¢(R) is Rflat. . Let ¢: R -+ S be a homomorphism of rings such that S is faithfully flat over R. Show that for any ideal I of R . .p-:l((I).S) =f.

8. 9. 10.

Give an example of a faithful and projective module which is not faithfully fiat. If K is a field, show that K(X) is not a faithfully flat K[X]module. An R-module M is caned flat for N if for every submodule N' of N, the natural map M ® N' -+ M ®N is injective. Show (i) If M is fiat for N, M is flat for any quotient module of N. (ii) If M is fiat for each N", then M is fiat for EB '}; N".

"

for each finitely generated ideal I of R, the natural map M ® I -+ M given by x ® a -+ ax is injective. (iv) Let R be a Bezout domain, i.e. an integral domain in . which every finitely generated ideal is principal. Show that M is fiat if and only if (it is torsion-free, i.e, ax = 0, a E R, x E M implies a = 0 or x = o. (v) Deduce that Q is fiat as Z-module.

.~~:

.~.~.

LOCALISATION

31

Proof: Let I be the collection of all ideals orR different from R. containing 1. Then 1: oF '" as 1 E 1:. By Zorn's Lemma: .1: has a maximal element m, which is a maximal ideal of R containing I. CHAPTER II

Corollary 1: If aE R is a non-unit, there exists a maximal ideal m of R containing a. Proof:

LOCALISAnON

Take 1 = Ra. ,

Proposition 2: Let/: R -+ S be a homomorpliism of rings. If Q is a prime ideal of S, then P = 1-1(Q) is a prime ideal of R. The first part is concerned with the study of ideals. We start with prime and maximal ideals which correspond geometrically to irreducible varieties and points over an algebraically closed field. We then study ideal operations such as sum, product, extension, contraction and taking radicals. . The second part is concerned with the study of local rings. The ring of meromorphic functions at a point P. in an Affine variety is a local ring and this' ring reflects the local properties of the variety at P. This ring is generalised to the concept of a ring obtained by localisation with respect to a multiplicatively closed set. The localisation operation which is defined also for a module is shown to be well behaved with respect to quotients. tensor products and exact sequences. Some applications of localisations are given in the study of projective modules as locally free modules. 2.1. Ideals

We recall some preliminary notions about prime and maximal ideals. An ideal P of R is a prime ideal of R if P =J: Rand ab E P implies a E P or b E P. Olearly P is a prime ideal if and only if RIP oF 0 and is a domain. An ideal m of R is called a maximal ideal of R if m oF Rand for any ideal I with m c I c Reither m = I, or R = I. Clearly an ideal m is a maximal ideal of R, if and only if Rlm is a field. In particular any maximal ideal is a prime ideal. Proposition 1: Let I be a proper ideal of R. maximal ideal m of R with t c: m.

Then there exists a

Proof: The given map

I induces a natural ring .homomorp~sm

1: R/P -+ S/Q which is injective.

If Q is a prime Ideal, SIQ IS a domain and so is RIP. Hence P is a prime ideal of R. Corollary 1: If R is a subring of Sand Q is a prime ideal of S, Q n R is a prime ideal of R. Corollary 2: An ideal P containing an ideal 1 is a prime ideal in R if and only if P/I is a prime ideal in RfI. , Proof: Let S = R/I, so that I: R -+ R/I is s.urjecti~e. Then. the induced map 1: R/P -+ S/Q where Q = P/~, IS ~ Iso~orp~lsm. Hence P is a prime ideal of R if and only If Q IS a pnme. Ideal ofS. ' Example: It is not true that an invers~ imag~ of a maxi~al ideal is a maximal ideal. [f i: Z -+ Q is the inclusion map, the Ideal {O} is maximal in Q but not in Z. Proposition 3: Let 11'I., ... , 1.' be ideals of Rand P a prime ideal containing

n I,.

Then

1-1

r :» lit for

some i.

Proof: If I, ¢: P for each i, choose a, E 1" a,¢ P. Then a = '1C a, E n I" a ¢ P, a contradiction. I

Corollary:

If P

=

n' [" P prime, then P = I, for some i.

1-1

32

LOCALISATION

COMMUTATIVE ALGEBRA

Proof: By Proposition 3, P ~ I, for some i. Since Pc IJ for all j, we have P = I,. Proposition 4: Let I be an ideal with leu P" P, prime. Then . 1-1 I c P, for some i. Proof: The proof is by induction on n. If n = I, there is nothing to prove. Assume the result for (n-I) and let leu P,. If I is contained in an (n-l) union of P,'s, induction :;;lies. If not, I ¢ }~/J for all t. Choose a, E I, a, ¢ PJo jO/; i, (1,,;;; i";;; n).

33

Proof: Let a E N(R) so that an = 0 E P for any prime ideal P. This implies that a E P. Conversely let a E R be non-nilpotent. We construct a prime ideal Po ofR with a¢ Po. Consider the set S = {anI n ~ O}. Let 1: be the collection of all ideals of R which do not intersect S. Since a is non-nilpotent, (0) E 1: and hence 1: 0/; 4>. By Zorn's Lemma, ~ has a maximal clement Po' Clearly a¢: Po and it is sufficientto show that Po is a prime ideal. Let a'¢Po and b¢P o, so that (Po+Ra')nSo/;c!> and (Po + Rb) n S 0/; rf>. Choose mI' m. such that am, = PI + Aa' a~d am' = PI + f4b, Pl. P9 E Po and A, fl.E R. Then a,",+m. =ps+A!-La b, Ps E Po which shows that a'b ¢ Po.

P, for each t.

Definition: The intersection J(R) of all maximal ideals of R is called the Jacobson radical of R.

Then the element a = 1: a1 ... a'-I' a'+1 ... an, is an clement of

Example: The nil radical N(R) C J(R), as every maximal ideal is prime. The inclusion is a strict inclusion for the ring R = k[{X]} the power series ring in X over a field k, as N(R) = O. Now J(R) = (X) as an element fER is a unit f has a non-zero constant term.

UP,. '-1

If for 'some i, a, ¢ P" I ¢ n

I not in

UP" '-1

Assume that a, E

'-1

a contradiction.

'

Definition: An clement a ER is called nilpotent if a" = 0 for somen;> 1. Proposition 5: The set orall nilpotent elementl of R is an ideal ofR.

Proof: Let I be the set of all nilpotent elements. If a, b E 1 then m a"=O=b for somem, n. Bybinomial Theorem, (a+b)m+n= 1: a'bl. '+J-m+n

Each term on the right hand side is zero as either t> n or j;> m. Hence a + b E 1. Clearly -a E I and ra E I for all r E R. Thus 1 is an ideal in R. DeftDition: The ideal of nilpotent clements of R is called the nil radical of R. . ExamPles: (i) Any integral domain has nil radical zero. (ii) Zm has nil radical zero ~ m isa product of distinct primes. (iii) For any R with nil radical N(R) the ring RIN(R) has nil radical zero. Proposition 6: The nil radical N(R) of R is the intersection of all prime ideals of R.

Proposition 7: An' element a E J(R) if and only if I unit, for all b E R.

+ ab

is a

Proof: Let a E J(R) and let l+ab be a non-unit for some b E R. Then there exists a maximal ideal m of R with I +.ab E m. Since a E J(K), we have a E m which implies I E m, a contradiction. Conversely assume that I + ab is a unit for all b E R. Suppose a ¢ m for some maximal ideal m. Then m + Ra = R, so that 1= ao-ba, ao Em, b E R. This implies that ao = I + ba is a unit, a contradiction as VI 0/; R. We now consider various ideal operations and study the relations between them. Given two ideals I and J, their sum- is defined to be the ideal I + J = {a + b Ia E I, b E J}. The product IJ of I and J is defined by U ~

6:'-1 a,b, Ia/ E I,

b, E J,

n

> I}.

Clearly IJ is an idea) contained in In J and not in general equal to InJ. However if I+J~R, then U=InJ for if l=a+b. a E I, b E J, then any x E In J can be expressed as x = xa

+ xb E

IJ.

34

COMMUTATIVE ALGBBRA

Definition:

WCALlSA1l0N

Proof: Take R = Z, I, = m,Z (1 I).

Proof: The proof is by induction on n. The result is already shown to be true for 11 = 2. Assume the result for (11-1). Consider 11'

Prool:

R. I of. j.

I ••...• 1•• which are mutually comaximal.

+

write I = u, +

Vit UI

a-I

+ u, u Ell' then x E

If x

'TI:

n

*

Corollary 1:

Proof: Since Kef f = position.

I

0
n or j:> m 1 E 1+ J. Hence R=I+J. t.e, 1 and J are comaximal.

vI

vI.

v]

2.1. EXERCISES

vI vI

1. 2.

Definition: Let/: R -+ S be a homomorphism of rings. If 1 is an ideal in R, the ideal generated by/(/) in S is called the extension of 1 under I- If J is,an ideal in S, f- 1(J ) is an ideal in R. called the contraction of J under f, An ideal J of S is called extended ideal if for some ideal I of R. its extension I' = J. An ideal 1 of R is called a contracted ideal. if for some ideal J of S, its contraction Je = I.

3.

4.

Examples: (i) Let R be a subring of S and j : R -+ S the inclusion map. The extension of an ideal I of R is the ideal generated by I in S and contraction of an ideal J of S is J n R. (ii) L: Z -+ Z[i] the inclusion map. Z[i] the ring of Gaussian integers. The extension I' of the ideal 1 = (2) which is a prime ideal in Z is not a prime ideal in Z[i] as 2 = (1 + i)(I-i) E I' but 1 + i and I - i do not belong to 1-. Proposition 11: Letf: R -+ S be a ring homomorphism. Then the mapping I -+ 1- is a bijection between the set C of contracted ideals in R onto:the set E of extended ideals in S. Proof: We first show that for any ideal 1 in R and any ideal J in S, we have (i) I'" "'" I' and (ii) J'" = Je. Clearly 1 C 1" and J:::) J «. Hence I' c and applying the second relation to J = I' we have I' :::) 1'''. This proves (i). The second relation is proved similarly. These relations show that the map C -+ E given by 1 -+ I' is bijective.

37

Let R be a ring in which x' = x, for all x E R. Show that every prime ideal of R is maximal. Let R[X] be the polynomial ring and f(X) E R[X] given by f(X) = aD + a,X + ... + a.X". Show that (i) fis nilpotent if and only if all the a, are nilpotent. (ii) fis a unit if. and only if aD is a unit and a,(i;;;>- 1) are nilpotent. If P isa prime ideal in R. show that its extension PIX] in R[X] is a prime ideal in R[X]. Is the same true of maximal ideals? Why? . LetRIlX]] be the power series ring in X over R andf(X)ER[[X]) given by f(X) = aD + a1X + .., + a.X· + .... Show that (i) f(X) is a unit if and only if aD E R is a unit. (ii) !(X) is . nilpotent implies that each a, is nilpotent (i ;;> 0). Let S = R[[X]] be the power series ring in X over R. Show that every maximal ideal of S is of the type (m, X), the ideal generated by X and m where m is a maximal ideal of R. If a E R is a nilpotent clement. show that 1 + a is a unit. Deduce that the sum of a nilpotent. element and a unit is a unit. Foran R-module M and submodule N of M, define ru(N) = {a E R I anM c N. for some n;;;>- I}-. Show that rM(N) is an ideal (called the radical of N in M). Show further that rM(N) = v(N: M). If 11' I••...• In are pairwise comaximal ideals and M is an R-module. show that the natural map M -+

~

'-I

Mll,M given

by x -+ (x + 11M• . " x + I nM). x E M is surjective. Show further that MIIM cs: 7t MII,M where 1 = n I,.

I....

I

9.

+
"

Theorem 1: Let R be a local ring. Any finitely generated projective R-module is free. .

Coker (v) - - + Coker (w) where f*, g*, 4>* and 1jI* are natural maps induced by the maps!. g, 4> and <jI respectively.

Proof: Let M be a finitely generated projective R-module. By Proposition 4. there exists a minimal generating set {xl" •. , x.} for M. Chose a free module F of rank n with basis {eb ..• , e.} and define an R-Iinear map f; F -+ M. by fee,) = x" 1 .;;; t c; n. If K = Ker f, we have an exact sequence

Proof: Since the diagram is commutative, f(resp t/J} maps Ker u. (Coker u) in Ker v (Coker v). This defines the map f* (resp t/J*). Similarly g* and 1jI* are defined. To define a: Ker(w) ...... Coker (u), let X E Ker (w). Choose y E M with g(y) = x. Then IjIv (y) = wg (Y). = w(X) = 0 so that v(y) = 4>(z) for some unique zEN'. Define 8(x) = Z + u(M'). a is well defined for if g(Yl) = x, then y-Yl' E Ker g = Imf so that y - Yl = f(t), t E M. Then v(y- Yl) = vf(t) = .pu(t). If V(Yl) = .p(Zl), then .p(Z-Zl) = .putt). This implies that z-z, = u(t) so that Z + u(M') = Zl + u(M} The verification of

o ...... Now K

c

f

K ...... F---+ M -+ 0

mFfor if YE Kwith Y fey)

a,

= l:, a,e"

(*) a, ER then

= ,~ aix, = O.

If some ¢ m, it is a unit, and then Jet will he a linear combination of the remaining x/s, contradicting the minimality of {Xl" . ,, X.}. Hence K emF.

1

the exactness of the sequence is a routine computation and is left as an Exercise. Theorem 2: Let R be a local ring with maximal ideal m and M a finitely presented R-module. If the canonical map

•

LOCALI~ATION

42

43

COMMUTATIVB ALGEBRA

2.2. EXERCISES

uM:m®M -+ M, R

1.

given by UM(a ® x) = ax, a E.m, x E M is injective then M is free. Proof: Let k=Rjm and choose a minimal generating set {Xl"'" X n} for M such that {I ® XI} E RIm ® M is a k-basis of RIm ® M. •

2.

R

Let F be a free module of rank n with basis {el.·.·. e.} and g an "R-linear a map g: F -+ M defined by gee,) -= x" I ~ i ~ n. If K = Ker (g), we have an exact sequence

o

-+ K

-.!.-.... F --.!.- M -+

3.

4.

O.

S.

This induces a commutative diagram

m®K -+ m®F -+ m®M

1 1 1 UK

O-+K ---+

U, •

-+ 0

6.

UM

F -- M

f g with exact rows. By Proposition 6, we have an exact sequence

r

3

7.

c'

Ker (UM) - - + Coker (Ug) ~ Coker (up) - - + Coker (UM)' Now Coker (UM) = MlmM O! Rim ® M. By the choice of {xl" .. , x.}, Coker (up) Coker (UM) ~s an isomorphism given by I ® e, -+ I ® X,. Hence Ker g* = 0, i.e. Imf* = O. Hence Im 5 = Ker f* = Coker (Ug). This implies that 8 is surjective. Since Ker U/4 = 0, by assumption, we have 1m 8 = Coker (UK) = O. i.e. X = mK. Since M is finitely presented. K is finitely generated. Hence K = 0 by Nakayama Lemma, i.e. M 0< F is free.

.x;

Corollary: Let R be a local ring and M a finitely presented Rmodule. Then the following conditions are equivalent. (a) M is free (b) M is projective (e) M is flat. Proof: Clearly (a) => (b) =>(e). To show that (e) => (a) assume that M is flat. Then the map m ® M -+ R ® M. given by a ® x -+ a ® x, a E m, X E M is injective. Hence the map UM is injective and the result follows from Theorem 2.

Let I be an ideal of R such tha t for all finitely generated R-modules M. the condition 1M = 0 implies M = O. Show that I c J(R) (Converse of Nakayama Lemma). Give an example of a ring R and a non-zero R-module M. (necessarily not finitely generated) such that J(R)M = M. An element e E R is called idempotent if e2 = e. Show that the only idempotents of a locai ring are 0 and 1. Let Ie J(R) and f: M -+N. a homomorphism with N finitely generated. If the induced map 1: MjlM -+ N/IN is surjective, show that f is surjective. A ring R is called semi-local if it has only finitely many maximal ideals. Show that (i) A finite product of semi-local rings is semi-local. (ii) Quotient of a semi-local ring is semi-local. Show that a ring R is semi-local RfJ(R) is a direct product of fields. Let M" be a finitely presented R-module and 0 -+ M' -+ M-+ M" -+ 0 an exact sequence of R-rnodules with M finitely generated. Show that M' is also finitelygenerated. (Hint: Express M as a quotient of a free module F of finite rank by a sub" module K and define a suitable map from K to M').

:z 3.

i

Localisation

Local rings are obtained quite often, by the process of localisation. We propose to describe algebraically the method of localisation. Let R be a ring. A subset S c R, is called a multiplicatively closed set in R if 0 ~ S, 1 E Sand ab E S whenever a E Sand bES. Examples: (i) R is a domain and S = R-{O}. (ii} P is a prime ideal of Rand S = R-P. (iii) Let {P,} be a set of prime ideal of Rand S = R- UP,. I

(iv) R is any ring and S is the set (except 0) of all non-zero divisors of R. Let M be an R-module. The localisation of M with respect to S is constructed as 'follows. Consider the set M x S. Define a relation ,.., on M x S by the condition, (x, s) ,..,(?, t) if and only if' Sl(tx-SY) = O. for some

44

COMMUTATIVE ALGEBRA

s, E S. M x S. element denoted M = R,

It is easy to verify that...., is an equivalence relation on The set of. equivalence classes is denoted by Ms. Any of M s being an equivalence class containing (x, s) is by the symbol (xis), x E M, s E S. In particular for the set R s is defined.

1

+ (bIt) =

(at

(als)(xlt)

= (axlst)

,a.)Ek·, let P={/ERI/(al, ... a.)=O}. Then.P is a prime ideal of R. If S = R-P, then R s is nothing but the ring defined in Example (ii), section 2.2. (iii) More generally let P be any prime ideal Rand S = R-P. An element (a/s) E R s is a unit ~ a E S. Hence the seC of nonunits of ~s = {(a/s).! a E R-S = P} and it is an ideal. i.e. R~ is a 10c~1 nng, The rmg R s for S = R-P is usually denoted by R». (IV) If S = {a"' n ;> O}, a E R non-nilpotent, R s is usually denoted by R a • .

x, Y EM, s, t E S.

a E R.

Proof: We first show that the operations are well defined. If (als) = (a' Is') and (bIt) = (b'lt') then s,(as' -a's) =0 and t,(bt'-b't) = 0 for some s" t, E S. These imply the relations s,/tI(at + bs)s't'--"(a't' + b's') st] = 0

PrClpositiClD 2:

Let/: R -+ Rs be the natural map given by

I(a)

=

(a/I).

Proof: Define g: R s -+ R' by g(als) = g(a) g(S)-l. We show that is well defined. If (a!s) = (a' /s'), then sl(as' -a's) = 0 for some lit E S. This implies g(s,) [g(a) g(s')- g(a') g(s)] = O. Since g(s,), g(s) and g(s') are units, g(a) g(S)-l = g(a')g(s')-l. i.e, g is well

•••

PrClpClsition 3: /[(als) ® x] modules.

i

t I

For any ring homomorphism g: R ..... R' such that g(s) is a unit for all SES, there exists a unique homomorphism g: Rs-+oR' such that K!=g.

i

gl(als) = g,[(a/l) (lis)] = g(a) g(S)-l = 'j(als).

a=(a1,

a, b E R, s, t E S.

and s,'t (s't'ab - sta'h') =0. Thus the addition and multiplication are well defined. The zero element is (O/s), s ES and the inverse of (als) under addition is (-als). The distributivity of addition over multiplication and the associativity of multiplication can be easily verified. Hence R s is a ring with unit element (Ill). Similarly the operations on M s can be shown to be well defined and that M s is an Rs-module. The ring R s has the following universal property.

g

(l) If R is a domain and S = R-{O}, R s is nothing bu. the quotient field of R. (ii) LetR=k[X" ... ,X.].kfield. Forafixed

The set Ms is an Rs-module for the operations (xis) + (Ylt) =(tx +8ylst)

defined. Clearly Ii is a ring homomorphism and 'if = g. Now is also unique for ifg1 is another map with gtf= g. then

and

+ bslst)

(als)(blt) = (ablst)

45

gl(l/s) = [8",(sll)]-' = [g,f(S)]-l = g(s)-'

PrClPClsitiClD 1: The set Rs is a ring for the operations (als)

LOCALISATION

The canonical map

I: R s ® M

-+oMs defined by

R

-+0

(axIs) is well defined and is an isomorphism of R _

s

Proof: We firstshow thatlis well defined. If (all) = (bIt), then sl(la-sb) = 0 for some Sl E S. This implies that s (ta-sh)x = 0 i.e. (~xfs) 0= (hxlt). Clearly lis surjective as/«lls) x) = (x/s). remains to show thatlis injective. Observe that any element of R s ® M is of the form lIs ® Y, s E S, Y E M, because any Z E R s ®R M can be written as z = '1X,al/sl) ® XI, a, E R. Sl E S, I . Xi E M. If s = ~Sl and t, = ~ sit then .

®

I Z

= I (alllls) ® XI I

I;

iFI

= II

lIs ® (a,tlx,)

=

lIs 18) Y

where Y = I a,/lx,. I

..... ,

. ~ow .if z = lis ® Y E Ker f. then 1«1/8) ® Y) = (Yls) =0. This ImplIes that SlY = 0 for some Sl E S. Now (lIs) ® Y = ~sllsls) ® Y = (l/ sliI) ® SlY = O. Hence I is injective. Clearly f IS an Rs-module homomorphism. Hence it is an isomorphism of Rs-modules.

.

46

LOCAUSATION

COMMUTATlVB ALGBBRA

o_

Proposition 4: Letf:M -+ N be a homomorphism of R-modu~es. Thenfs:Ms-+Ns defined by fs(x/s) = (f(x)/s) is well defined and IS a homomorphism of Rs-modules. Moreover (i) fs = Id whenf= Id and (ii) (gf)s = gsfs for any other R-homomorphism g: N -+ L.

M~

- _ M s --+

M~

47

_ 0

11' s,1

0_ Rs ® M' - R s ® M _ R

® M" _ 0

R

R

Hence the sequence

Proof: We first show that fs is well defined. If (XIS) = (y/t), then Sl(tX-SY) = 0 for some Sl E S.Then f(Sl (tx-sy» =Sltft X)slsf(Y) = 0, i.e. (f(x)/s) = (f(y)/t). Clearly fs is an Rs-module homomorphism asfs[(a/s)(x/t)} = fs(ax/st) = (f(ax)/st) = (af(x)/st) = (a/s)(f(x)/t) = (a/s)fs(x/t). The two relations (i) and (ii) arc clear from the definition offs.

plies that R s is R-Bat.

Proposition 5: Let M' ~ M ~ M" be an exact sequence of R-mQdules. Then the induced sequence

Pr~~®~=~®~®~=~®~®N

0_ Rs®M' _R s ®M _Rs®M" _ 0 is exact. R

Corollary 3:

R

This im-

R

For any two R-modules M and N, (M ® N)s=M s ® N s• R R.

R

R

R

R

R

C>!.Ms®N=(Ms®R s)® N=.M s ®(Rs®N)

, f. g." . t Ms-M s - Ms is exact.

. R

Proof: "Since gf = 0, we have (gfh = gsfs = O. Hence 1m(fs) c Ker (gs). Conversely let (x/s) E Ker (gs) so that (g(x)/s) = 0, i.e. S,g(X) = 0, for some 51 E S. Then g(slx) = 0 i.e. SIX E Ker g = Im f. Write SIX = fey) for some y E M. Then

R.

R

"R.

R

C>!.Ms®Ns• R.

Corollary 4: For any two R-modules M and N and a prime ideal P of R, we have (M®N)P C>!.Mp®Np• ; R

R,

Proof: Take S = R - P.

(x/s) = (SIX/SIS) = (/(Y)/SSl) = fS(y/SS1) E Im fe.

Hence Imfs

=

Ker gs and the sequence is exact.

Corollary 1: If N is a submodule of M then N s is a submodule of M s and (M/N)sC>l.Ms/Ns.

Proof: The exactness of the sequence 0 _ N _ M _ M /N -+ 0 implies the exactness of 0-+ N s -+ M s - (M/N)s-+ O. Corollary 2:

R s is a flat R-module.

Proof: Consider an exact sequence of R-modules 0 - M' - MM" _ O. Then the sequence 0 _ M~ -Ms -+ 0 is exact. By Proposition 3, M s R s ® M making the following dla-

=

gram commutative.

M; _

R

Propositiol! 6: Let Sbea multiplicatively closed set andf: R _ R s the natural map given by f(a) = (a/I). Then (l) Every ideal of R s is an extended ideal. (ii) The prime ideals of R s are in (I, 1) correspondence with the prime ideals of R not intersecting S. (iii) f preserves the ideal operations of taking finite sums, products, intersections and radical. Proof: (i) Let J be an ideal of R s and let 1= f-I(J) = {a E R I (a/I) E J}. Then Ie c J. Conversely if (a/s) E J, then (a/1) 0= (s/l)(a/s) E J so that a E I and (a/s) E I«, Hence J = I«. This shows that J is the extension of I. (ii) Any prime ideal of R s contracts to a prime ideal of R. Conversely let P be a prime ideal of R and consider its extension

48

COMMUTATIVE ALGEBRA

LOCALISATION

pe = PR s. Then P« = R s if and only if P n S #- I. Write b = (a/s), so that b" = (a"/sn) E Ie. This implies that SIan E I for SOme SI E S. Hence s~a" E I, t.e. Sla E 'Ill and b = (s,a/s,s) E (vi)e. Hence 'l/P=(vI)e. Corollary 1: The nil radical of R extends to the nil radical of R s• Proof: Consider I

= (0) in the

relationy'P = (vi')'.

Corollary 2: For any prime ideal P of the ring R, the prime ideals of R p are in (1, I) correspondence with prime ideals of R contained inP. . Proof: Take S

= R-P in (ii),

of R with InS = P', Q" 'D p .. = P. . Since Q n T =.p. we have QO n (R-P) =.p. i.e. QO c P and hence Q" = P. Theorem 1: Let/: R -+ S be a homomorphism of rings. If Sis a faithfully flat R-module. every prime ideal of R is the contraction . of a prime ideal of S. Conversely iff: R -+ S is a ring homomorphism such that S is flat over R and every prime ideal of R is the contraction of a prime ideal of S, then S is faithfully flat over R.

52 COMMUTATIVE ALGEBRA Proof: Assume S is faithfully flat over R. maps S!.. S

® sis given by (a) = l®a R

Consider the natural and ljI(a ® b) = ab,

Since IjI4> = Id and {J,I\I are R-homomorphisms, Im( (iv). Let Y-{f E RIMI is free of finite rank over R/}' For any maximal ideal m of R, Y ¢ m and hence Y generates the

i '-1

57

-

(ul)""

(Mr}",'

1

Um"

-----+

....

Mm"

where the vertical maps are isomorphisms given by Corollary to Proposition F, 2.3. Since m" E D(f), U",It is an isomorphism and this implies that the horizontal map above is an isomorphism for all maximal ideals m' of R I . This implies by Corollary to Proposition 9, 2.3 that Mr is isomorphic to Rj. i.e, MI is free of rank n over RI' (iv) => (i), For each i, choose a free Rji module L, such that Mji is a direct summand of L, and we may assume without loss of generality that all the L, are of the same rank. Then L = nL, is

,

o,fi for some fi E Y, a, E R.

free over S = nRji and M' = nMr, is a direct summand of L.

This implies (iv). . (iv) => (iii). Clearly M is finitely generated over R by Corollary to Proposition 2. Since IRfi = R, for each prime ideal P, there exists some i(l ~ i ~ n) such that fi E R-P. By Corollary to Proposition 8, 2.3, we have M p c:>!. (Mfi}PRII and hence Mpis free over R p of the same rank as Mr,. Moreover rank M p = n if and

HenceM' is finitely generated projective over S. Since M' c:>!. M I8l s

whole ring R. In particular 1 =

,

'

I

I

R

and S is faithfully flat over R, M is finitely generated projective over R. Corollary: Any finitely presented flat module is projective.

S8

COMMUTATIVB ALGBBRA

Proof: Follows from condition (ii) and Corollary to Theorem' 2 2.2. ' .If R is a local ring and M af.g. projective R-module then M is free, say of rank n. Then M p is also Rp-free of rank n so that the function P -+- rank M p is constant. However, in general, the rank function is only locally constant by condition (iii) of Theorem 1. It will be a constant if Spec R is connected. The following Proposition gives condition under which Spec R is connected,

i T

LOCAUSATION

Proof:

I

'1'\

Proposition 5: (Patching up of Localteations). Let R be a commutative ring, f l , I. E R with RJ. + R/I = R. Let Ml and M I be R" and R,.-modules respectively such that (M')/. (M.)". Then there is an R-module M with M, = M't, i = 1,2. If M l and M. are finitely generated (finitely presented, finitely generated projective) thenM is also finitely generated (finitely presented, finitely generated projective). Proof: Let CIt: (M1 )1. -+ (M.)" be the given isomorphism. Define and a map tf.: M l EEl M I -+ (M.)" by .p(X, y) = lX.(xI.)-Y/~where y" are the images of x and y in (M;)I. and (M I ) " respectively. If M = Ker .p, then we show that M is the required R-module. Consider the exact sequence

X,.

Proof: Let X be disconnected so that X = Y, U YI , Y.. Y. nonempty closed, Y, n Y. = rP. Write Yj = V(I,), i = I, 2. Then V(J. + II) = Y, n Y. = rP implies II + II = R. Let 1 = e1 + e . Now elel is nilpotent, i.e. ~e: = O. Replacing luI by R:~ and Re; we may assume that II n I. = O. Hence e,: e. are ~utually orthogonal idempotents. They are non-trivial as Yi =;6." ' I = I, 2. Conversely assume that e1 , ' . are mutually orthogonal nontrivial idempotentswith 1 =e1 + el' Let Y t = V(R(l-e,», i= 1,2. The relation 1 =0 l-:-e1 + e,(I -e.) shows that '" = V(I) = Y, n Y •. Also Y1U Y. = V(R(I- el)(I-el = V(O) = X. Since e, are. nontrivial, Y, tf. (i = I, 2), so that X is disconnected.

o -+ M

-+- M l

Ee M I

'"

----+ (M.)/t.

Localising at I" we have an exact sequence

."

0-+ M,,-+- M, Ee (MI )f1 - - (MI)!l'

Now rP!l restricted to (M.)/t is an isomorphism so that M" =: MI' Similarly.Tocalising at/l , we have an exact sequence

»

';1.

0-+ MI. -+ (Ml ), . Ee M. - - (MI ) "

which gives the isomorphism M,."""M•. Suppose now that M l and M. are finitely generated. Choose Xl' X•••• X. EM such that if M' is the submodule generated by {Xl' ... X.}, then I, = M l and = MI' Then for a~y prime ideal P, either.li ¢P or I.¢ P since Rll+RI.=R. Hence M p """ M p • thus M """M' and M finitely generated. If M l and M. are finitely presented, then there exists an exact sequence

Corollary 1: Spec R is connected ~ R has no idempotents other than 0 or 1. In particular, if R is a domain, Spec R is connected and the rank function is constant.

M

DefiDition: A projective R-module M is said to be of constant rank n if for every prime ideal P of R, M'1' is free of rank n over R p • Corollary 2: For an R-module M and n;;' I, the following conditions are equivalent. (i) M is projective of rank n over R. (ii) M m is free of rank n over R m for every maximal ideal m. (iii) M p is free of rank n over R p for every prime ideal P of R. (iv) For every maximal ideal m, there exists somelE R-m with M, free of rank n over R,.

Follows from Theorem 1.

=

Proposition 4: X = Spec R is disconnected .. 1 = el + e. where e1, e. are mutually orthogonal, non-trivial idempotents in R.

'*

59

Mr.

O-+K-+F-+-M-+O

-'

where K" and K,. are finitely generated and F is free of finite rank. By the above argument, K is finitely generated and hence M is finitely presented. , If M l and M I are finitely generated and projective, then M p is free for all prime ideals P and M is finitely presented. Therefore

60

LOCALISATION COMMUTATIVE ALGEBRA

(i) rank (M ® N) = (rank M).(rank N) (ii) rank M* = rank M follow by localisation at every prime ideal. Consider the set C of isomorphism classes of projective Rmodules of rank one. The operation on C given by [M] + [N] = [M ® NJ, where [M] is the isomorphism class containing M is a

M is finitely generated and projective, by Theorem 1. We now study properties of rank one projective modules.

Theorem 2: LetM be a finitely generated Rsmodule. The following conditions are equivalent (i) M is projective of rank lover R. (ii) M ® N is isomorphic to R for some R-module N. •

M*

R

well defined operation. Theorem 2 shows that C is an abelian group for this operation with [R] as the zero element and [M*] as the inverse of [M]. This group is called the Picard group and is denoted by Pic(R). The Picard group is important in the study of geometry of hypersurfacea.

R

Moreover if (li) holds, N is isomorphic to the dual module HomR(M, R).

=

Proof: (i) => (ii),

Let u: M ® M· ...... R be the natural map .

R

definedbYu(x,f)=f(x),XEM,fEM*. If m is any maximal ideal of R, (M ® M*).. "'" M .. ® (M*)... Since M is finitely preR

2.4. EXERCISES

R..

sented and R.. is a fiat R-algebra, it follows from Exercise 7,2.3, (M·).. is isomorphic to the dual (M:') = HomR..(M... R..). Thus the localisation map u.. : (M ® M*) .. ... R.. induces a map

1.

2.

,

is an isomorphism and hence u is an isomorphism. This implies (ii). (ii) => (f). By Theorem I, it is enough to sho';'" that M is free of rank one under the assumption that R is a local ring. Let m be the unique maximal ideal of Rand k = Rim. Given an isomorphism u: M®N ...... R. we have an isomorphism u: MlmM®N/mN R

k

R

R

Proof: Clearly M ® Nand M* are R-projective if M is R-projecR

+

i Rfj = '-1

5.

R-module. Suppose XI E Mft are such that XI and XJ have the same image in Mftfj under the natural mappings Mit ...... MIIIJ, and MJj ...... MJilp i =1= j, Show that there exi~ts a unique x E M such that canonical image of X in M Jj is XI (1 ,.;;; i:;;;"n). Show that if M and N are finitely generated projective modules of constant rank, the following modules are also projective of constant rank and the ranks satisfy the conditions as below. (i) rank (M EB N) = rank M + rank N. (ii) rank (M ® N) = (rank M).(rank N).

R

R

R

Let M be a projective R-module of rank n. Show that there exists a finitely generated faithfully nat R -algebra A such that the scalar extension A ® M is free of rank n over A. Let (Ji)l where

1-1

4.

"",'(M® N)® M* "",R ®M* cy-M*. R

=;

R

® N "'" (M ® M*) ® N R

R I and P

P, is a projective R, module of rank n" and the n, are distinct. Let R be a domain with quotient field K and Man R-module. Show that the following conditions are equivalent. (i) M is finitely generated and projective over R. (ii) M is finitely generated and M .. is free over R .. for each maximal ideal m. (iii) M is projective and dimg K ® M is finite. .

3.

...... k, Thus M/mM has rank one as k-vector space Hence Mis , R-cyclic. The annihilator of M also annihilates M ® Nand hence zero. Thus M is isomorphic to R and it is free of rank one overR. Finally, if M ® N"", R, then N CY- M* under the isomorphisms N"", R

= ; 1-1

u..: M .. ® (M".)· ...... R... Since M", is free of rank one over R.., u.. R..

Let P be a finitely generated projective R-module. Show that Rand P can be decomposed as R

R

tive, The relations

61

Rand M an

62

COMMUTATIVE ALGEBRA

= (rank M).(rank N). rahk(AM) = ( ~ ). where n= rank M.

(iii) rank HomI!.,M. N)

(iv)

6.

If A is an R-algobra and P is a finitely generated projectlve. R-module -of rank n. show that A ® P is finitely generated

CHAPTER In

R

7.

A-projective of rank n. Let f: R -+ Sbe a r.ing homomorphism. Show that the natural map Pic(f): Pic(R) -+ Pic(S) given by [P)-+ [S ® P] is weU de-

NOETHERIAN

RINGS

R

•.' 8.

·fined and is a group homomorphism• Let M be a finitely generated R-module. is called basic if and only if its

An element x E M

i~~ge in p~p is non-zero

An affine variety obtained as the common zeros of polynomials of an ideal in k[XI • Xz.....X,J. k field, can be obtained as the commOn zeros of a finite number of polynomials. ' This is due to a Theorem of Hilbert which says that every ideal in k[XlO .... X.),.k field, is finitely generated. Rings with this property are called Noetherian rings. Every ideal in such a ring can be written as a finite intersection of primary ideals. This is the algebraic analogue of expressing an affine variety as a finite union of irreducible subvarieties. In such'il.decomposition, there are embedded and non-embedded components. The latter are shown to be unique while the former may not be unique. Noetherian rings are best studied in a general set up by studying Noetherian modules. These have the property that every submodule is finitely generated or equivalently any strictly increasing chain of submodUles is finite. Modules with the corresponding property for decreasing chains, lead to the notion of Artinian modules. Artinian rings are shown to be nothing. but Noetherian rings in which every prime ideal is maximal. For finitely generated modules over an Artinian ring. we define the concept of length which generalises the concept of dimension of a vector space.

for

every prime ideal P. Show that for a finitely generated pro-' jective R-module M. x E M is basic if and only if it is unimodular.

3.1.

1

lSoetherianmodules

Proposition 1: Let M be an R-module. The following conditions are equivalent. (i) Any rion-empty collection of submodules of M. haa a maximal element. (ii) For' any increasing sequence of submodules of M.

1 64

NOETHERIAN RINGS

COMMUTtJlVB ALGBBllA

M1 C M. c ... c M. c ..., there exists some integer m such that Mil ""' M .. for all k > m. . (iii) Every submodule of M is finitely generated.

Proof: (i) ~ (ii), Let any increasing sequence of submodules of M. say M 1 C M. c .... c,M• .... be given. Consider the collection :E :;=. {M.}. This collection hll'l a maximal element M",. Then M.;= M", for all k ;;;, m. . (ii) .. (iii). Let N be a submodule of M. Choose any Xl E N. If RX1 =1= N. choose x. E N. x. ¢ Rx1 • If RX1 + Rx. is not equal to N continue this process. By assumption (ii), after a finite number of steps we have RX1 + Rx. +....+ Rx, = N. Thus N is finitely generated. (iii) .. (ii), Consider any increasing sequence of submodules of M say M I

C

Me c ...c M. c.... Then their union N =

UM.

is a submodule of M which is finitely generated by (iii). If {Xl' ~., X,} is a generating set for N. choose m such that N = {Xl' X X,} c M",. Then MII;= M", for all k;;;. m. (ii) ~ (i). Let}; be a-non-empty collection of submodules M. Choose any M 1 E li. If M 1 is not maximal. choose M E M 1 C M.. IrM. is not maximal continue this process. After

. .

of

65

is also a submodule of M. iC?vis Noetherian. Now every submodule L of M" is isomorphic to MI/M' where M 1 is a submodule of M. Since M is Noetherian. M I is finitely generated. Hence L is also finitely generated. This implies that M" is Noetherian. Conversely assume that both M' and M" arc Noetherian. Identifying M' with a submodule of M. we have M" "'" M/M'. For any submodule N of M. we have N/NnM' C! N + M'/M',a submodule of M". Since Mil is Noetherian, it is finitely generated. Since M' is Noetherian NnM' is. finitely generated. These imply that N is finitely generated and hence M is Noetherian.

•

Corollary 1: If {M,h..,... arc Noetherian, then EB }; M, is Noethe1-1

rian.

Proof: I Consider the exact sequence 11-1

II

0-+ ffi }; M,-+ EB }; M,-+M. -+ 0 '-1

,-]

where the first map is the inclusion and the second map is the projection onto M n' The result follows by induction on n.

I.

a

=1= finite number of' steps, there exists someM", E I which is maximal for otherwise condition (ii) will be violated. •

Corol1ary 1: Any homomorphic image of a Noetherian ring is Noetherian.

DefiDition: An R-module M is called Noetherian if it satisfies any one of the above equivalent conditions.

Proof: If R is a Noetherian ring and I an ideal in R, R/I is a Noetherian R-module and hence also Noetherian R/I module. Hence R/I is a Noetherian ring.

DefiDition: A ring R is called a Noetherian ring if the R.module R is Noetherian.

Corollary 3: Let R be a Noetherian ring and M a finitely generated R-module. Then M is Noetherian.

Examples: (i) Any field R is a Noetherian ring. (ii) Any finite ring R is Noetherian. (iii) Any principal ideal ring is Noetherian.

Proof: Since M is finitely generated, M is isomorphic to a quotient of the free module R», By Corollary I, Rn is Noetherian and hence M is Noetherian.

Proposition 1: Let 0-+ M' -+ M -+ M" ~ 0 be an exact sequence of R-modules. Then Iv! is Noetherian if and only if both M' and Mil are Noetherian.

Proposition 3: Let M be a Noetherian R-module and S a multiplicatively closed set in R. Then the R s module M. s is Noetherian.

Proof: Assume M is Noetherian.

Since every submodule of M'

Proof: Let·N'be a submodule of M s andN={x E M l(x/I)E N'}. Then N is a submodule of M and N s = N'. Since M is Noetherian,

66

1

COMMUTATIVE ALGEBRA

(,) X} Then {(XI!i),· .. , (xk/ l )} { say N is finitely generated'H bYMxI isN·~~t~~rian.

\

. N' 0 ver R s· ence s generate , . . and S a multiplicatively • If R is a N oetberian ring Corollary. . closed set. the ring R s is Noethenan. If R is a Noetherian ring, I , (H"lbert's Basis Theorem). m. h Teore h' the polynomial ring R[Xl is Noet enan. Proof:

r

t

InK.

:.:'ir/,-.k.

.

• A finitely generated algebra Corollary 2 , R is Noetberian.

X

1>....... Xl•

IS

A over a Noetherian ring

. . X ]!I The result 1• • .. , • : • bra then A R[X Proof: If A IS afg· R-alg e 'd Corollary 2. Proposition 2. now follows from Theorem I an .

=

3.1. EXERCISES are submodules of d Ie (i) If M1 an d M 2 that 1 Let M be an R-mo u are Noetherian, show . M such that M!M1 a~d MIM. MIMl M• is Noethenan.

n

(ii) If I = Ann (M) and M is a Noetherian R-module show that RII is a Noetherian ring. Show that if M is Noetherian and N is finitely generated N is Noetherian. R-module then M ® R _

4.

for some n ~ I. Deduce that nil radical is nilpotent. . Show that every ideal in a Noetherian ring contains a produce of prime ideals.

5.

Let

1>"'

R be a Noetherian ring and f(X) = ~,a,X'. power series in a. is nilpotent. Show that ! is nil-

X over R such that each

Noetherian. Choose f, E • WI d 'b {fi fik} is contain..ed iiSI. .d I I' generate Y 1>" .• f R[X] II = max II" The' ea h K is the R-submodule 0 WeshowthatI=I'~InK:': ereLetfEi. sotbat !=bX,.+· ... generated by {I. X, X '· ..·fX Ifm;;> n-I. write b =},~al+'" bEJ. Ifm';;;;II-I.th~n E I nomial g=f-},lkX" .....:... a }, E R and conSider the po Yh t by successive operations Then deg g';;;; m-l. so t a less than or equal to n-l. Jilk ake the degree 0 f g I' + InK of this type we can m h E I' Hence 1= . 1"his implies!-h E InK, ~r s7; is Noe~herian, InK is finitel.y Since K is finitely generate an R[Xl Since l' = {fl.... .!k}, I IS R and hence o v e r , b . generated over Xl Hence R[Xl is Noet enan. finitely generated over R[ . . Noetherian, the rmg R[

~

67

3. Show that if lis an ideal in a Noetherian ring then (v I)" c I

I

highe~t det~e;:::~e~t~:it:I;~enerated,say by {a l •· ... jk} ~:I~~ is an Idea 0 I itb fi = a,X·' +.... a, E a .,~

Corollary 1: If R is Noetherian.

2.

I

id al of R[Xl. Consider the set J of the ents of I together with O. Then!

Let I be a n~n-zero/

NOIlTFlllRIAN RINGS

potent. Show that a ring in which every prime ideal is finitely generated is Noetherian. (Hint: Let I be the collection of ideals which are notfg. If I :P ",. choose a maximal element of I and show that it is a prime ideal). 7. Let R be a ring such that for all maximal ideals m, the ring RIO is Noetherian. and each non-zero clement of R is contained in only finitely many maximal ideals of R. Show that R is Noetherian. 8. If I is a finitely generated ideal of R such that the ring R] I is Noetherian, show that R need not be Noetherian. Show that under the additional assumption that 1 is nilpotent, R is Noetherian. 9. Let R be a ring and A a faithfully flatR-algebra. Show that if A is Noetherian, then R is also' Noetherian. 10. Let A be a finitely generated R-aIgebra generated by {Xl' X.,....x.} and M a finitely generated A-module. If Xx>xa, ... , X. E'::: AnnA. M, show that M is a finitely generated R-module. 11. Let A be a finitely generated algebra over a field k generated by {Xl' ... , X.} and I the ideal of relations between {Xl' .... X.} with coefficients from k. For any extension field K of k, show that the map! -+ (!(x l ) , .... /(x.» is a bijection between the set of k.homomorphisms of A in K onto the set defined by V(I) = (a., ... ,a.) E K' I f(ax> ....a.) = O. IE I}. 6.

V

3.2. Primary decomposition

by

If M is an R-module and a E R, the map ~.: M -+ M defined = ax. is R·Hnear.It is called the homotbety defined bya.

~.(x)

68

CoMMUTATIVE ALGBBRA

DefiDition: A submodule N of M is called primary if N *- M and , for each aE R, the homothety Ao : MIN~MIN is either injective or nilpotent. An ideal I of R is called a primary ideal if it is a primary submodule of R. It is clear from the definition that N is a primary submodule of M if and only if ax EN, a E R, x E M implies either x E N or aRM C N for some n;> 1. l.e. a E rM(N) = V Ann (MIN). Thus the set of all a E R for which Ao:~/N - MIN is not injective is an ideal equal to rM(N). This ideal is a prime ideal, for if a, b ¢ rM(N), then A", At, are injective and then AOb = Ao'l\b is injective, i.e. ab ¢ rM(N).

r ,

.lI

Definition: If N is a primary submodule of M and P = rM(N), then N is called P-primary. If I is an ideal of R. then clearly r8(I) = VI. If N is a proper submodule of M and P = rM(N) then N is P-primary if and only if ax E N, a E R, x E M implies, x EN or aE P. Examples: (i) R = Z and I ... (p"), P prime. I is a P-primary ideal where P = (P). (ii) Let R = k[X. YJ, k field and I = (X. Y2). Then I is Pprimary where P = (X, Y). (iii) Power ofa prime ideal p. need not be P-primary. Let R

= kIX. Y. ~. k field and P = (XY-

(X,

2). Then P is a prime ideal in

I "

However any power of a maximal ideal is primary. More generally we have the following Proposition.

VI = m a

Corollary: For any maximal ideal m, all its powers m'(i m-primary.

~

1) are

Example: A primary ideal need not be the power of a prime ideal. Let R = k[X, YJ, k field and I = (X, Y2). Then m = (X, YJ is a maximal ideal, with V I = m and so lis m-primary. But I *- m for any i;;;. i. ' Let H be a submodule of M. A decomposition of the type N = N, n N 2 n ... n N, where N , (1 ~ i < r) are primary submodules of M is called a primary decomposition of H in M. The primary decomposition is said to be reduced if (i) N cannot be expressed as an intersection of a proper subset of {N1 , N., ... , H,} and (ii) N , -are P,-primary with all the P, distinct (1 1. Hence ck + A'b = 1, for some A' E R. This implies that a = ack + A' ab E 1, i.e. I is m-primary.

vI

Ker~ocKer~c..... cKerA:C ......

'./

70 COMMUTA'IlVB ALGBBRA

NOETHERIAN RINGS

Since MIN is Noetherian there exists some i for which Ker A~ = Ker A~+1 = ..... If rP = A~. then Ker rP = Ker .p2. This implies that Ker ( 1. as P, is finitely generated. Now )

P7i )

r

J

e n N, = (0) so that 'It P71 e P" This implies that I_I 1=1 PI e P for some i. For any a E P, we have ax = 0 and this implies that the homothety Aa:MIN, -+ MINI is not injective and hence nilpotent, i.e, a E PI. Hence P = P,. Conversely we show that PI E Ass(M), (1 ~ i ~ r). We show for instance that PI E Ass (M). Since the primary decomposition is reduced, there exists some x E N 2 n .... nN" x¢N1 • Since N I is P,"primary. there exists some n » 1 such that P'ix e NI and P'i,lxq:N,. Let YEP'i-'x, withyr/iNl' Then PlY E rl N ,=0 (

'It

1_1

X

1

+

and hence r,c: Ann (y). Conversely if ay=O, aER, then "a:MINI~ MINI is not injective and hence a E PI' Thus PI = Ann (y) and PIE Ass (M).

I

Corollary 1: (First Uniqueness). Let M bea finitelygenerated module over a Noetherian ring R. If'N = ;; N I is a reduced primary de-

-I

composition of N, N , being Prprimary, then the P, are uniquely determined by N.

I

1

I

Proof: Clearly (0) =

n

(NtlN) is a reduced primary decomposition '=L of (0) in Mllf. N'[N being Prprimary. Hence Ass(MIN) = {PI'"'' P r } is uniquely determined by N.

72

COMMUTATIVIl ALGEBRA

NOBTHIlRIAN RINGS

Corollary 2: With the same assumption as in Theorem 2, Ass(lI1) is a finite set and M = 0 if and only if Ass(M) = ,p. Proof:

Fol1ows from Theorem 2.

DefiDition: Let M be an R-module. An element a E R is cal1ed a zero divisor of M if there exists some non-zero x E M with ax = O. Proposition 4: Let Rbe a Noetherian ring and M a finitely generated R-module. Then the set of zero divisors of M is equal U P. to

.. ,

73

Proof: ClearlyF --. P s is a bijection of C onto the set of prime ideals of R s . If PEe n AssR(M). then P is the annihilator of x E- M. x =1= O. and hence P s E AIISR/Ms). Conversely assume that P is finitely generated and P s E AnnR/Ms). We show that P E AnnR(M). Let P s be the annihilator of (xlt) E Ms. x EM, t E S and let P be generated by {oJ' a., ... , an}. Then (adl)(xlt) = 0 shows that for each i, S,O,X = 0, s, E S. If s = 1tSIo then sax = O. for all a E P, so that P C Ann(sx). Conversely if b E Ann(lx) then bsx = 0 so that (bl\) E P s and hence b E P. Hence P = Ann(sx) E AssR(M).

peAss(M)

Theorem 3: Let N be a P-primary submodule of M and P' any prime ideal of R. Let M' = M p,; N' = Npo andf: M -+ Up, the canonical map. Then

Proof: Let a E P for some P E Asd..M). Since P is the annihilator of some non-zero x E M, we have ax = 0, i.e. a is a zero divisor of M. Conversely let a be a zero divisor of M so that there exists some x E M, x =1= 0, such that ax = O. Let 0

=

n

1-1

(ii)

primary decomposition of 0 in M. Then x ¢ N, for some i; If P, is the 'prime ideal corresponding to N ,• then ax = 0 E Nt, x ¢ N" implies that a E Pi' . DefiDition: Let N=

Proof: (i) We have M'/N' = Mp,/Np' ~ (M/N)p », By Proposition 5, ASIIR~,(M'/N') "'" AssR(MjN) n C. where C is the set of prime ideals of R contained in P'. Since N is P-primary. Ass(M/N) = {P} and ifP ¢ P', we have ASSR~,(M'/N') = '" so that M' = N'. (ii) Since Ass(MjN) = {P} and PcP'. the multiplicatively closed set R-P' does Dot contain any zero divisor of M/N. . Therefore the natural map (M/N) -+ (M;N)p. ce M'/N' is injective. This . implies that f-l(N') = N.

nN, be a reduced primary decomposition of

1=1

N. N i being Pi-primary. N, is called a minimal primary component if the corresponding P, is a minimal element of the set {PI' p ...... P,}, i.e, P,:P P J for any 1 =1= i, (1 1. From Theorem 3, we have (N,)pl = M p1 for i > 1 and f- 1[(N1)p,] = N l wherefis the natural map M -+ M h

.

.

But N p , =

n(N,)Pi =

'.1

(Nl)pl' Hence N 1 = f-J(N p,) depends only on N and PI and not on the decomposition.

74

NOETHERIAN RINGS

CQMMUTATlVB ALGEBRA

since P' ::) y'AnnR (M) ::) AnnR(M). (iii) =>- (i). Let P:::> AnnR(M) and suppose M» = O. Since M is finitely generated. there exists some s 1ft P with sM = O. This is a contradiction as P ::) AnnR(M).

We now consider some applications of primary decomposition. Proposition 6: Let R be a Noetherian ring and M a finitely generated R-module. Then

Proof:

Let M

n

P

= V AnnR(M).

P

e

*'

0 and let 0 =

Ass(M)

nN, be a reduced primary decom-

'-I position of Oin M.N, beingP,-primary. Then Ass(MJ={P

Corollary: Ass(M)eSupp(M) and the minimal elements of Ass(M) and Supp(M) are the same.

-. .,

PI .... P,}. If a E y'AnnR(M). !hen a·M = O. for some n ~ I. as M is finitely generated. Hence 'A.: MIN, -.. MIN, is nilpotent for each t. I

(1";;; i";;; r) i.e. a E

n

P,. Conversely if aE

nPh

•


- (ii), Let P E Supp(M) and suppose P:tl P, for any i. Then P:p np,. i.e. P:p AnnR(M). Hence Ml' = 0, a contradiction. I

. no'

If P ::) P', P' E Ass(M), we have P::) Annll(M).

2. 3.

4.

Proof: Let Ass(M) = {Ph Ps,'''p.} so that

-.n

Rr

L p =1= 0 and K p =P O. This proves (ii).

3.2.

I. Proposition 7: Let R be a Noetherian ring and M a finitely generated R-module. For any prime ideal Pof R. the following conditions are equivalent. (i) P E SupP(M). (ii) P::) P' for some P' E Ass(M). (iii) P::) AnnR(M).

(ii) =>- (iii).

*' 0

i

t

S.

6.

EXERCISES

Let R = k[X. Y, ZJ. k field, PI = (X, Y). r, = (X. Z) and m = (X. Y, Z). Show that for the ideal I = PIP-a. the decomposition l=p~nPanml is a reduced primary decomposition. Show that for an ideal I. I = v'T; implies tbat all the primary components of I are minimal. Let R -+ R[XJ be the inclusion map. Show that a reduced primary decomposition of I in R. extends to a reduced primary decomposition of /[X] in R [Xl. Let R = k[X1 , X..... ,X.l. kfield and 1=(X1 , X 2, ... ,X,) (I";;;i";;;,,), Show that powers of I are primary ideals. Letf:M -+ M' be a surjective homomorphism. If N' eM' a primary submodule, show thatf-1(N' ) = N is a primary submodule of M and rM(N) = rM'C,N'). Let M be an R.module. N a submodule of M. ·Prove that Ass(N) e Ass(M) e A~s(N) U Ass(MjN)•

76

GOMMUTATIVE ALGEBRA

7.

Let M be an R-module such that for every multiplicatively closed set S e R, the kernel of M -+ M s is either (0) ot M. Show that to} is primary in M. Let P be a minimal prime ideal of Rand K = Kerf. where j: R ~ R p is the natural map. Show that (i) K is a P-primary ideal and it is the smallest P·primary ideal. (ii) The intersection of all K, when P runs through all the minimal prime ideals of R is contained in the nil radical of R.

8.

I

T

3.3. Artinian modules

NOI!THI!RIAN RINGS

77

(iv) R = Z is not an Artinian ring as the sequence of ideals (n) ::J (n") -:::> (nS) -:::> ..... is not stationary, nEZ, n > 1. (v) R = k[X], k field is not Artinian, as the sequence of ideals (X)::J(X")-:::>..... (X·)-:::> ..... is not stationary. Butk[Xl isa Noetherian ring. (vi) Let R = k[X], X., ... " X., J, the polynomial ring in infinite number of variables X.. X.' , X., ... over a field k, Then R is neither an Artinian nor a Noetherian ring. Proposition 2: Let 0 ~ M' ..... M ~ M" ..... 0 be an exact sequence of R-modules. Then M is Artinian if and only if both M' arid M" are Artinian.

Proposition 1: Let M be an R-module. The following conditions are equivalent. (i) . Any non-empty collection of submodulesof M has a minimal element. (ii) Any decreasing sequence of submodules of M of the type M o -:::>M] -:::> M 2 -:::> ... -:::> M. -:::> ... is stationary, i.e. M k = M k+] = ... for some k:> O.

Proof: Identifying M' as a submodule of M, any strictly descending chain of submodule of M' (or M") gives rise to a strictly descendiug chain of submodules of M. Hence M Artinian implies both M' and M" are Artinian. Conversely assume that both M' , and M;' are Artinian and let M = M o ::J M, ::J M. ::J ....•• be a decreasing sequence of submodules of M. Consider the sequence MIM' ::J (M] + M')IM' ::J (M. + M')/M' -:::> ••••• Since M/M' 0< M" is Artinian, there exists some i for which M 1 + M' = M'+l + M' = ~ . .. Since M' is Artinian, the sequence M' n M ::J M' n M 1::J M' nM. ::J .... is stationary. There exists some integer which can be assumed to be i (by choosing the maximum) such that M'IlMI = M'nM,+] = ..... ' Since

Proof: (i) => (ii). Consider a decreasing sequence of submodules Mo::J M] :::> M.::J ... ::J M.::J ..... Then the collection ~ = {M.}~o

has a minimal element M,. Then M 1 = M,+1 = .... (ii)=> (i). Let ~ be a non-empty collection of submodules of 'M. Since ~ =I',p, choose any M o E~. If M o is not minimal, there exists some M 1 E~, with M] e Mo. If M] is not minimal, =l= there exists M 2 E ~ with M 2 c u ; This process will stop after a =l= finite number steps yielding a minimal element of ~ for otherwise condition (ii) will be violated.

M'+l eM" Ml+lnM'

= M,nM' and

M i +1

+ M' = M,+M'

Definition: An R-module M is called Artinian if it satisfies any one of the above two.equivalent conditions.

we have by an easy computation that M, = Artinian.

Definition: A ring R is called an Artinian ring if it is an Artinian ' R-module.

Corollary 1: If {M,hoC'''. are Artinian R-modules, then EB ~ M, is Artinian. '

Examples: (i) Any field k is Artinian. (ii) Any finite ring is Artinian. (iii) Let G C QIZ consist of all elements whose orders are powers of a fixed prime p. Then it can be shown that any proper subgroup of G is finite. Hence G is an Artinian Z-module. It is not Noetherian over Z.

Ml+l'

Hence M is

I

I

-+

Proof:

Consider the exact sequence n-l

O~EB ~ I-I

\

,.

M,-+fB

~ M, ..... M......o, '_I

where the first map is the inclusion and the second map Is the

I

!

NOETHERIAN RINGS

78

79

COMMUTATlVB ALGBBRA

projection onto the n-th component. The result follows by induction on n, Corollary 2: Any finitely generated module over an Artinian ring is Artinian, Proof: Let M be a finitely generated R-module. Write M"", F/K where F"", R" is a free module of rank n. Since R is Artinian R-module, so is F, by Corollary I. This implies that M"'" F/K is Artinian. Corollary 3: If I is ideal in an Artinian ring, the ring R/I is an Artinian ring. Proof: Since R is Artinian, R/I is an Artinian R-module. Since it is also an R/I-module, it is an Artinian R/I-module l.e, an Artinian ring. Proposition 3: Let R be lin Artinian ring. Then every prime ideal of R is maximal.

Proposition 5: If R is an Artinian ring, the nil radical N(R) is nilpotent. Proof: Let N(R) = I and consider the sequence of idealsl::J 12::J ... ::J1"::J .... There exists some i for which I' = 1i+1 = .... = J (say). If J = 0 then I is nilpotent. If J 1= O. we obtain a contradiction as follows. Let I = {K IK ideal in R with KJ ¥- O}. Now I ¥- .p, as JI = J ¥- O. Hence I has a minimal element Ko• Now K o is principal, for if a E K o' with aJ ¥- 0, then Ra C Kg and hence Ra = K o, by minimality. Since (aJ)1 = aJ' = aJ ¥- 0, by minimality aJ = Ko = (a). This implies that a = ab for some b E J. But bE J c 1 = N(R), so that b" = 0 for some n ~ I. Hence a = ab = abo = .... = abo = 0, a contradiction. Corollary: Proof:

In an Artinian ringR, the Jacobson radical is nilpotent.

Follows from Corollary to Proposition 3.

Theorem 1: (Structure of Artinian Rings) An Artinian ring is uniquely isomorphic to a finite direct product of Artinian local rings. Proof: Let ml' ml.....m, be tbe distinct maximal ideals of R. Then

Proof: By passing to the quotient, it is sufficient to show that an Artinian domain is a field. Let R be an Artinian domain, a E R, a¥- O. The sequence of ideals (o.)::J (0. 1) ::J.... ::J (a") ::J...•, is stationary, so that (a') = (a'H) for some i. If a' = ba ' H , bE R, than I = ba, as a' ¥- O. Hence R is a field. Corollary: For an Artinian ring, the nil radical is equal to the Jacobson radical.

J(R)

= n" m, =

'"

1=1

k

~

1-1

(' )k =

m, is nilpotent. Hence '" m, 1

0 for some

I. By the Chinese -Remainder Theorem, the natural map

R ~ ~ R/m~ is an isomorphism. 1-1

-

Since R/m7 has a unique prime

ideal m,/m7, it is a local ring. Since R is Artinian, so is R/m7. Thus R is a finite product of Artinian local rings. To prove the uniqueness, assume that R

= ~

R , where each

1=1

Proposition 4: An Artinian ring has only finitely many maximal ideals. Proof: Consider the family:E of finite intersections of maximal ideals of R. Since I =i= .p and R is Artinian, :E has a minimal elem.. Now for any maximal ideal m ment. say I = ml ml of R, m ml m, c I, and is equal to I by the minimality of I. Hence I c m and this implies that m, c m for some t, (I.e;;; i.e;;; r) i.e, m, = m. Hence the only maximal ideals of Rare mit milo •• , m,.

n n... n n n... n

R , is an Artinian local ring. Since every prime ideal of an Artinian

ring is maximal, each R, has a unique prime ideal. Let "'I : R _ R , be the i-th projection and I, = Ker '1 VI ::::> ... ::::> Vn = {O} is a composition series of V. (ii) Let G (a) be a cyclic group of order 6 and H the subgroup generated by aa. Then G ::J H::J {e} is a composition series

=

ofG. (iii) An R-module M may not always have a compositions series, The Z-module M = Z has no composition series. Definition: Given a composition series of M.say M =Mo::JM1::J... ::J Mn = {O}. the number nis called the length Qf the composition series. Example: In the first' example above. the length of the composition series is 11. the dimension of the vector space: The following Theorem shows that if an R-module M has a composition series. its length does not depend on the composition series.

8.

Show that R=~ R, is an Artinian ring if and only if each R, is

Theoreni 1: (Jordan Holder Theorem) Let M be an R-module haviJ:lg a composition series of length n. Then any other composition series of M also has length n and any decreasing chain of submodules can be extended to a composition series of M.

9.

an Artinian ring. Let R be an Artinian ring with N(R) ,. O. only finitely many ideals.

Proof: Let 1(M) denote the least length of a composition series of M. If N is a submodule of M. we show that leN) ~ I(M) and for

'-1

Show that R has

82'

COMMUTATIVE ALGEBRA

a proper submodule N of M we show that leN) < I(M). Let M = M o ::::> M 1 ::::> ... ::::> M, = {O} be a composition series of Mofleastlength 1=I(M). Then N=NnMo::::>NnMl::::>"'::::> N n M, = {O} is a decreasing sequence of submodules of N. For each i, the natural map NnM;/NnM/+1 _ M//M'+I is injective. Since M,jM'+1 is simple, either Nn M/ = NnM'+1 or NnM// Nn M'+1 is-simple. This gives rise to a composition series of N of length ~ I by omitting the repeated terms. Hence leN), the least length of any composition series of N is ~ I = I(M).' Suppose leN) = leM). Then, for each i (0 ~ i ~ I-I), Nt> MtlNnM'+I'*O and is isomorphic to M;/M/+1" Since M, = 0, we have N n M'~I = 'M t-1· This implies in turn ivn M'-a = Mt~., etc. and finally NnMo=Mo,i.e.N=M. Hence for a proper submodule N of M, leN) < I(M). . Now consider any composition series of M of length k, say M = M o::::> M 1 ::::> ... ::::>.M" = (0). This implies 1= I(M) > I(M1) i> ... > I(M,,) = (0). Hence k. Since t is the least length of a composition series of M we have I ~ k, Hence all the composition series of M have the same length t = 1(M). Now consider 'any decreasing chain of submodules

I>

M=Mo::::> M 1::::> ... ::::>Mm={O}.

=1=

=1=, =1=

If m < I(M), it is not a composition series of M and hence for some i, M,/M'+1 is not simple, i.e. there exists submodule L, with M, ::::> L ::::> M,+l' Consider the new decreasing chain, by adding

=1=

=1=

, NOETHERIAN RINGS, 83 Proof: Assume that M has a co .'. n any strictly increasing or decrea;POSltlO series of length n, Then M has length ~ n by Theorem I In~ sequenc.e of submodules of and an Artinian'module T . ence M IS both a Noetherian is both Noetherian and ·Arti~I·PanrOVes!hecMon~erse, assume that M IS . has a proper maximal submodule M . Nmce . No etheri enan It and hence it has a proper maxi~ I o~ M 1 IS a also Noetherian, this process, we have a decreasin a ~u mfodule M.. Continuing M = M ::::> M g ch~In 0 submodules. =1= 1 ~ M a ~ .... SlDce Mis Arlinian, M" = (0)' f~r some k; Then M =M ::::> M '. non series of M. 1::::> '" :J M" = (0) IS a composi-

°

°

ProposltJon 2': ,Let 0 .... M' I'M" g " --l> - - l > M" 0 be ,sequence of R-modules. Then I(M) = I(M') I(M;;). an exact

+

Proof: M is Noetherian (Arli' ). H DIan ~f and. only ifboth M' and Mn both ~(M') and I(M") II '11' ence (M) IS finite if and only if re mte Th It"· one of the modules has lenath e re a Ion IS clearly true if any lengths . ~e finite. Let M' ~ M~OCI~ ;!sume now, that all the composition series of M' wI'th '(M') k 1 ::::> ... ::::> M" = (O) be a '1- .. ' II = andM"=M" " ::::> M, :- O.a composItIon series of M" with '(M" ° ::::> M 1 ::::> ... the senes !' ) = I. Consider are Noetherian (Artinian)

M =g-l(M") ::::>g-I(M;)::::> ... ::::>r1(M;)

L to the chain, M==Mo::::> M 1::::> ... ::::>M,::::>L::::> M ,+1::::> ... ::::> M m = {O}.

=1=

\

~

1('1 "", •

If m

+ I (iii). Every P E Supp (M) contains some P EAss (M). (iii) => (i). Choose a decreasing chain .M = M n~ Mit-I ~ ',' •• ~ = {O} with M,/Mt_ l ~R/P" P,-pnme (l::;;; i ~ n). ~1Dce {P 10 •••• , Pn} C Supp (M) and every P E Supp .IS ~axlmal, we have that R/P, is a field (I::;;; t c; n). ThIS Implies that IR(M,/M'-I) = I. (1 ~ i::;;; n). Hence IR(M) < 00 by Proposition 2.

u,

(ft!>

1

NOBTHBRIAN RINGS

85

Corollary 1: Let M be a finitely generated R-module over a Noetherian ring R with Is(M) < 00. Then Ass (M) = Supp (M). ""' Proof: If P E Supp (!vI), then P ~P" E Ass (M) and hence p = P" E Ass (M). Hence Supp (M) = Ass (M). Corollary 2: Let M be a finitely generated module over a Noetherian ring Rand P a prime ideal of R, Then M p oF 0 is of finite length over R» if and only if P is a minimal clement of Ass (M). Proof: We have by Proposition 5, 3'.2. AssRp (Mp) = {Qp I Q E Ass (M), Q c Pl.

Now 1s,.(Mp) < 00 if and only if every clement of AssRp(Mp) is maximal. Since R p is local with unique mllllmal ideal PR p , we have JRp (Mp) < 00 if and only if there exists no Q E Ass (M) with QcP. But MpoFO*PESUPp(M)*P~P', P'EAss(M).

=F

HenceMp-:j:.O and lRp(Mp)


(ii) by Proposition 3,3.3. (ii) => (iii) is obvious. (iii) => (I) By Theorem 2, We have IR(R) < implies (i).

00

and this

Proposition 5: A ring R is Artinian*length of R as an R-module is finite. Proof: If lR(R) < 00, clearly R is Artinian. Conversely assume that R is Artinlan. Then J(R) is nilpotent by Corollary to piot

position 5,3.3. Since J(R) .

(n m,)n == I

'II:

,

== , nI m" m,

•.

maximal and J(R)"=

m: = 0, (0) is a, product of maximal ideals, say .

0= mimi' . , .11Ik, m, not necessarily distinct. Consider the sequence

86 COMMUTATIV!l ALGEBRA

NOI!THI!RIAN lUNas

5.

of ideals R = mo::J m1::J mimi ::J •.. ::J m1mB.•. mlc-1::Jm1ma... m" = O. Since R is an Artinian R-module, m 1 , •• m,_Jm l ••• m, is also an since it is also a~ Rlm,-module Artinian R-module (I ~ i ~ k) it is an Artinian Rlm,-module, i.e. a finite dimensional vector space over RIm,. Hence it has finite length as RIm, module and also as an R-module. This implies th.at MR) ~i).

If at satisfies the relation

a..,.: a1at·-1 +

+ a; = 0;

a, E R,

then 11." = -alat°-l-aIOtIl-I-ao, so that I, «. «I, .... , ato- 1 generate R[«] as an R-module. (ii) ~ (iii). Take R' = R[a.] (iii) (iv), Let R[a.] c R' c S, where R' is a subring of S finitelygenerated as an R-module. Then M = R' is an R[a.]-module, which is faithful because xR' = 0, x E R[«] implies xl = x = O. (iv) => (i). Let M be a finitely generated R-module which is faithful as R[Ot]-module. The result follows from the following Lemma by taking 1= R.

*

Lemma: Let M be a finitely generated R-module which is faithful as an R[at]-module and I an ideal of R such that «M c 1M. Then a. satisfies a relation of the type at° + OtatoH + ....

+ ao = 0, af E 1.

90

COMMUTATIVE ALGEBRA

INTEGRAL BXTENSIONS

Proof: Let M be genera ted by {Xl>

aM C 1M C '~1 u; we have rxx, = system of equations

.

7

(3/}«-0I}) x)

XS • • • • •

IOI}Xh

x,} over R.

Since o

on E I. Consider the

Corollary 1: Let

Clt10

.•••

+

0,

= 0, 1 E;;; i:S;; n.

= 0,

Definition: If the integral closure of R in S is the whole of S then S is said to be all. integral extension of R. Proposition 1: Let ReS c T, be ring extensions. If Sis integral over R, and T is integral over S, then T is integral over R.

)

Proof: If oc E T, then oc satisfiesa relation of the typAn+ 0IOC·-1+ ... + o. = 0, aj E S. Then oc is integral over R' = R[a1• a., .... an], so that R'[oc] is a finitely generated R'-module. Since S is integral over R, by Corollary I to Theorem 1. R' is a finitely generated R-module. Hence R'[Gt] is a finitely generated R-module. Since R c R[«] C R'[rx] C T, CIt is integral over. R, by Theorem 1.

Let A be the determinant of the matrix (30 « - 01) . Then Ax = 0 1 E;;; IE;;; n ~d this implies that A = 0 asM is a faithful R[«].m~dule: By expanding A, we have CIt'+ 111«n-l+

91

aiEl (1 E;;; iE;;; n).

«s, ... , «, E S be integral over R.

Then

Corollary: If closed in S.

R[«" CIt••...• «.] is a finitely generated R-module.

R is the

integral closure of R in S,

R

is integrally

Proof: Let oc E S be integral over R. Since R is integral over R. oc is integral over R and hence CIt E R. Thus R is integrally closed in S.

Proof: The proof is by induction on It. For n = 1. the result follows by Theorem 1. Assume the result for (n-I) so that ~ ["1, «" ... , «0-1] is a finitely generated R-module. Now «. is Integral over R and hence over R [0'1' rxl. • ••• 1Xn-1]' Hence R [«t.~, '; .. , .«,] is a finitely generated R [1Il1•....• «.-1] module and this Implies by transitivity that it is a finitely generated R-module.

Proposition 2: (i) Let S be an integral extension of R, J any ideal of S and 1= JnR. Then S/J is integral over R/I. (ii) For any multiplicatively closed set Tin R, ST is integral over R T•

Corollary 2: The set of elements of S integral over R is a subring of S containing R.

o

Proof: If «, () E S are.integral over R. then R [oc. ~] IS a finitely generated R-module. Since e ± ~, and IX~ lie in R[oc, ~], the result follows from Theorem 1.

Proof: (i) Let i = CIt + J E S/J. Clearly R/I is a suhring of S/J. !f« satisfies a relation of the type•. «' + a101.· - 1 + ... + a. = 0, OJ E R, by passing to the quotient modulo J, and identifying R/l as a subring of S/J, we have i" + iiI in-I + ... + ii. = O. ii, E R/l. Hence S/J is integral over R/I. (ii) Let ( (ii) by Corollary to Proposition 1. (ii) => (iii) is clear. _ To show that (iii) => (i), let R ~ the integral closure of R in its quotient field K and i: R ~ R the inclusion map. Then

Proposition 4: Let ReS be extension of rings and R the integral closure of R in S. Then the integral closure of R[X] in SIX] is

RlX]·

98

INTEGRAL EXTENSIONS COMMUTATIVE ALGEBRA

Proof: Since R is integral over R, R[X] is integral over R[XJ (Exercise 7, 4.1). Conversely let I(X) E S[Xj be integral over R[X]. It -, satisfies a relation of the type In + gdn-l +...+ gn = 0; gj ER[X]. Choose m to' be greater than n, and the degrees of g,(l ~ i ~ n). If heX) = I(X)-xm, then

I and

(h + xmt+ gl(h + Xm)n-1 +... + gn = 0 i.e. hn + h1hn--1 +...+ h« = 0, where hn = Xmn + gIXm(n-l j +...+ gn E R[X]. This implies that -hW-l + h1h n-a + .::+ hn- l ) = h. E R[Xj. Now hn has coefficients in R, i.e. integral over R. Applying Proposition j to the pair of polynomials on the left hand side, we have heX) E R[X]. This implies that I(X) = Xm

+ heX) E

R[X].

Corollary 1: If R is an integrally closed domain then R[X] is also integral1y closed. Proof: Let K be the quotient field of R. Let IX E K(X~ be integral over R[Xj so that it is integral over K[X]. Since K[X] IS a unique factorisation domain, it is integrally closed so that IX E K[Xj. Since R[ Xj is integrally closed in K[ Xj, IX E R[ X). Corollary 2: If R is an integrally closed domain, then R[Xw"'Xn] . is integrally closed. Proof: Follows from CoroIlary 1. In Theorem I, 4.2, we studied 'the going up property for an integral extension, i.e. an increasing chain of prime ide~ls in R, can be lifted to a corresponding chain in S. The following Example shows that a similar property for decreasing chain of prime ideals is no longer valid without additional assumptions on Rand S. Example: Let R=Z and S=Z[Xj/l where 1=(2X, xa-~). Then P/ = (2, X -I).where X = X +lis a prime ideal of.S Iy.mg ~b?n P = 2Z and there is no prime ideal of S contatned In Plying above (0):

99

We.propose to study the going down property for ring extensions. Definition: Let R be an integrally closed domain with quotient . field K wid L a normal field extension of K. Let G be the group of all K automorphisms of L. If S is the integral Closure of R in L, S is called a normal extension of R with the Galois group G = G(L/K). Proposition S: Let R be an integrally closed domain with quotient field K and S a normal extension of R with Galois group G = G(L/K). Then (i) G is the group of R-automorphisms of S. (ii) Two prime ideals P/ and Q/ of S lie over the same prime ideal of R if and only if there exists some a E G with a(P') = Q'. Proof: (i) For any aEG=G(L/K)clearlyo(S)=S, so that 0 is an R-automorphism of S. Let 0 I S = or I S for a, T E G. Since L is the quotient field of S we have 0 I L = TIL, i.e, a = or. Moreover any R-automorphism of S can be extended to a K-automorphism of L. This proves (i). (ii) If there exists 0 E G with o(P') = Q/, then clearly P' and Q' lie over the same prime ideal of R. Conversely assume that P' and Q' lie over the same prime ideal P of R. We prove the result first, when G is a finite group. If Q' -=F 0 (P') for any o E G, then there exists a E' Q' with a ¢ a(P') for all a E G. Let b = 'It a(a). Then a(b) = bfor any a E G and hence b is purely aeG

inseparable over K. i.e, bm E K for some m :> 1. Then Q' n R = P contains 1Jm since a E Q/ and R is integrally closed in K. But P' n R = P does not contain bm since P' is prime and o(a) ¢ P' for all a E G. This is a contradiction. To complete the proof, we have to consider the case when G is infinite. Consider the set :E of all pairs (La, a,,) where L" is a normal extension of K and a" E G(LJK) such that La eLand a..(P' nL,,)=Q' nL". Introduce an order-s; in:E by defining (L,.,a..)
x;

Xf". ..x:"

is

another

«;

is a submodule of a free R-module. S is also R-free with basis {flJft,· ... fm}(m,.;; PI). Now {fi. !.,..·.fm} being a basis of S over R.

'-I

If M' =

1

'-I

=

~ m,oc,.

x:·

monomial occurring in/, then it is easy to verify that w{M»w(M') if and only if (ocl' ocf... ·.oc.) > ( (iii) is clear. (iii);> (i). Let M(R: M) = I an integral ideal of R. Then Mm(Rm:Mm) = 1m for every maximal ideal m of R. By assumption Mm(R m:M m) = R m so that 1m = Rm for each maximal ideal m. Since t c: R, this implies 1= Rand M is invertible. I

Proposition 3: Let R be a local domain. Every non-zero fractionary ideal of R is invertible if and only if R is a DVR. Proof: Let m be the unique maximal ideal of R. If R is a DVR, m = (I), for some 1 E R. If M is a fraetionary ideal, then aM t: R,

124

COMMUTATIVE ALGEBRA

for some a E R, a 0/:- O. Then aM = (t r), so that : M = Ra is principal where a. = tria. Hence M is invertible. Conversely assume that every fractionary ideal of R is invertible. In particular every integral ideal is invertible and hence finitely generated. Thus R is Noetherian. By Theorem 2, 5.2, it is sufficient to show that every non-zero ideal of R is a power of m. Suppose this is not true. Let I 0/:- .p be the collection of non-zero ideals of R which are not powers of m. Let 1 be a maximal element of :E. Then 10/:- m and hence 1 em. Since m is invertible this implies m-11 c R . 11' =1= • .=1= t.e. m- IS a proper integral ideal of R. Now 1 em-II, as x E 1 1 can be written as x = a.- ( a.x), a. E m. If I = m-J1, then 1= ml and by Nakayama Lemma, 1= O. Hence t c: m-11 and by maxi. 1

mality m- l is a power of contradiction.

111,

125

DEDEKIND DOMAINS

i.e. I is a poter of m which is a

Theorem 3: Let R be a domain. Then R is a Dedekind domain if and only if every non-zero fractionary ideal of R is invertible. Proof: Assume that R is a Dedekind domain a~d let M be a non-zero fractionary ideal of R. Then for any prime ideal P =F 0 of R, M p is a fractionary ideal of R p and since R p is a DVR, M p is invertible. Moreover M is finitely generated as R is Noetherian. Hence by Proposition 2, M is invertible. Conversely assume that every non-zero fractionary ideal of R is invertible. In particular every integral ideal is invertible and hence finitely generated. Thus R is Noetherian. We now show that for every prime ideal Po/:-O, in R, R p isaDVR. ByProposition 3, it is sufficient to show that every integral ideal of R p is invertible. Let J be an integral ideal of R p and 1 = R n J. Then I p = J and since by assumption, I is invertible, I p = J is also invertible and thus R p is a DVR. It remains to show that every non-zero prime ideal P of R is maximal. Let P be a non-zero prime ideal of Rand m a maximal ideal of R containing P. Then PRm is a non-zero prime ideal of;Rm and since Rm is a DVR, P R m = mRm , i.e. P =m. If R is a Dedekind domain, the non-zero fractionary ideals of R form a group for multiplication. The quotient of this group by the subgroup of principal fractionary ideals is called the ideal class group of R. This group is important in number theory.

Proposition 4: Let R be a Dedekind domain with only finitely many prime ideals. Then R is a principal ideal domain. Proof:

Let Pi> p., ... , P, be the non-zero prime ideals of R.

If

= p~, then m = m 2 in R p , where m = P,Rp , and by Nakayama Lemma, m = 0, a contradiction. Choose tl E P, - P,' (1 ,.;;; i ,.;;; s)

P,

so that m = (t,). Let 10/:-0 be an ideal of R and write I

= PIn....

P;'. By Proposition 8,2.1, there exists some aE R with a == t;',

s; s). Then a EP~,_p:'+l. If the ideal (a) is decomposed as a product (a) = P~l .. "p;" then clearly r, = for each i. Hence I = (a) is principal.

(mod p;'+l) (1";;; t

n,

Corollary 1: Let R be a Dedekind domain and I a non-zero ideal of R. Then every ideal of R/I is principal. Proof: Let I

= P~l, ... , p;r, P, prime

ideal and let S

= R- UP,. 1

Since 10/:-0, So/:-R* and R s is a Dedekind domain (Exercise 1,5.3). Since R s has only finitely many maximal ideals P;R s (I ,.;;; i < r) it . is a PID. Hence every ideal of

~:

is principal.

But

~: "'" (R/I)s,

S = (S+ 1)/1. Now S consists of units in R/I as for any s E S, ($) I is not contained in any prime ideal of R. Hence R{Il!!t (R/I)s and R/I is a principal ideal ring.

+

Corollary 2: Let R be a Dedekind domain and I a non-zero ideal of R. Then I is generated by atmost 2 elements. Proof:

Choose a E I, a::f= 0 and consider R{(a). By the above

Corollary, R/(a) is a principal ideal ring so that the ideal 1= 1/(a) is generated by some b = b + (a). Then I is generated by a andb. Theorem 4: Let R be a Dedekind domain with quotient field K and L a finite extension of K. If S is the integral closure of R in L, then S is a Dedekind domain. Proof: We first assume that LIK il separable. By Proposition I, 4.4, S is a finitely generated R-module and since R is Noetherian, S is also Noetherian. If Q is a non-zero prime ideal of S, Q nR is a non-zero prime ideal of R (Corollary 2, Proposition 1,4.2), Hence Q is maximal in S. Since S is integrally closed in L, S is a Dedekind domain.

126

DEDEKIND DOMAINS

127

COMMUTATIVE ALGEBRA

We now consider the general case. There exists a subfieldL' of L such that Ke L'C L with L'/K separable and L/L' purely inseparable. If R' is the integral closure of R in L'. then R' is a Dedekind domain. Since the integral closure of R' in L is the same as integral closure of R in L, we may assume without loss of generality that L/K is purely inseparable and ch K=p>O. Let S be the integral closure ofR in L. If /l E S, then /l satisfies the minimum polynomial of the type X" - a E K[X] (e > 1) and by Corollary 2 to Proposition 3, 4.3, a E R. Since L/K is finite. [L:K] = p" = q for some n ~ 1 and clearly S = {at ELI /lq E R}. Let Kq-l -:::;L be the field obtained by taking q-th roots of elements of K. Since the mapping x ~ x q is an isomorphism of Kq-l onto K. mapping the set SI = {/l E Kq-l I xq E R} isomorphically onto R, SI is a Dedekind domain. Clearly SI n L = S. It remains to show that S is a Dedekind domain. Let I be a non-zero-integral ideal of S. Since S1 is a Dedekind domain. IS. is invertible. . Hence

1=

};, a,b,. a, E

I. b, E (SI: IS.).

e(Q/P) = v(}(PS(}) where v(} is the discrete valuation of L corresponding to S(}. By Theorem 2 if we write PS = Q~' Q;• ... Q;', Q" prime. then Q•• .. . Q, are precisely the prime ideals of Slying above P and e(Q'/ P) =. e,. 1 ~ i ~ r, -

Theorem 5: (Ramification Formula). Let R be a Dedekind domain with quotient field K. L a finite separable extension of K and S the integral closure of R in L. Let P be a non-zero prime ideal of R, Q,(I ~i~r) the prime ideals of S lying above P. Then I eJi = [K:

L1. where e, = e(Q,f P) and Ji =

,

f(Q,,'P).

We first prove the following Proposition. Proposition 5: With the same notation as in Theorem 5, assume further that T = R - P. Then (i) - Sr is the integral' closure of Rr in L. (ii)

(iii)

This implies

, a1 b~, with b1 EKe L.

1=I

Q1Sr; .... Q,Sr are the prime ideals of. Sr lying above PRT.

e(~':::)'= e(Q,/P) and

!(~;:) =!tQ,IP). I ~ i~ r.,

Rewriting the above relation as 1 = %a,c, where c, = a1-1b1.

,

we have c,I e b1ls «: SI as b,Ie SI'

Also

c,l e L as b~ E L.

Hence c,IeSlnL=S. i.e. C,E(S:I). This implies that I(S: 1) = S, as 1 = I a,c,. Hence I is invertible in Sand S is a Dedekind domain. Let R be a Dedekind domain with quotient field K. L a finite separable extension of K and S the integral closure of R in L. By Proposition 1,4.4 it follows that S is a finitely generated R-module. Let P be a non-zero prime ideal of Rand Q a prime ideal of S lying above P. Then S/Q is finite field extension of RIP and its degree is called the degree of Q OVer P and is denoted by f(QIP). The ramification index of Q over P denoted by e(QIP) is defined by

Proof: (i) Since S is the integral closure of R in L, Sr is the integral closure of Rr in L by Proposition 1. 4.3. (ii) If Q' is a prime ideal of Sr. then Q' = QST where Q is a prime ideal of Sand Q' lies above PRT if and only if Q lies aboveP. (iii) R TIPRTc>«lllP)7'. where Tis the image of R-P under the projection R ~ RIP. Since P is maximal. T consists of units of RIP so that

RriPRT~ RIP.

Similarly ~~T~ SIQ,.1

~ i ~ r, Let

PS = Q~'Q;• .. . Q~' be the unique decomposition of PS as a product of prime ideals. Then PST = (Q1ST)"'" (Q,Sr)" and hence

e(~t:) = e, = e(Q,IP). Moreover

128·

COMMUTATIVE ALGEBRA

DBDBKIND DOMAINS

Proor or Theorem 5: By Proposition 5. it is sufficient to prove the result after localising at P. We may therefore assume without loss of generality that R is a DVR with maximal ideal P. Since R is a DIP. by Corollary 2, Proposition 1.4.4. S is a free R-module of rank [L: K]. Hence SIPS is also free over RIP = k of rank [L: KJ. It is therefore sufficient to show that I e,f, = dim" (SIPS). We

129

have e(Q,jP)=val (t)andjtQ,/P) = I. since k is algebraically closed. It follows from Theorem 5 that I

vCt) = [L: k(t)]

veE

,

have a welldefinedhomomorphism S ~ SIPS"

SIQ~I

a homomorphism S" ; (SIQi') . . This map is surjective by the lal

Chinese Remainder Theorem. Its Kernel is

n Qi' =

'-I

SIQi'

~

2.

This also is an isomorphism as k-spaces.

It is therefore sufficient to show that dim" SIQ!' = e,f,. But

1.

PS.

I

Hence SIPS ~ ~ SIQi'.

5.3.

and therefore

SOI_ ~ Sal. Qj'SQI PSQI

But by Proposition I. 5.2

. = e(QIIP) [SIQI: k]

.

Corollary: Let k be an algebraically closed field of characteristic O. t an indeterminate over k and L a finite field extension oC k(t). Let }; be the seaof discrete valuations I' oC Llk such that I'(t) > O. Then }; is a finite set and }) I'(t) = [L: k(t». 6.

m.n

I'

(i) R is a Dedekind domain. (ii) For every maximal ideal m of R, there exists no ideal, I#: m and m2 such that ma C I C m. (iii) For every maximal ideal m of R. the set of m-primary ideals is totally ordered for inclusion. (iv) For every maximal ideal m of R. every m-primary ideal is a product of prime ideals.

and the proof is complete.

Proof: Let K=k(t), R = k[tl,t) and P = (t)R. Let S be the integral ' closure of R in Land Q, (I ~ t ~ r)the prime ideals of S lying above P, with 1'01 the corresponding discrete valuations on E. Then 1'01 E }; for 1 ~ i ~ r. Conversely if I' E I. then k[t] C R" the valuation ring corresponding to I' and t E m, the maximal ideal of R. so that R cR•. Hence S cR.. If m' = S. then wi is a prime ideal of S containing (t)S = PS. Hence m' = QI for some i and R. = SQ,' This shows that }) = {I'Qu 1'0...... va,}. Now we

Let R be a Dedekind domain and S( #: R*) a multiplicatively closed subset of R. Show that R s is a Dedekind domain. Le~R be a Dcdekind domain and I(X) E R [X] be written as I(X) = I a,X'. Define the content e(/)· of I to be the ideal

generated by. (00."", .... a.). Show that e(lg) = e(f) c(g), f,gER[X]. . 3.-' Let R be a Dedekind domain with quotient field K and L a subfleld oC K such that R is integral over RnL. Show that RnL is a Dedekind domain. 4. Show that in a Dedekind domain the notions DC primary ideal; irreducible ideal and prime ideal are equivalent. . 5. Let R be a Noetherian domain. Show that the following conditions arc equivalent.

=e,fj

veE

EXERCISES

Let R be a Dedekind domain and T( #: R*) a multiplieativmy closed subset of R. Show that the map M -+ MT is a homomorphism of the group of' fractionary ideals of R onto the group of Cractionary ideals of RT and the Kernel consists oC fractionary ideals oC R which intersect T.

COMPLBTIONS

131

{SnR.} is a filtration on S called the induced filtration on S.

l'v) I~ ~ l~ CHAPTER VI

Completion, like localisation is an important tool in geometry. It is an abstraction of the process of obtaining power series from polynomials, By successive application of localisation and comple-: tion, it is possible to reduce some questions in geometry to the study of the power series ring, Completion is well behaved with respect to exact sequences for Noetherian rings and modules. An important result concerning completion is the Krull's intersection theorem which describes the part"of the ring annihilated by completion. Krull's theorem and the preservation of exactness are consequences of' a basic result known as Artin-Rees lemma. Completion is best studied by introducing filtered rings and the corresponding graded rings. A graded ring is an abstraction of the polynomial riog in several variables where the gradation comes from the degree of monomials. Graded rings and modules are useful in projective algebraic geometry.

.Examples: (i) Let I be an ideal in R and let M, N be R-modules. Any R-homomorphism f: M ~ N is a homomorphism of filtered modules with respect to the I-adic filtrations on M and N. . (ii) The natural projection p: M ..... MIN is a homomorphism of filtered modules. Definition: A graded ring R is a ring R which can be expressed as a direct sum of subgroups {R.}, i.e. R = EB ~ R. such that

Definition: A filtered ring R is a ring R together with, ~ family {R.}.;oo of subgroups of R. satisfying - the conditions (i) R o = R (ii).R.+l C R. for all 71 .;> 0 (iii) R.R", c R",+. for all m. n ;> p.

~'. ~

1i:"R", c. (Z ...

J Ul.. ,h

R",

q~~ .

Definition: A map f: M ..... N is called a homomorphism of filtered modules if (i)fis an R-module homomorphism and (ii)f(M.) eN. for all 71 ;> O.

Filtered riogs and modules

.;00

l~ ~ t;l.~ .

Examples: (i) For any ringR. define a filtration by setting R o= Rand R. = 0, n ~ 1. This filtration on R is called the trivial - filtration. .(ii) Let I be an ideal in R and let R. = I·, 71 ;> O. Then {R.} is a filtration on R called I-adic filtration. (iii) If {R.} is a filtration on Rand S is a subring of R. then

e:t

Examples: (i) Let M be an R-module and let R have trivial filtration. Theo M also .... trivial filtration defined by M o = M and M.= 0, 71 ~ 1. ~l""" I), 1"\VtM ~ (Ii) Let [be an ideal of R and consider the I-adic filtration on R. Definll the I-adic filtration on M by setting M. = I·M. Then M is a filtered R-module. (iii) Let M be a filtered R-module and Nan R-S'libmodule of M. The filtration 1M.} onM induces a filtration {N'.} on N where N. = N n M.., n ~ Or It also induces a filtration on the quotient module MIN. where (MIN). = (M. + N)IN. Let M and N be filtered modules over a filtered ring R.

COMPLETIONS

6.1.

fl'VI,

Definition: Let R be a filtered ring. A filtered R-module M is an R-module M together with a family {M.}.;aoo of R-submodules of M satisfying (i) M o = M (ii) M.+ 1 c M. for all n ~ 0 (iii) R",M. c M",+. for all m, 71 ~ O.

f 1

R.R.. c Rm+. for a~1 m. n ;> O. • I \. D r 6J.1 R - nuU ' ~£rI..l ~ R o '> 6- S...b/'I'v' -- O. n>O

t

Xl' X., .., X. over a field k and m = (Xl' X., ..., X.). Consider the m-adic filtration on R. Clearly f E md if and only if order J-. ·M", (~~v.,. Ro ~9IAbl,o'Vl.vJwJll-. Examples: (i) Let R = K[X1 " .. , Xn] , the polynomial ring with gradation defined as in Example @) above. Then R considered as an R-module is graded. (""'!"vv..o and {Rn}n>o respectively. Let grn(M) = Mn/MO+ l and gT(M) = EEl I sr, (M;. ·>0

/~

of M with itself j times, and To (M) = R is a graded ring. (See 5xercise 8, 1.3). (B~~"-"- -t.-t..e "J~-\-v.-c:. ....A,),t"Vt,\

Ji>(,

.,v'L.

I,;o!~ A ~""lc--\.h/'" V'--i

i", t"\-t-.; b..

Then gr(M) has a.natural gr(R).module structure given by ,(a + R.+J (x + M m+J) = ax+ Mm+n+h aeR.. xe Mm. This module is called the associated graded module of M.

1>0.

Remark: The elements of R n or M n in a graded ring ora module are caned elements of homogeneous degree n. Let M be a graded R-module. A submodule N of M is called a graded submodule if N = EEl 1: N n where N. = NnM.. In this case the quotientR-

PI:oposition 1: Let R be a filtered ring. M, N filtered R-modules and f: M -+ N, a homomorphism of filtered R-modules. Then f induces a natural map gr(f):gr(M) -+ g,(N). which is a homomorphism of graded gr(R) modules such that (i) gT(ld) = Id and (ii) gr(hf) = gr(h) gr(/) for another homomorphism of filtered modules

";;;'0

module M/N is also graded.

h:N-+ K.

Definition: Let M and N be graded modules over a graded ring R. A map f : M -+ N is called a homomorphism of graded modules if (i) fis an R-homomorphism of modules and (ii) f<M.) eN•• n;> O.

Proof: The R-homomorphism f: M -+ N satisfies f(M.) C N n , Oand hence induces a maPl.: M./MnH -+ Nn!N.+ l defined by f. (x + Mn+t> = f(x) + N.+IJ X EM•. Wedenote this map by gr.(f) and extend it to a map gr(f) = EEl 1: gr.(/): gr(M) -+ gr(N).

Definition: Let R be a filtered ring with filtration {R.} ... e- Let gr.(R) = R./Rn+l and gr(R) = EB I gr.(R). Then gr(R) has a n>O

natural multiplication induced from R given by l~

If{

133

(a + Rn+J) (b + Rm+l) = ab + R",+.+l, a E R. o b eRIII •

-;;:T:::-his"'"""m'""a"'k~e~s~ into a graded ring. This ring is'called the associated graded ring o'f R. Examples: (i) Let R be any ring and IE R a non-zero divisor. Consider the (/)-adic filtration on R. Then gr(R) is naturally isomorphic to the polynomial ring

~) [Xl. where Xis

the canonical

image of I ERin grJ(R) = (/)/(/1). (ii) Let R = k [[.1"10 X., .... X.]] the power series ring in

~_:>

Ii

1

.

n~O

Clearly gr(/) is a homomorphism of graded gT(R) modules and satisfies conditions (i) and (ii). Rema..k: The following Example shows that gr(/) may be zero with f being zero. Example: Let R = Z. 1= (n), n> 1 and M = N = Z with the I-adic filtrations. Then f: M -+ N given by .Itx) = nx, x E Z is non zero but gr(/) : gr(M) -+ gr(N) is clearly zero. Proposition 2: Let R be a filtered ring. M, N. filtered R-modules andf: M -+ N a homomorphism of filtered R-modules. If the induced map gr(f): gr(M) -+ gr(N) is injective then is injective. M. = (0), . provided

n

r

11-0·

Proof:

By assumption, the map gr.(f): M./M. H -+ N./N. H is

134

COMPLETIONS COMMUTATlVI! ALGIIBRA

Proposition 4: Let M be a finitely generated filtered R-mod.ule over a Noetherian ring R with an I-filtration. The following conditions are equivalent. (i) The filtrat!on on M is I-stable. (ii) If R* = EB 1: I·. and M* = EB 1: M.. the graded R*·

injective for all n ~ O. Hence M.nf- 1(Nn+t) e M. w n ;> O. We prove inductively that f-l(N.) C M. for all n ~ O. Clearly f-I(No) e Mo. Assume that f-'(N.) c and let x Ef-1(N.+J. Since Nn+leN., xE/-1(N. )e M . so that XE/-l(N.+I) nM.cM.w

u,

00

Mn+l' Since n M.=(O), we havef-i(O) cf-I(nN.) o en M. = {O}. i.e. f is injective.

i.e·f-l(Nn+l)

c

Remark: With the same assumptions as in the above Proposition, gr(f) is surjective need not imply thatfis S~ithout addi_. tional as~umptions on the filtrations~])~_ ~..

r-

-,

!Ex.~·. k[xl ~ k([~l '"Proposition 3: Let R = e 1: s, be a graded ring. The following d>O

conditions are equivalent. (i) R is Noetherian. (ii) Ro is Noetherian and R is a finitelygenerated Ro-algebl'll. Proof: (i)~· (ii). Now Ro ~ RJI where 1= EB 1: ~ is an ideal d>O

in R. Hence Ro is Noetherian. Choose a finite generating set for I consisting of homogeneous elements {aI' .... ar} where a, is homogeneous of degree n/ (1 ~ i ~ r). Let R' be the Ro subalgebra of R generated by {a,l. We show by induction on n that R. c R' for all n :>0. Clearly Ro C R' and assume that Rd e R' for all d ~ n - I. Let a E R. (n > 0) so that a E I. Hence a = 1:J..,alt where 71/ is homogeneous of degree n - n, < n. By induction ~ E R' (I ~ j..-;; r) so that a E R'. Hence R = R' and the proof is complete. (ii) ~ (i) follows from Corollary 2 to Theorem 1, 3.1. Definition: Let M be a filtered R-module with filtration {M.} and I an ideal in R. The filtration {M.} is called an I-filtration if 1M. c M.+1 for all n ~ O. Definition: Let M be a filtered R-module with an I-filtration {M.}. The filtration is called I-stable if there exists some m such that for all n ~ m, 1M. = Mow EXample: The I-adic filtration on M is I-stable.

135

n~o

n>O

module M* is finitely generated. Proof: If N. =

?

e io Mit

then N. is finitely generated over R

as each M/ is finitely generated. Define M: = M o EB M I tB ... €a M. elM. EB PM. ff) .... Clearly M: is a finitely generated R*-module as N. is a finitely generated . R-module. Hence M* =- U M: is finitely generated over R* if and only if .>0

M* = M,: for some m. t.e. Mm+k = IkM", for all k ~ I. the same as the condition for the filtration to be I-stable.

This is

Proposition 5: (Artin-Rees Lemma). Let M be a filtered R-module with an I-stable filtration. Assume that R is Noetherian and M is R-finitely generated. Then the filtration induced by M on a submodule N of M is also I-stable. Proof: If {M.} is the given filtration of M, clearly {Nn M.} is a filtration on N. Let . R* = EB }; I'. M* ,,>~

=

ff)

I M. and N* 11>0

=

EB 1: N. 11;>0

where N. = M n N.. Since R is Noetherian and I is finitely generated, R* is a finitely generated R-algebra and hence R* is Noetherian. Since {M.} is I-stable, M* is finitely R* generated by Proposition 4. Hence N* is also finitely generated over R*. i.e. {N.} is I stable. Corollary 1: Let R be a Noetherian ring, I an ideal in R, M a ftDitely generated R-module and N a .submodule of M. Then there exists some m such that 1"'+kMnN=Ik(I"'MnN) for all k~O.

Proof: Apply the Artin-Rees Lemma for the I-adic M.

filtrat~n

on

136

CO~UTATlVE ALGHBRA

6.1. EXERCISES

Definition: The inverse limit of the system {M•• 6.} is an R· module M together with R homomorphisms {.Ii} fi: M -+ M" with 6/+lfi+1 =j,(i;;' 0), such that Mis universal for this property. i.e. if M' is another R-module with natural maps g/ : M' -+ M, satisfying 6'+1 gi+l=g/ (i ;;. 0). there exists a unique R-linear map X: M' -+ M with loA = g/ (i :> 0).

~ R. be a graded ring and R+ = EEl ~ R" the _>0 , _>1 .aubriB~ of R. Show that if M is a graded R-module with R+M = M. then M = O. 2. Let M be a graded module over a graded ring Rand N an Rsub module of M. Show that the following cOnditions are equivalent. (i) N is a graded submodule of M. (ii) If x E N, all the homogeneous components of x arc in N. (iii) N is generated by a set of homogeneous elements. 3. Let M = EEl ~ M. be a graded R-module with trivial gradation

1. Let' R = EEl

jJ~

Proposition 1: The inverse limit of {M•• 6.} exists and is unique up to isomorphism. Proof: The uniqueness is clearly a consequence of the universal property. To prove the existence. consider the product N = 1t M/

.>0

4.

5.

/

on R. Show that if M is finitely generated over R, then there exists some m such that M" = 0, k;;. m. Give an example of a bijective homomorphismf: M ~ N of filtered R-modules for which gr(f) : gr(M) -.. gr(N) is neither injective nor surjective. Let M be a filtered R-module, N a submodule with induced filtrations on N and MIN respectively. Show that the'exact"

and the submodule MeN defined by M

M

gr({)

6.

7.

IIi'lO

ideal in R. Show that I is a prime ideal if and only if x ¢ P and y ¢ P implies xy ¢ P, for all homogeneous elements x, y in,R.

:> O}.

The inverse limit of the system {M •• 6.} is denoted by +-- M•• n

,

..

Completion

The study of completions require some preliminary concepts on inverse limits. We restrict ourselves to countable inverse systems. Definition: An inverse system of R-modules is a collection of Rmodules {M.}~ and homomorphisms {6.} where 6. is an R~homo­ morphisms 6.: M. ~ M.-I (I,l;;' 1).

.'~

Example: (I)

,

Let M.= (X.)' k field and 6.+1:Mn+I ...... M. the

natural map induced by identity map on k[X]. Every element of Mn+l is a polynomial over k of degree utmost n and its image in M. is the same polynomial with the last term truncated. Hence

,~ n

6.2.

= xu i

lim

gr(p)

gr(N) _ _ gr(M) - - ) 0 gr(MIN) ~ O. Let R be a filtered ring and M a finitely generated R-module. Show that the filtration on R induces a natural filtration on M and that gr(M) is a finitely generated gl(R)-module. Let R = EEl ~ R. be a graded ring and I = EEl }: I. a graded ,. ... 0

== {(x,) I 6/+I(x/+1)

Let jj: M ~ M, be the restriction J)f the i-th projection to M. Clearly 6/+1 fi+l = li(i ;> 0). Consider an R-module M' and maps g/: M' ...... M i satisfying 6/H g/+1 = g/ (i;;' 0). Define X: M' ~ M, by setting J.(x) = (g/ (x». x E M'. Clearly A is R-linear and li A= g/ (i ;;;: 0).

sequence 0 -+ N ......!....... -!!.-.. MIN ~ 0 where i is the inclusion and p is the natural projection induces an exact sequence

o -+

137

COMPLBTIONS

M. can be identified with k [[XU. the ring of formal power

series in oX over k, (ii) Let R' = R[XI , X X.] and I = (Xl..... X.). If M. = R'll· and 60+1: M.+ 1 M. is the natural map induced by

u

the identity map, ~ M. ~ R[[X1 • X....., X.]], the ring offormal n

power series in Xl' X...... X. over R. Let M be a filtered R·module. The filtration {M.} on M defines a topology on M compatible with the abelian group struc-

138

COMPLETIONS

COMMUTATIVE ALGEBRA

ture of M for which {Mnl'is the fundamental system of neighbourhoods of (0). It is called" the topology induced by the filtrnion {Mn } · Proposition Z: Let N be a submodule of a filtered module M. In the topology induced by the filtration we have

N=

n(N+ M

n=O

n) .

Coronary: The topology defined by the filtration is Hausdorff if

nu, o .

=

then x" - X~.E M•• n ;> 0 and {x.} is equivalent to the Cauchy " . lim M sequence {x n } . Hence the mapping ex: -;;--- M. -+ M given by A.

= {x.} is well defined. We show that « is an isomorphism. Clearly « is R-Hnear. If «(y) = {x.} = 0 then Xn -+ 0 and an easy argument shows that Xn E M n for all n. This implies that Yn = 0 for all n, i.e. Y = o. Consider now any Cauchy sequence {z,,} in M. Choose inductively a subsequence {x.} of {znl such that x.+t - X. E M" for all n, Ify. = X n M. and Y = (Yn) then «(y) = {zn}. Hence« is surjective and the proof is complete.

«(y)

+

Proof: x¢N if and only if some neighbourhood of x does not intersectN, t.e, (x + Mn)nN=,p, for somen, ;.e.x¢(N+Mn) for somen.

and only if

139

(0).

Proposition 3: Let {M~. O:}. {M", a,,} and {M:'. O:} be inverse systems of R-modules such that there exist R-linear maps {/n}.>o. {g.}.>o such that r f". K. ." • 0-+ M n _ _ M. M n -+ OIS exact for each n and the diagram

Proof: The topology is Hausdorff if and-only if

{O} = {OJ =

0 .... M;+l -+ M nH

nu;

S:t1l

o

Let M be a filtered R-module with filtration {Mn}n;;;lQ. Under the topology (uniform structure) defined by the filtration, M admits a Hausdorff completion M. It is the set of equivalence classes of Cauchy sequences of elements of M modulo the equivalence . relation given by (x n ) ' " (Yn) if for each m. there exists some no such that X n - Yn EM",. n ;;> 1/0. The following Proposition shows that the completion can be obtained as an inverse limit. Theerem 1: Let M be a filtered R-module with filtration {Mnland lim M completion M. Then M = ~ M . •

A

A

n

II

~

= _lim II

M

-

•

M • Y E M with Y = (Yn). Choose Xn E M n with X n + M n ... Yn. Now {x.} is a Cauchy sequence in M because x. - x'" E Mn• for n ;> m. We map Y onto {x.}, and the mapping is well defined because. if we choose x~ E M with + M• ... y", Proof: Let

JY.I

x:

o ....

M:

0'+1

!

-+ M.

....

M~'+l

-+

0

....

M"n

....

0

18~1

is commutative for all n, Then (i) the sequence 0-+

(El~m ~)-+(~Ij:

Mn)-+-

(.!i: M:)

is exact. (ii) If

o -+

0: is surjective for ali n, the sequence (/~m M~) -+ U~ M.)-+ U M; )-+ 0 i :

is exact.

= 'l':Mn and dM : M -+ M be defined by • dM(x n) = (x.-a"+1(Xn+1)). Then d M is R-linear and

Proof: Let M

Ker did =

lim _ II

M..

This implies {In} and {g.} induce

140

COMPLETIONS

COMMUTATIVE ALGEBRA

Corollary 2:

R-Iinear mapsj= 1Cf. :M' = 1CM~ .... M and g = 1Cg. : M -+ M"

•

= -rrM;; and

•

induced filtration on M. is a submodule of

f 0 -+ M' -+

g

M

M"

-+

-+

0

f

g

M

-+

M"

-+

-+

0

~ o -+ M.

By Proposition 6, 2.,2, we have an exact sequence

o -+

Ker dM '

....

Ker dM -+ Ker rIM" .... Coker elM'

0:

which proves the relation (i). Assume now that is surjective, for alln. Then dM' is surjective, Coker dM' = 0 and (ii) is proved.

o -+ M'

-+

M -+ Mil

Since n is fixed .M'' :

= 0 for all m ;> n, i.e. Mil has discrete topology.

M" and this implies ~

nf-l(M.+ 1)

6~+1

1

-+

~

....

M.+t

~ -+ 0

g(M.+ 1)

1 1

0;+1

M M.

Mil -+ g(M.)

-e- 0

By passing to the inverse limit, the sequence ~

o-+ M' -+ M A

is exact by Proposition 3 as

This

'

{M.} on M and it == M.

Proof: Clearly {M.} is a filtration of Mas {M.} is a filtration on

6.+1

M' 0-+ M nf-1(M.)

"" M .

• M.,

M.

By Corollary 2, we have

"

.... Mil -+ 0

0: is surjective for each n.

Z.,

ot

it under a natural isomorit.

phism making the following diagram commutative

such that the following diagram is commutative M'

-+ 0

~M. r:(Z.)=M"

induces a filtration

M' M Mil 0-+ M' nf-l(MIf} -+ M. -+ g(M,,) -+ 0

-e-

If

and Mil = M/M. we

A

-+ 0 is exact.

M'

"" MM for all n.

Corollary 3: Let M be an R-module with filtration {M.}.

Proof: For each n, we have an exact sequence of R-modules

o -+

" (M) M.

-+ M -+

M.

to the

t.e.

Hence Mil ""'" Corollary 1: Let 0 -+ M' .~ M ~ M" -+ 0 an exact sequence of R-modules and {M.} a filtration on M with induced filtration {M' n f-1(M.)} ,on M' and filtration {g(M.)} on Mil. If completions are taken with respect to these filtrations, the sequence

Sf and ~

= M.

Proof: Applying Corollary I to M' have the exact sequence

dNl dM"!

dM'l M'

M. of M. with respect

it is easy to check that the follow-

ing diagram is commutative

0 -+

Let M be an R-module with filtration {M.} and

completion M. Then the completion

If

141

M

if

- - -+-.

Mn+l

I I

M"

8+1

I I

4.

Hence

lim

M

lim

Sf

M:

--"""'~-If

M.

If

~

t.e. M""",M'

142

J

COMMUTATIVE ALOBBRA

6.2. EXERCISES 1. Two [-fi~rations {Mo } and {M.'} on M are said to be equivaI~t if there exists an integer k such that M O +k c M.' and MO+k c M o for all n ;> O. Show that two such equivalent filtrations define the same topology on M. 2. Let R be a Noetherian ring, M a finitely generated R-module and N a submodule of M. If [ is an ideal in R, show that the two filtrations on N, viz. {ION} and {N n ["M} are equivalent. Deduce that the completions of N with respect to these two filtrations are isomorphic. 3.

,

Z

Z

Let p be a fixed prime and let 10 : p-Z -+ -pnZ be the natural map defined by lo(D = pO-I. Let group map from EB :

R-+ S,

1= ~ M,IM.+l is finitely generated OVer RfI. This implies a

that sr, (M) is finitely generated over 8rl (R). : F -+ M, given by ." [(a,)] = 1:a,x, is a homomorphism of filtered R-modules.

is Noetherian. Since R is complete, f· = O. By applying " 0 ,. ¢;. • Corollary 1 to the ring .J{ and the module M = R, we get that .J{ IS Noetherian. .

I

_

The associated graded homomorphism gr ("') : gr (F) --I>- gr (M) is surjective as the {YI} generate grI(M). Hence by Proposition 2, i:i'''': M is surjective. Consider the commutative diagram

n

Corollary 3: If R is Noetherian, the power series ring R [[Xl' X,. ..•• X,J] is Noetherian. Proof: Since R [Xl' X" "', Xnl is Noetherian, its completion for the (Xl' X" ... , X.)-adic topology, R[[Xt • X" "', X.]] is Noetherian.

COMPLBTIONS

152

Proposition 3: Let R be a Noetherian ring, I an ideal contained in the Jacobson radical of R. If grl (R) is a domain, then R is a domain. Proof: Let a, b EO R. a:t= 0, b:t= o. By Corollary 2 to Krull's Intersection Theorem,

(rI· = O.

Assume now that g.(X) and h.(X) have been constructed satisfying the required conditions. Write [(X) - g.(X) hnCX) =

d

topology.

If [(X)

_.

= };

'-0

a,X' E R[XJ and a, = a, + mE Rim =k,

1

!<X) denotes the polynomial f<X) = iitx' E k[X]. AssumeJtX) . ~o E R[X] is a monic polynomial such that there exist lX(X) and ~X) relatively prime monic polynomials in k(Xj of degrees rand (d-r)

respectively with/(X) = lX(X) ~(X). Then there exist monic polynomials g(X), h(X) E R[X] of degrees rand (d-r) respectively

= lX(X),

such that i(X)

h(X)

=

~(X) andJtx)

= g(X) heX).

Proof: We construct inductively, sequences of polynomials

of degrees utmost d-r and r respectively such that )(l

Since g.(X) = lX(X) and Ji.(X)

_

= «(X), ]i.(X) =

gn+l(X)

[(X) - g.(X) h.(X) E m"R[X], n;;;' I.

and Let n

=

I.

Since 1(X) = lX(X) ~(X), there exist

I

h.+1(X) = h.(X) +}; ).,,(X).

,

Clearly g.+J and h.+1 are of degrees utmost" and (d-r) and

in+! = i. = at, 7i.+! = Moreover

I>: gn+l

such that ~(X)

= «(X), Jit(X) = ~(X),

[(X) - gt(X) ht(X) E m ReX] deg gt(X)

=

r, deg ht(X)

h.+1 =[ - (g.

=d-

r.

Ji. = ~.

+ 7' ).,ojI,) (h. + 7).,rf>/)

g.h. - ~ A,X' + ~ ~X' - g.71.,,

_ h. };"Iq., -

1 1

}1 A/A q.~

/.1

=};~ (XI - g., - h.ojI,) - }; A,A1 1jI,j.

,

Since X' _

',I

g."" -h.IjI, Em R[X] and A/ Em·, it follows that [_ g.+! h.-tl E

gt(X), ht(X) E R[X]

and ~,(X), ojI,(X) E R[X]

= g.(X) +}; ).,ojI,(X) and

I

~(X)

= ~(X)

~(X) (0 ,;;:; i';;:; d)

XI - ",,(X) g.(X) - ojI,(X) h.(X) Em R[X'J.

Let

in R[X] of degrees almost rand (d-r) respectively, such that g.(X)

= q;,(X) lX(X) + fJ I and Ho{X) 0< M.

X has

the .property

that

Examples: (i) If M is projective, choose ~o = M, e = Id and X. = 0, PI Then X is a projective resolution of M. (ii) Let R = Z and M any abelian group. Choose a free abelian group Xo and a surjection e: Xo -+ M. Let Xl = Ker c. Then Xl is also a free abelian group and

>}.

I

f

M' _

0

Since X o is projective and e' is surjective," there exists an R-linear map Fo: Xo - X;; withe'Fo =/e. Assume inductively that F,:X,_X; have been defined for all i, 0 ,.;;;; i ,.;;;; n satisfying F'_ld, = d;Fi (i";;;; n), with F_l =/, do = s, d~ = s'. Consider the diagram

d'-..... + l

X'+l

""

d.' X'_

X.

--+

l

I "F.d'+ IF. -.-,

r:

is a projective resolution of M. X~+!

Proposition 3: Every R-module M has a projective resolution.

"'\.

x: -;?

•

0-+ Xl - ..... X o --+ M -+ 0

1\

"

__

cr.+!

l

,

'\.~ x:

_-+

• cr.

~

X;;-l

160

HOMOLOGY COMMUTATIVE ALGEBRA

Observe that Im (F.d.+l) C 1m d~+l = Ker d'. because d.'F.d.+ 1 = Fn-l d•d.+l = O. Consider the diagram Xn+l

s:

IF,,dn+l

The projectivity of X.+ 1 gives anR~linear map F.+l: X~l-+~~+1 by induction. such th at d'• +1F.+1 = F.d.+l and the proofis complete , S We now prove the uniqueness of Fup to homotopy. upposewe have another mapping of complexes G: X -+ X'. such that fe = c'Go' We construct a homotopy s = {s.}.;oo between F and G inductively as follows. Consider the diagram £

~

X,

X.

-4-

lFo-Go

I

t

X'l --+ X'o--+ E'

a\

Since .'(Fo - GJ

= c'Fo -

.'Go = f. - fE = 0, Imd'l = Ker.' :::> Im(Fo - Go)'

Since X is projective, there exists R-.linear map so: Xo -+ X'l such o that d1'so = Fo - Go· _ . . Assume inductively that {s,} have been defined for I, 0 ~ I ~ n satisfying d;+18, + s,-A = F,-G, (i ~ n) with 8-1 = O. To define 8.+1' consider the diagram tf.+1

X.+J ---+

X.+l

kO~.

\

dn+l

.+2

--+ d'"H

X'

.+1

X.

/\'._0. ---->-

S.+l:Xn+l -+ X~H such that d~+J8.+1

= F.+1

- G.+! -s.d.+!.

Remart: The map F:X -+ X' is called a lifting of the map I:Y -+ M' to the corresponding projective resolutions: Jf/= Id, , F can be chosen to be identity and if g: M' - M' is another R-linear map with a lifting G: X' - X', then GF is a lifting ofg/. CerolI..,= If Z aJ;Id Yare two projecti;veresolutions of M, then ,

.~~~ Jlc)~otopleanyeqaivaJeD,t.

':~~it~ Th~~m land the re~rk above.

F G lbeorem2: Let 0 -+ X' --+ X _ X' -+' 0 be an exact sequences of complexes, i.e. for each n, the sequence of R-modules F.

O-+X~_

G.

X. --+ X:-+O

ia exact. .Then there exists a connecting homomorphism

a.: H.(X')

-+ Hn-t (X')

auch ·that the sequence -+ H.(X')

~

H.(X)

~

H.(X")

...!..'-+

H._l(X') -+ ...

is exact.

!"roof: Since the given sequence of complexes is exact, we have a .:!lOmmutative diagram

t

t,/ X'

Since X.+ 1 ts projective, there exists an R-Hnear map

Thus 8. is defined for all nand s = {s.}.>o is the required homotopy between F and G.

X~+1 ~~ d~+1 (x:.+J -+ 0

a1

161

---+

d',,+1

X'.

We observe that Imd~+J = Ker d~l :::> 1m (Fn+l- Gn+l - 8.dn+J. because d' (F.+I - G.+l - s.d.+ l) = (F. - G.) dn+l ~ - (F.~G. - Sn-ld.) d.+1 = 0•

X". -+ 0

tla"•

x:.:"l

-+

0

with exact rows. l'his induces a commutative diagram

162

HOMOLOGY

CoMMUTATiVB ALGBBRA

X'.

B·F)

X.

-+

Xli..

B.(X') n

~

x:.'

ZII-](X")

0 ~ Z._I(X') -+ ZIt-I(X) with exact rows. where the vertical maps are induced by the boundary operators. i.e,

if,,:

B~X) -+ Zn-I(X)

««

i" + B.(X» = d.«.. Clearly Ker 1. = H.(Xj. and Coker a" = H.-I(X). By Proposition 6, 2.2, we have an exact sequence

is given by

Ker ti. -+ Ker J" -+ Ker ii"" ~ Coker 'ii'" -.. Coker tl" -+ Coker

...!:-.. is exact for all n

- respectively by

e("'o, x'J=fr.' x'o+Ofcri'., d"(x',,, x'")=(d'"x'"+IIl,,x",,, d""J(',,).

. ;' C9nditiOlil -

.(ii) -

> O.

Remark: The exact homology sequence derived above satisfies the naturality condition (see Exercise 4, 7.1).

e: X o -+ M, d,,: X" ~ X"_I

Thoco~dltion.. 4-~d" =0, 11 ;> 1 and e d" = 0 'i~pose the following '. (I)

H"-I(X') .... Hn-l(X) ~H"-'-I(X")

Oil

tho GI".

. . .reso Iution ' X exists a projective sequence of complexes 0 the diagram

0

....

X' 0 e'

I

,j..

0 -+ M' Is commutative.

• --7

, F .... x ~

F

o --7

f

--7

Xo

£1 M

en

r.' GIl = "'od"I'

Since the left half of the diagram (*) is commutative, the commutativity of the right half imposes the condition (iii) e" =~. We show by induction that {Of,,} can be chosen to satisfy (i), (ii) -and (iii). Since X o is projective. 1Il0 can be chosen to satisfy (iii). To ~efine «I' consider the diagram

e

M"" Oof Mand an exact

X

--!!-7

Go --7 X"0 a" g

--+

X" -+ 0 such that

-+

(*)

I

1f"

-+

0

X' o

-fe' -~

Since sOd"l =glllod''t = 0, [m( 1) are defined giving rise to the complex ' d

a

X..,: ... X. ~ X"-l -+ . '" -+ Xo _ _ M -+ O.

Tho encb1es. of the sequence of complexes 0 -+ X'-+ X ~ X" -+ 0 giftS the eXactness of 0 -+ XM -+ XM -+ x'~ -+ O.

164

COMMUTATIVB ALGBBRA

Since X' and X" are projective resolutions of M' and M" respectively the complexes at the extremes are acyclic. This implies that XM is also acyclic (see Exercise 2, 7.1) showing that X is a projective resolution of M.

F*

,

i

•.• -+

G*

0

H~(X') -~ Hn(X) ~-+ H~(X") -~ HIl-I(X')

1

~«

II

1'.

."I IX,.".

I •"

«'·

Definition: A projective resolution of an exact sequence of Rmodules 0 -+ M' -+ M ~ M" -+ 0, is an exact sequence of complexes 0-+ X' -+ X -+ X" -+ 0 where X', X, X" are projective resolutions of M', M and M" respectively, such that the diagram

X'0 c'

I

c

~

M'

0 -+

Xo

-+

-+

I

~

M

X"0

-+

-+ 0

E" r

...~

-+

M',' -+ O

"':

J

7.2. Derived functors An additive covariant (contravariant) functor is a correspondence which associates to each R-m04ule M an R-module T(M) and to ea,ch R-linear map f: M -+ M', an R-linear map

1'(f>: T(M) -+ T(M') (T(f): T(M') -+ T(M»

7.1.

4.

...

is commutative.

(i)

T(ld) =

(ii) T(gf)

fl.' 0 -+

F --+

I

~ Y'

G

X --->-

'"

= T(f) T(g»

X" -+ 0

Eumples: (i) Let N bea fixed R-module.

t

Y --+

i-

Y"

isa covariant additive functor. (i~) For a fixed R-module N, the functor T'(M) =HomR (M, N) and T (f) = HomR (/, IN),f: M -+ M', is an additive contravariant functor. . ~e now ~efine the left derived functors L"T (n ~ 0) of an additive covanant functor T. Let M be an R-module and

Jf ~ M a projective resolution of M. Applying the functor T OD t~e

complex X X'

.... -+

-+ 0

The functor

T(M) = M~N, and T(f) """f® IN, forf: M -+ M ',

I fl."

fl.1 --+

(T(gf)

= T(f) + T(g).

EXERCISES

Show that a complex is acyclic if and only if the identity map of X in X is homotopic to the zero map. Let 0 -+ X' -+ X -+ X" -+ 0 be an exact sequence of complexes. II any two of the three complexes X', X, X" are acyclic, show that the third is also acyclic. Show that a finitely generated module over a Noetherian ring admits a projective resolution by finitely generated projective modules. Prove the following naturality condition for exact homology sequence. Given a commutative diagram 0 -+ X'

u,

= T(g)T(f)

(iii) T(I + g)

3.

(Y')-+ n-)

satisfYing

Remark: The above Theorem shows the existence of such a projective resolution;

2.

(J'~

"\

r

is commutative.

1.

oIJ:

. -+ '"

tX.._ 1

•.• -+ Hn(Y') --+ Hn(Y) --+ H IY") --+ H

0 -+

165

HOMOLOGY

X

we have the complex

n

d. d) --+ lf~-l ... X) -:--+ Xo -+ 0

oIJ

where the rows are exact sequences of complexes and the vertical maps are mapping of complexes, the induced diagram

T(X): ...

-+

T(Xn)

T(~

TOe. ) -+ \ JI-l

.••

The n-th homology of the COmplex T(X)

l1(X) 1

T(d l )

--;)0-

1'(X), 0 0 ....

166

COMMUTATIVB ALGEBRA

HOMOLOGY

167

(ii) LoT=- Tif T is right exact. (iii) For any exact sequence of R-modules, Let/: M -+ M' be an R-Hnear map and X, X' projective resolutions of M and M' respectively. By Theorem 1, 7.1,fcan be lifted to a mapping of complexes F: X -+ X' which induces a mapping of complexes T(F): T(X)-+ T(X'). By Proposition 1,7.1, T(F) induces maps on the homology T(F):: H. (T(X» -+ H. (T(X'» (n

Proposition 1: L.T(M) and L.T(f) are well defined and are independent of the projective resolutions. Moreover L.T( n ;> O} are additive covariant functors. Proof: Let X and Y be two projective resolutions of M. By Corollary to Theorem 1,7.1, X and Yare homotopically equivalent. Hence T(X) and T(Y) are also liomotopically equivalent (Exercise 2, 7.2). This implies by Corollary to Proposition 2, 7. 1 that Hn(T(X» "'" Hn(T(Y» for all n, i.e. L.T(M) is unique. up to isomorphism. Let I: M -+ M' be an R-Iinear map and X, X' projective resolutions of M, M' respectively. Choose a lifting F: X -+ X' off. Then F is unique up to homotopy, i.e. if G: X -+ X' •

.

... _L.T(M')

-!_

M _0, there is an exact sequence U

L.T(f)

~

L.T(g)

L.T(M) _ _

8. , L.T(M·)~Ln-IT(M)

-+ ... which has the naturality property. Id

;> 0)

L.T(/) : L.T(M) -+ L.T(M') (n ;> 0).

t.e.

0_ M' ~ M

nl)

is another lifting of I then F "'" G. Then T(F) "'" T(G} and by Proposition 1, 7.1 T(n: = T(G): for all n, i.e. L.T(/) is well defined. It remains to verify that LnT(n ~ 0) are covariant functors. Now L.(Id) = Idforif/=ld:M -+M, we can choose X' = X and F=Id. Lei! X', X and X" be projective resolutions of M', M and Mil respectively. If F: X-X' is a lifting off: M -+ M' and G: X' ~X" is a lifting of g: M' .. M then GF: X-XU is a lifting of gf: M _ M and L.T(gf) = T(GF): = T(G): T(F): (T covariant) = L.T(g)L.T(/). The verification that the functors LnT(n ;> 0) are additive is routine and is left as an Exercise (Exercise 3, 7.2). U

Proof: (i) If M is projective, 0 -+ Xo - _ M -+ 0, is a projective resolution of M where Xo = M and X, = 0, i ~ 1. If I: N .. K is the zero map, condition (iii) of the functor shows that T(f) = O. In particular if N = 0, IN = 0, and hence T(IN} = IT(N) = 0, i.e, T(N) = O. Hence X, = Ofor i ~ i implies T(Xt) = 0, i ~ I, i.e, . LnT(M) = 0, n ~ 1. (U) Assume now that T is right exact and let .. X. -+ ..... Xl ..!.~ Xo ~-+ M .. 0 be a projective resolution of M. Since T is right exact, the sequence T(dl ) __

T(Xt )

T(o)

T(Xo) -

=-

T(M)

.

T(M) .. 0 is exact. Hence

T(Xo)

Ker T(E)

=

T(Xo) ImT(dl )

=L

T(M). 0

(iii) Let 0-+ M' ...!.. M -!._ M"_O be an exact sequence of R-module. By Theorem 3, 7.1, it has a projective resolution F

G

0-+ X' - .. X ~ X" .. O. F.

G.

For each n, ..

0_ X'. __ X n _ Xu. -+ 0 IS split exact. By Exercise 8, 7.2 the sequence

o -+

,

T(F.)

T(G.) ,

II

T(X.) --+ T(X.) - _ T(X .)_ 0 is exact and splits, i,e, we have an oxact sequence of complexes

U

o ..;. T(X') T(~

T(X)

~

T(X") -+ 0

This induces an exact homology sequence 8•

Theorem 1:

Let T be a covariant additive functor.

(i) Lnr(M) = 0, n ;> 1 if M is projective.

Then

... -+ H.T(X') ..... H.T(X) -+ H.T(X") .... H.-IT(}(') -e- :..

168

HOMOLOGY

COMMUTATIVB ALGBBRA

The naturality condition is a consequence of the naturality of the homology exact sequence. We now define the right derived functors R"T(n ~ 0) of an additive contravariant functor. Let M be an R-module and

•

IX --c>- M a projective resolution of M. complex X d.

X: '" -+ X" --c>- Xli-I we have the complex 'T(X)

~

... Xl

We now consider the left derived functors of the tensor product functor. Let N be a fixed R-module. Consider the functor T given by T(M) =M® NandT(/) =I®IN for I:M -+ M'. R

Since T is a covariant additive functor, LnT(n ~ 0) are defined and we denote L"T by Tor~( ,N). . From the definition of LoT, it

Applying T on the dl --c>-

169

is clear that if M is an R-module, X _ ..... M -+ 0 a projective resolution of M, and X ® N is the complex

Jf ----+ 0

R

T(d J)

T(d.)

T(X); 0 -+ T(Xo) --+ T(XI) -+.~.-+ T(X"_I) --+ T(X,,)-+ ...

The n-th homology of the complex T(X) i.e.

then TorR (M, N) = Ho(X ® N).

s, (T(X)) = K;~ ~~~:t) is defined to be R"T(M). Let/: M -+M'

•

i.e.

~

-+

M' is an R-Hnear map, .'

F=F®N:X®N .... X'®N. R

R

R

The induced map on the homology

.F: :Tor: (M, N) -+ Tor: (M', N)

(n ~ 0)

R"T(/) : R"T(M') -+ ROT(M) (n

IfI: M

R··

X, X' projective resolutions of M and M' respectively.! can be lifted to a mapping of complexes F: X -+ X' which induces a map

be an R·Hnear map, and X, X' projective resolutions of M and M' respectively. Then I can be lifted to a mapping F: X .... X' of complexes which induces a map T(F) : T( X') -+ T(X). Since T(F) is also a mapping of complexes, it induces mappings on the homology T(F): : H" (T(X'» ..... H" (T(X»

.

0)

is the map Tor: (f, N). From Theorem I, it is clear that Tor: (M, N) = 0, n ~ 1, when M is projective, for all N, and Tor: (M, N) """- M ® N, for any two

The following are analogues of Proposition I, Theorem 1 and can be proved on the same lines.

R

R-modules M and N. We will now show that for any R·Hnear map t/>: N.~ N ', there exists an R-Hnearmap

Proposition 2: R"T(M) and R"T(f) are well defined and are independent of the projective resolutions. Moreover R"T(n:> 0) are additive contravariant functors.

Tor: (M, .p): Tor: (M, N) - Tor:(M, N').

If X is a projective resolution of M, ,p induces a mapping of complexes G: X® N .... X® N', which in turn induces a mapping

Theorem 2: Let T be a contravariant additive functor. Then (i) R"T(M) =-0, n ~ I if M is projective (ii) ROT"", T if T is left exact (iii) For any exact sequence

.

o -+

f

M' - .... M _

g

M"

~

R

on the homology

G:: H" (X® N) ..... H" (X®N'), i.e.

0, of R-modules there is an

exact sequence

R

1 R"T(g)

....... ROT(M") - - + RnT(M)

R"T(/) --c>-

which satisfies the naturality condition.

/

R"T(M') ~ R"+IT(M') -+...

Tor: (lIf,

R

,p): Tor: (M, N) .... Tor: (M,

Clearly the correspondence ,p ..... Tor~(M, (i), (il) and (iii) of a covariant functor.

N').

,p) satisfies the conditions

-

170'

COMMUTA TIVB ALGBBRA

HOMOLOGY

f

Proposition 3: Let 0 -+ M'

--+

+

g

M --+ M" <jI

4

it is clear that if M is an R-~odule and X -"_ M _ 0 a projective resolution of M and HOmB(X, N) is the complex

0

.

and 0 -+ N' --+ N --+ N" -+ 0 be exact sequences of R-modules. Then there exist exact sequences R

Tor. (f, N)

I

HomR (X, N): 0 -+ HomR(Xo, N) -+ Homll (X1,N) -+ ... HomR(X.. N)

. R

... -+ Torn (M • N) ------+ Torn (M, N) Tor. (g, N)

.R

- - - - - . - + Tor;. (M", N) -

and

Tor. (M.

R

+)

a. Il

... -+Tor;;(M,N') - - - - . - + Torn (M, N) Tor. (M. ;; 1 for any R-module M and pdRM = 1, if M has torsion elements. Hence gl-dlm R = 1. Similarly if R is a PID which is not a field, gr. dim R = 1. In particular gl-dim k[X] = I, k field. (i)

If R = Z. and M

= 2z., pdRM =

Proposition 1: Let M be an R-module. if and only ifExti (M, N)

=

Ext~+'(M, N)

= 0, i;;;'

(iii) Ext,;r(M, N)

1 for all N

= 0, for all N

(iv) If O-+K"':'l-+XII-l -+ X1I-1 -+ ... -+ Xo -+ M -+ 0 Is exact with X, projective (0 :>;; i:>;;

over all R-modules M.

(iii)

(ii)

/I

~ 1). then Kil-l is projective.

where the supremum is taken

M

EXlUIlpJes:

The following conditions

(i) pdRM:>;; n

p_

... -+Z. - - Z. - - ... _Z. - _ Z. - _ Z. - _ 2Z._ 0

Definition:

-+ Extle(M, N') = O. It follows that;ji is surjective, i.e. there exists some g E HomR(M, N) with t.\i(g) = <jig = f.

Theorem 1: Let M be an R-module. are equivalent.

Consider the complex

175

00

and gr. dim Z.

=

00.

Then Mis R-projective

0 for all R-modules N.

Proof: li):> (ii). If pd.(M) :>;; n, then M has a projective resolution X of length n and using this resolution to compute Ext, we have Ext;+' (M, N) = 0, i 1, for all N, as X.+, = 0, i:> 1.

>

(Il)

:>

(iii) is clear.

(iii) :>(iv). Assume that Exti+ l (M, N) = 0 for all N.

Consider an exact sequence i·-l

O-+K..-l - - +

with X,(O

~

X

d.-l

11-1

--+

X.

dl

&

•

-Xl -~ X o --+Jl.(-+O

11-1-+ ...

i:>;; n - 1) projective.

n is sulRcient to

show that K1I-1 is projective

i.e. Ext1 (KII- 1' N) = 0 for all N. 1(M, N).

We show that Exti (K"-ll N)

=. EXli+

The given exact sequence splits into short exact sequences Proof: If M is R-projective, clearly Ext; (M, N) = 0, n;;;' 1 for all N (Theorem 2, 7.2).

Conversely assume that Exti (M, N) = 0 for

To show that M is R-projective, consider an exact sequence ,.; oj. O-+N - - N - _ N" -+ 0 and an R-lincar mapf:M -+ N". It is sufficient to find some g E HomR(M, N) with ljIg = f. Consider the exact sequence (Proposition 4, 7.2)

L

Hom.(M, N)

--t. Hom.(M. N")

•

i

d

1 1 O-+Kl --+ X, --+ Ko _ 0

all N.

0-+ Hom.(M, N')

io

O_Ko --+ X o --+ M_O

........................................., /._1

d._

1 O-+ KII-l --+ Xn- 1 --+ K.- 2 -s- 0

where

Ko = Ker e = Imd" and K,=Ker d, (1;;n -1)

~.~

."','"

~,,';\.'-";,

176

.

COMMUTATIVE ALGEBRA

HOMOLOGY

"0

6:

The exactness of 0 -+ Ko -----+ Xo -----+ M -+ 0 implies the exactness of -+ Ext; (Xo• N) -+ Ext; (K o' N) -+ Ext~+l (M, N) -+ Ext~+l (Xo• N) -+ ... and the two terms on the extremes vanish as X o is projective.

(n ~ 1).

=

EXtR (Ko' N) EXI;+1 (M. N). (n ~ 1).. Using the same argument with the second exact sequence

Hence

o -+ KI -+ Xl -+ we have we have

Ko -+ 0

ExtR(Kg. N) """ Ext R- 1 (KI • N).

Continuing this process

Ext;tl (M, N) = Ext; (K" N) = Ext~-I(KI' N) """ Ext;;a (K a, N)

1

... """ Ex t (K_I,N)·

Hence Ext;+l (M, N)

=0

implies Ext~ (K....l • N) = 0, for all.N,

i.e. K.-l is projective. (iv) => (i). Construct a projective resolution of M as in Proposition 3, 7.1 and stop at the (n - 1)-th stage giving rise to an exact sequence 0-+ K.-l -+ X.- I -+ ... -+ X o -e- M -+ 0 with Xl (O.r;; i (iv) The given exact sequence induces an isomorphism, as in (ii) above, viz.

Ext;+l (M, N)

O!.

for all R-modules M .

Extli (M, T._,),

. ~y a~sumption (iii) we have Ext1t' (M, N) = 0 for all M. This ~mpbes . Exth (M, T.-I ) = 0, t.« Til--, is injective. '(IV) => (I) Construct an injective resolution of N and stop at the n-th stage yielding an exact sequence O~N~QO~QI~ '" ~Q.-I~T._I~O

with

Q,(O";;;;i";;;;n-l) injective.

. By (iv),

T._, is injective and hence idRN ,.;;;; n.

Corollary 1: idRN ,.;;;; n ~ Ext;+l (M, N) Corollary 2:

0 for all M.

=

st. dim R = Sup idR(N), over all R-modules N. N

Proof:

~ IdR (N)

Corollary 3: gl. dim R

= 0 for all M, N,

00,

To prove the second relation, consider a pr~jective resolution X 'ofN d.

Since P is projective, the complex X®P: ... ~X.®P~ ... ~Xo®P~N®

R

R R R Hence To~ (N, P) =H.(X@P)=O,n;;;. 1.

< n for

then gl. dim R= 00. Assume therefore

any R-module N.

o~ N ~ Qo~ ... ~ Q.-I -+ Til--, ~ 0,

R

O~K~F~M~O.

This gives rise to the following exact sequences 8.

Consider an exact

o=

with Q, injective

0= Tor: (N, F) ~ Tor: (JV, M) ~ Tor=_, (N, K)

T.-J

isexaet.

Let n;;;. 1 and assume the result for (n - 1). Express M as the quotient of a freC? module F with Kernel K so that we have an exact sequence

Tor~

(F, N) ~ TorR" (M, N) ~ Tor:_ l (K, N) ~ Tor:_ 1 (F, N) if.

~ Tor=_l (N,

(0";;;; I " n-l). Now as in Theorem 3 Ext;+l (RII, N) O!. Extl(RII,

P~O

We now prove the Proposition by induction on n. For n = 0, the result is true as

that SfP pdRRII";;;; n. It is sufficient to show, in view of Corollary 2 that idR (N) sequence

d,.

X: ... ~X.~ ",XI ~Xo~N~O.

R

1

1

[

O~Xo~P~O,Xo=P.

M®NO!.N®M.

= Sup pdRRII, where the supremum is

If Sup pdRRII =

Proof: We firstobsorve that if P is projective, then To~ (P, N) = 0 .(n ~ 1) for all N and Tor: (N, P) = 0 (n ;;;. 1) for all N. The 'first relation follows by computing Tor using the resolution

,.;;;; n for all N.

taken over all ideals I of R. Proof:

Proposition 5: Fo~ R-modul~s M and N, there exists an isomorphism Tor: (M, N) c: (i).

Let 0 -? N' -+ N be an exact sequence of R-modules with N' and N finitely generated, It is sufficient to $ow that

188

COMMUfATIVB ALGBBRA

189

HOMOLOGY

o ~ N' ® M R

~N

® M is exact.

Proof:

R

The given sequence induces an exact sequence

N'

o ~ N'n[kN ~

o ~ R ~ R ~ R/[ ~ 0,

N

is exact,

N

N

(*)

7.3. EXERCISES -,

1.

(N' nlmN) = N' nIkN for all k;;;' m.

Let P denote the image of(N'nImN) ®M ~ N' ® M. N'

~~kN

:=

idRM

.

193

Mill' We claim thatYl' YI'.... Y,

is a generating set of M" over R a• If Y EM". write Y .... Ea,x, and 1

Proof: Iff is a polynomial function of degree r, then there exilts some g(X) E Q[X] ofdegreersuch thatf(n) =g(n), n» O. (i.e. for all n greater than or equal to some fixed no)' If heX) = g(X + 1)g(X). 6.f(n) = h(n), n > > 0, showing that 6.f is a polynomial function of degree (r:-l). The converse is proved by induction on r. If r = 1 6.f(n) =/(n

+ 1) -

fen)

= qo E

Q (n»

= ~[(n + l)'-n'J + hen) where heX) E r

of degree ~ (r-2). If .p(n) = f(n)-

aon' , then

. r

function of degree

C"

b, ERa'

Q[X] is

Colollary: IR.(M,,)
Pr(M. n) >P..(M. n). This shows that Pr(M. ) and P..(M. ) have the same degree.

then

n>O

X (grr(M). n) = /R/r (

Proposition 4: The degree of the polynomial function Pr(M. n) doe. not depend on the ideal of definition I. .

Thus

II!..M'/M~")

is a polYnomialfunction.

(Z:J

(*)

196 COMMtlTATIVB ALGEBRA DIMENSION

By Artin Rees Lemma. 1M'. = M~+l for all n > no so that l-'+·M'

which implies

c

M~+ •• =, l·M~. c InM'

lR(I.:;'~') > lR

i.e. P,(M'. n + no)

(M:J >

homogeneous component of A = k [Xl. Xl..... X,]. Then A. Co!

m"lm·+1 as k·spaces so that

lR

(l~')

;> lR (M~' );> P,(M', n)

(**)

"+'"

This shows that the polynomial functions P,lM'. n) and l,c(M'IM'n) have the same degree and the same leading coe;fficient. Hence P,(M';n) -IR (M'IM'.)

= R(n)

is a polynomial function of degree less than deg lR(M'IM'n) which is less than or equal to d(M) = deg P,(¥, n) by (*). Since by (**). R(n) 0 for n > > 0, the leading coefficient of R(n) is non-negative.

X(grm(R), n) = t, (m.lmn+l) = lk(A.) = (Exercise

deg P,(M". n) ~ deg P,(M. n)

8.1). Since n -+

'X (grm(R), n)

== (n + r

-

r-I

is a polynomial

n~

1)_

MBn)

(*)

Choose a non-zero homogeneous element b E Bd of degree d. Then bAn c Bn.+d' n ~ 0 so that

h (B.",,);;;. t, (hAn) = lk(A.);> lk (B n).

deg P,(M'. n)";;; degP,(M, n).

(mn::l )

r-I

o -+ B. -+ A. -+ m·lmn+l -+ 0., This implies

always. Hence by Theorem 2. we have Let R be a Noetherian local ring with maximal ideal m generated by r elements. We saw in the Corollary to Theorem 1 that

C: ~~ I)

+ r - 1)

we have for each n an exact sequence of k-spaces

Corollary: If M' is a submodule of M. then d(M') '" d(M). Proof: If MIM' = Mil, we have

I,

(n

function of degree r - I (Exercise 2, 8. I) the result follows. Conversely assume that .p is not an isomorphism so that ker.p = B oF O. B has an induced gradation B = EB 1: B n and

>

'X (grm(R). n) = lRI,.

197

Thus lk(B.) and lk(A.) = (n

+r -

1)

are polynomial functions r-l of the same degree (r - 1) and the same leading coefficient. The equation (*) now gives deg 'X «grm(R), n) < r - I, a contradiction to the assumption. Hence e is an isomorphism.

is a polynomial function of degree ~ (r-l). The following Theorem gives conditions under which the degree is equal to (r-l).

Corollary:

Theorem 3: Let R be a Noetherian local ring with maximal ideal m generated by all all ••• , a, and Rlin = k, The graded k-algebra homomorphisme : k[X1 , Xl' ...• X,] -+ grm(R) given by

Proof: Since X (gr.. (R), n) = t1P m(R, n), the result follows from Proposition I.

.p(Xi) =

ii,

=

a, + m

l

Assume that

.p is

(I E;;; i E;;; r)

an isomorphism and let A. be the n-th

isomorphism if and only if deg Pm (R, n) = r.

8. I.

is an isomorphism if and only if deg X(grm(R). n) = r-l. Proof:

.p is an

1.

EXERCISES

Let S=R[XI , .... X,] be the polynomial ring in Xl' XI' .... X, over R. Show that if Sn is the n·th homogeneous component

198

DIMENSION

COMMUTATIVE ALGEBRA

The following Example shows that even if R is Noetherian. dim R can be infinite.

of S consisting of homogeneous polynomials of degree n. S. is

(n + r- 1).

_a free R-module of rank

Example: Let R = k[X1o XI' .... X•• .. .J the polynomial ring in Xv XI' .... X..... k field and {III} an increasing-sequence of positive integers satisfying nI+1-nl:> ",-11/-1 for all i,

r-l

2.

'Show that if r;?: 1. n

~( :

) isa polynomial function of

degree r. Deduce that if R is Ar1:inian, /R (S.) is a polynomial function of degree r - I. 3. If f is any polynomial function of degree r, show that there , exists ao• a10 .... a, E Q such that g(n)=ao+a·G)

+a G)+.. -+ l

Let

S.

a, ( : )

Let M be a finitely generated module over a Noetherian local ring R and a E m, a non-zero divisor on M. Show that d(M/aM) == d(M) -1. Let M be a finitely generated module over a Noetherian local ring R with maximal ideal m and Show that d (M)

M

its m-adic completion.

P~l

Let R be a commutative ring with 1. By a chain of prime ideals of R we mean a finite strictly increasing sequence of prime ideals of R of the type Po C PI C p •• ... . C p •. II

=1=

=1=

is called the length of the chain.

DefiDitioD: The Krull dimension of R is the supremum of alI lengths of chains of prime ideals of R. Krull dimension of R is denoted by dim R. Examples: (i) If R is Artinian, dim R

=

O. (Proposition 3. 3.3)

(ii) If R is a Dedekind domain. dim R = 1. (iii) If R = k[X1• XI' .... X •• .. .J. k field. then dim R = 00 as (Xl) c (Xi; XJ C ... c(X1• XI' .... X.) (n;?: 1) are chains of prime

=1=

=1=

=1=

ideals of arbitrary length.

,

Xnl+h"', X. I +1 } and S = R- UP,.

P~l

Coht P, where the Irifand Sup are taken over all prime ideals P~I. We now generalise the concept of Krull dimension for an ' R-module M.

8.2. Kroll dimension

The integer

= {X.,.

Definition: Let P be a prime ideal of R. Then the height of P denoted by ht P is dim R p • The co height of P denoted by Coht P is dim RIP. Forany ideal I, we define htl = Inf hi(P) Bnd Coht I = Sup

= d (M).

=1=

P,

Then Rs is Noetherian as each localisation of R s at its maximal ideals is Noetherian and every non-zero element is contained in utmost finitely many maximal ideals (Exercise 7. 3. I). Clearly dim R s = 00 as there exist in R s chains of length n'+I-nl- and these are unbounded. However we shall show later that for a Noetherian local ring R. dim R.«; 00.

= f(n) (n > t> 0). 4.

199

•

I

DefiDition: dim M = dim (An:CM)) if M =1= 0 and dim M = -1 ifM=O. A prime ideal P of R contains Ann(M)if and only if PE SupP(M). Sup Coht(P) = Sup Coht(P) as the Hence dim M = PESuppCM)

PEAssCM)

minimal elements of Supp(M) and AsJ{M) are the same. Example: Let M be a [g. R-module. Then dim M = 0 if and only if every P E Supp(M) is maximal. This is 'equivalent to the condition that /R(M) < 00. (Theorem 2. 3.4). Let R be a Noetherian local ring with maximal ideal m and M af.g. R-module. Since SUPP(M/mM) ={m}, /R (M/mM) < 00 and there exists a least integer r such that /R (M/(au " " a,)M) (i). Under tho assumption (ii), IRp is an ideal of definition of R p generated by n elements so that ht P = dim Rr .:;; n. (i) * (ii), By (i) dim R p .:;; n so that there exists an ideal of definition of R p generated by nelements say (aJs) , ., .• (a.!s). S E R-P. The ideal I generated by {a•• .. .. an} satisfies (ii).

Proposition 1: Let R c S'be an integral extension, J an ideal of S and I = J n R. Then (i) dim R = dim Sand Coht I = Coht J (il) ht(J) 0:;;; ht(I) and equality holds if further Rand S are domains and R is integrally closed.

Corollary 6: (Krull's Principal Ideal Theorem). Let R be a Noetherian ring. a E R which is neither a unit nor a zero divisor of R. Then every minimal prime ideal of (a) has height 1.

Proof: Since ReS is integral. Theorem 1, 4.2 shows that any chain of prime ideals of R can be lifted to a chain of prime ideals of S. Hence dim S ;> dim R. Now. any chain of prime ideals of S contracted to R gives a chain of prime ideals of R. Hence dim R ;;;.dim S. i.e. dim R = dim S. Since RJI c S!J is an integral extension Coht I = dim Rl] = dimS!J = CohtJ. (ii) Let J be a prime ideal of S. Then 1= J n R is a prime .ideal of R and for any chain of prime ideals of S contained in J. its contraction to R is a chain of prime ideals of R contained in I. Hence ht(J) 0:;;; ht(I). If J is any ideal, and 1= J n I, choose a prime ideal P of R containing I with ht(P) = ht(/). Since RII C SIJ is an integral extension, there exists a prime ideal Q ofS containing J which lies above P. Then ht(J}':;; ht(Q} < ht(P) = ht(!). To prove equality, assume that Rand S are domains and R is integrally closed. Consider first the case when J is a prime ideal

Proof: Follows from (ii) => (i) of Corollary 5 for n = 1 and observing that if ht(P) = 0, P consists of zero divisors of R. Corollary 7: Let R be a Noetherian local ring with maximal ideal m and a Em a non-zero divisor. If R' = RI(a). then dim R' = dim R-l. Proof: Since a E m is a non-zero divisor. dim R > O. Taking M = R in the proof of last implication of Dimension Theorem. we get that dim R' < dim R, i.e, I + dim R' 0:;;; dim R. Conversely let aa.· ... a, E m be such that their images in R' generate an

a..

204

DIMENSION

COMMUTATlVB ALGBBRA

* *' . . *

of S. If 1 =0 J PI 1 = Po

n R, I

is a prime ideal of R and for any chain p. of prime ideals of R, there exists a chain

hof prime ideals of S lying above H (Theorem I , 4 .3). H ence t(J) hf(I). Now let I be any ideal of Sand Q any prime ideal o~ S con~aining I: Then P = Q n R J 1 and ht(Q) >0 ht(P) ~ hr(I). Since Q IS an arbitrary prime ideal containing I, ht(J)~ ht(I).

>

Proposition 2: Let ReS be an extension of rings such that S is flat over R. Then the going down theorem holds for the extension ReS.

P~oof: Let P' e P be prime ideals of Rand Q a prime ideal of S lying above P. Then Sa is flat over R p and R p -7 Sa being a local homomorphism (i.e. mapping the unique maximal ideal into the unique maximal ideal) Sa is faithfully flat over R p • Hence by Theorem 1.2.3. there exists a prime ideal Q* of Sa lying above P'R p • If Q' = Q* n s, Q' is the required prime ideal of S lying above P' ,nd Q' e Q. \

Coronary: Let ReS be an extension such that S is faithfully flat over R. Then dim S ~ dim R. Consider a chain of prime ideal of R, Po e PI e ... e P r • S. =1= =1= =1= I~ce ISfaithfully flat over R, there exists a prime ideal Qr of S lying above Pr and the above chain can be lifted to a chain of prime ideals of S. Hence dim S dim R. If P is a prime ideal of R, we have clearly the inequality ht P + Coht P :e;; dim R. We now give an Example of a prime ideal P in a Noetherian ring R for which ht P + Coht P < dim R. Proof:

S·

>

Example: Let S = k[[X, Y, Zjj the power series ring, k field and R = S/1 where 1 = (XY, XZ). Let X, Y, Z denote the images modulo 1 of X, Y, Z in R. Since X e (X, Y) e (X, Y, Z) is a =!=

=!=

chain of prime ideals of R, dim R ;?; 2. By Corollary 7 to Theorem I, dim~ ~ 2. Hence dim R = 2. Since I = (X) n (Y, Z), we have (0) = (X) n (Y, Z). Now P = (Y, Z) is a prime ideal of ht 0 in

R and RIP c:< k[[X]] is of dimension I.

Thus ht P

205

+ Cohr P = I

< dim R. Theorem 2: Let R be a Noetherian domain. Then R is integrally closed if and only if it satisfies the following two conditions. (i) For every height one prime ideal P, Rp is a DVR. (ii) The associated prime ideals of a non-zero principal ideal are all of height I.

Proof: Assume R is integrally closed. If P is a prime ideal of ht 1. Rp is one dimensional. If R is integrally closed, so is R p • Hence R p is a DVR (Theorem 2, 5.1). To prove (ii), assume that P is an associated' prime ideal of I = Ra, By localisation; PR p = m is an associated prime ideal of a principal ideal (b) in R p = R'. This implies R' is a DVR, for let

e (b). Then lImlr I e R' and M-l ¢ R'. Now ;\mb-l ¢ m for if ;\mb-1 e m, then M-l is integral over R' h ¢ (b) with >.m

(Theorem 1, 4.1) and this is a contradiction as R' is integrally closed. Hence >.mb-1 = R'. There exists t E m with Mb-1 = 1. For any x E m we have x = 1· x = (htb- 1) x, i.e. m = (t) showing that R' is a DVR. Since dim R' = 1. ht P = 1. Conversely assume that R satisfies conditions (i) and (ii). We show that R =0 n Rp • If CIt =0 alb E Rp for all P, with ht P = I.

.'1'-1

then a E bRp , for all P E Ass(b) by (ii). This implies a E bR, i.e, at E R. By (i) each R p is integrally closed and hence R is integrally closed. Corollary:

Let R be a Noetherian integrally closed domain. Then R=

n

IIp-l

Rp•

Proof: R is Noetherian integrally closed domain implies condition (ii) which implies R = n R p from the proof of Theorem 2.

.t

1'-1

206

DIMENSION

207

COMMUTATIVE ALGEBRA

8.2.

EXERCISES

may a'ssume that R is a field. Then S = R [X] is a PI D and clearly Q~Q.

1.

2. 3.

4.

Let ReS be domains, S integral over Rand R integrally closed. Let Q be a prime ideal of S and P = R n Q. Show that dim Rp = dim SQ. Let 1 c J be ideals such that J is not contained in any minimal prime ideal of I. Show that COhI(I) ;;;, 1 + Coht(J). Let ~ be .Noetherian such that R has only finitely many height 1 prime Ideals. Show that R has only finitely many prime ideals and they are all of height ~ J . Let Pc Q be prime ideals in a Noetherian ring. If there exists one prime ideal P' with PcP' c Q. show that there

Proposition 2: Let R be a Noetherian ring, S = R[Xl, I an ideal of Rand J = IS. If P is a minimal prime ideal of I, then Q = PS is .a minimal prime ideal of J. Proof: By passing to the quotient modulo I, we may assume that 1 = O. If Q is not a minimal prime ideal of S there exists a, prime ideal Q' C Q. As Q'nR c QnR = P, we have Q'n R =P, and

=t-

by Proposition I, Q' = PS, a contradiction as Q' :1= Q.

. In inflni . =F =l= exist mte Iy many such prime ideals. 5. Let R be a Noetherian local ring, M finitely generated Rmodule. a E m. Show that dim(MlaM) ;) dim(M) - 1 and equality holds if a is not a zero divisor of M. 6. Let R be a Noetherian local ring. Show that either R is a domain or every principal prime ideal of R has ht O. 7. Let R be a Noetherian ring and 1 an ideal of ht O. Show that I consists entirely of zero divisors. Show that the converse is also true if (0) has no embedded primes. 8 . Let M be a finitely generated module over a local ring R with

Proposition 3: Let R be a Noetherian ring, S = R [X]. P a prime iclCal of R and Q = PS. Then ht P = ht Q.

m-adic completion M. .Show that dimR M = dimIl M. Let f: R -)- S be a local homomorphism of local rings R and S, and k = RIm. Show that dtm S ~ dim R+ dim (k® S).

Proposition 4: Let R be a Noetherian ring and S = R[X]. Then dim S = 1 + dim R. .

9.

R

8.3.

Dimension of Algebras

We study in this section dimension o~affine k-algebras t:e. finitely generated k-algebras, which are domains. We begin with the study of dimension of polynomial algebras. ' Proposition 1: Let R be a ring, S - R[X] the polynomial ring in X over R. If Q c Q' are prime ideals of S lying above the same

'*'R, then Q -= PS. prime ideal P of Proof: By passing to the quotient by PS, we may assume that P = 0, i.e. R is a domain. By localising with respect to P, we

. 1'rGOf: If ht P = n; by CoroUary. S to Theorem I, 8.2, there is an ideal! of R generated l1y n elements such that P is a minimal prime ideal of I. By Proposition 2, Q is a minimal prime ideal of J "'" IS and since J is also generated over S by n-elements, ht Q ~ n = ht P: Conversely for any chain of prime ideals

Po C PI C ;.. C p. = P of R. there is a chain Qo C Ql C .. , Q. = Q =I- '*' =I9'*' of prime ideals of S where Q, = P, [Xl. Hence ht Q ;;;, ~I P.

hoof:

Let Po C PI C ... C p. be a chain of prime ideals of R.

. = P,S'*'(0 Ideal y,-Al r 1. we claim that there exists . an . 1, 1 with ht I' = h-l. This is so because. there eXists a p~e id c. I P"""' I with ht P = h. I f Po C P Ie ... C PCP-PIS r.-l hI ea oJ, =l= =l= =l= =l= Proof:

r:'

a chain of prime ideals in P of. len~th~. the~ l' ~ 1 Pr.-~ has ht (h-1). By induction, there exist Y to Y I' .... Y a satlsfymg (I) ,an~ (ii)forl'. LetR·=K[y·I.Y·I.· ... Y'a]. Then l'nR'= '~hYIR. By Proposition 1,8.2. ht(I' n R') = ht(I') = ~- ~ and h~(l n ~') = htI=h. Hence there exists somelEI n R. wlth/lf; 1 n ~. By . 1"0 0 Y' h. Y'HI. .... y') replacmg/by \. 0•.... •• we may assume Without loss of generality that IE Kfy' h. Y~+l;"" y' a]' By Exercise 7, 4.4,

Proof: By Normalisation Theorem.R is integral over a polynomial algebra k[YI' Y•• ..., y.] where n = dim R. Let F and L be the quotient fields of Rand k[yl....Y.) respectively so that FIL is algebraic. Hence Tr. degkF= Tr, deg"L == n. Theorem 2: Let R1 and R. be affine k-algebras over a field k and ' P a minimal prime ideal of R 1 ® R Then k

Coht (P)

Proof:

I.

= dim R1 + dim R~.

By Normalisation Theorem. there exist polynomial algebras Slover k such that SI C R 1 and S. c R. are integral extensions. Let X b L" K •• L. be the quotient fields respectively of 8 1" R l • S. and R I. Consider the commutative diagram

.'Sx -ID~

210

DIMENSION

COMMUTA11VB ALGEBRA

o

1

o~

KI

isa minimal prime ideal of S containing 1+ Ker .p and its image i2 under IjI in S is a minimal prime ideal of S containing IjI(Ker .p). SinceKer.pisgenerated by n-eJementsX, 0'1 - I 0 X,(l"';;i or;;; n). its imageundercjl is also generated by a-elements so that ht(Q) .,.;; n. .Lot Qa be a minimal prime ideal of S contained in Q so that O[Q. has height utmost n in, S!Qo' By Theorem 2

0

o ~ SI ® S.

1

® K, k

~

~

-1

R 1 ® R,

LI

211

r

--

\

dimSIQo = dim RIPI + dim RIP•.

® L, k

= I, 2), L I ® L, is free over K I ~ x; k Hence no non-zero element of SI ® S. can be a zero divisor of k R ® R • Since P is a minimal prime ideal of R I ~ R•• its conI 1 tra:tion Q to SI ® S. is zero for if Q¢O. there exists x E Q. x::l= 0

Sinceht (QIQJ = dim (SIQo)- dim (SIQ). we have

Since L, is free over K, (i

n ;;,'hl ("QIQo) ~ dimRIPI + dim RIP. - dim RIP I

i.e.

.

Coht P >Cohl PI + Coht p.- n.

~; .'~

.

k

such that x is a zero-divisor of R 1 ~ R.. Coht P

~ CoM Q = dim SI ® S.

Hence =' dim SI+ dim S.

8.3. EXERCISES •

= dim R I + dim R I • We now prove the algebraic an~logue of the geometric ~esult concerning the dimension of the ir.re~ucible components of intersection of two irreducible affine varleties,

'Ih(or(m 3: Let PI and P, be prime ideals of the ~olynomial. r~ng R=k[X • X ••.•. , X,,]. k field and P a minimal prime ideal containing I PI + Pt. Then CoM (P);;' Coht(P 1) + Coht (P.) - n, .Proof: Consider the map

bE R

;~nerated

+: R ~ R ~ R

given by +(a ® b) 0= abo

Clearly.p is ak-algebra homomorphism with ~er q, by elements of the type a ® I - I ® a. a E R. for If I a, ® b, E Ker.p. then J a,b, = 0 and I

. I

J a, ® bi = I (a, ® 1 - I ® a,) (l ® b ,).

,

,

Let S =R®R. S=RIP 1®RIPI and <jI: S~S-the tensor product k kWh of the natural projections R -+ RIPI and R -+ RIP.. e ave an -

tjI-

exact sequence O~l~S~ S~O where " 1=Kerljl=PI®REB R 0 P •• SinceP is a minimal prime ideal of Rcontaining PI +p•• Q=.p_l(P)

Let P be a prime ideal of ht h in R = k [XI' X•• "" XnI. k field. Show tha~ PR p is generated by h elements in R p . Show also that if P is a maximal ideal of R. P is generated by n-elements. 2. Let R be an affine k-algebra. k field. Show that all maximal chains of prime ideals of R are of the same length equal to dimR. 3. "Let R be a ring and let S consist of all IE R [X] such that the coemcients of I generate the unit ideal. Show that S is a multiplicatively closed set in R [X] and dim R [x]s = dim R. 4,'0 If R is a Noetherian local ring, show that 1.

dim R [[X]]

=

1 + dim R

(Hint: If d = dim Rand 1 =(al • at .. " ad) is an m-primary ideal, then (al • a•• ." ad, X) is an (m, X) primary ideal in .R ([Xl].). "S., Let m' be a maximal ideal of R ITXl] whose image under the Qtural map tIt:.R [[XJ] ~ R given by ~ (X) = 0, tit IR = Id is m.' . <J" Sbow that R ITXl].., ""R.. [[Xl]. (see Exercise S. 2.1) ., s. If. R is a" Noetherian ring show that dim R = Sup dim R..

w1lere thi supremum is taken over all the maximal ideals of R. Deduce from Exertises Sand 6 that if R is Noetherian. _dim R[X,• X...... X,J] =n + dim R.

212

COMMUTATIVe ALGEBRA

DIMENSION

8.4. Depth

Let R be a Noetherian local ring with maximal ideal m and M a finitely generated R-modulc. We know from the dimension Theorem that there exist a" ... , a. E m (n = dim M) such that

(. al • a., M ..., an) M hal finite length.

~~~;fIttn( M) =s(~);> s(M)-1 =dimM-l. ··'·'·"~L.· . aIM '. 1 M . . N

Example: If R is a Noetherian local ring of dim d, and ai' ... ,0(/ is a generating set for an ideal of definition of R, then au "', ad is a system of parameters for R. In particular if

lN

~~~duction dimN~ dimM-(t-l) and dim(

then XI' XI' ... , X. is a system of parameters for R. Theorem 1: Let M be a finitely generated R-module of dimension n over a Noetherian local ring R with maximal ideal m, If

a., ''', a, E m, then dim (

Ql'

M ) M ;> n - , and equality holds 0 .••

,0,

if and only if ai' .... a, is a part of a system of parameters for M. Proof: Let I and J be ideals of Rand N = MIIM. The composite of the maps M -+ (1+ J)M.

:it- = N

Hence MI(I

-+ ~ is surjective and has Kernel

+ J)M ~ J:v. Inparticu1arifl=(alO""

M

N

a'_l)

M

and J = (01), then (01,... ,a,)M 0/. -N a, where N = (0 1" .. ,a,_JM' We now prove the Proposition by induction on t. If

a" a.....a, E

MM = N are such that (4 N _)N has finite a 1, ••• ,a,

length. (t least). then ( M ) has finite length so that a,-au a.,..., a,

s(M) If R = K[X. Y. Z]; then X. Y(I-X), Z(I-X) is an R-sequence. Proposition 1: Let M be an R-module. The following conditions are equivalent. 0l,

On

(ii)

a.,

O, is an M-sequence and a,+1.... 0n is an

Proposition 3: Let M be a finitely generated module over a Noetherian local ring and a...... a,an M-sequence. Then 0 ...... 0, is a part ofa system of parameters for· M. Proof: The proof is by induction on t. divisor of

if and

. fOt. :... Ot_,

where n = dim M.

0 1, ...0,

N

=

=

dim M - 1.

By Theorem

M

(all ...• 0'-1) M

zero

I, {Ot}

=n-(t-l)

Since a, is not a zero divisor of

.!!- = dim N -

M • dim (a1> ... , 0'-1)M a.N

N

M

a,N =- (alo ...• a,) M

The result now follows from the definition of

M-sequence and by repeated use of the isomorphism for 1 = (01"", a,) and J = (0,+1). (1 .,;;; i";;; n-I). There are many similarities and differencesbetween M -sequences and systems of parameters. We have seen in the Example at the end of Theorem I that the members of a system of parameters can be zero divisors. However this cannot happen in an M-sequence. We have also seen that if a1, .... at is a part of a system of para-

~

= 1. 0 1 is not a

If I > I. by induction, is a part of system of parameters for M and hence dim

M )M sequence, for all i, (1 ,;:;; i";;; n).

N

dim

If t

isa part of system of parameters for M.

Proof: If 1 and J are ideals of Rand N = M/IM. then we have M

a, au

: Clearly ht(1)";;; t by Corollary 5 to Theorem 1. 8.2. To wht(l) ~ t, consider a minimal P E Ass(l). Then P::J P', P' E -t5A;a(OI ..... 0'-1) and P ::J P' since a, is not a zero divisor of ~{,:. 9= . , R )' i.e. a, ¢ P', at E P. Hence ht P;> ht P' + 1. By (OJ,••••, a'-l induction ht P';;Jo t-I so that hi P ~ t, i.e. hi 1;> t.

is an M-sequence.

(i)

(1 + J)M "'" IN'

0 10 O2.... ,

I.

.

An M-sequence for R = M is called an R-sequence. The condition (ii) for i = 1, means that a1 is not a divisor of M.·

(

215

DIMENSION

COMMUTATIVB ALGBBRA

l. But

so that

M = n -(t - 1) - 1 = n - I. (01, .... 0,) M ,.bY· Theorem I. 0 10 ... , at is a part of a system of parameters I

~.

.

.:, ..:_) t: . .~

dim

.

216

DIMENSION

217

COMMUTATIVE ALGEBRA

'~'.," .: 3: ,Let Robe a Noetherian ring, M a finitely gener~ted Then any two maximal '. '"x Ceiicontained in I have the same length.

'tr~ /. all ideal ofR with 1M #' M.

The following Example shows that a permutation of an Rsequence need not be an R-sequence. Example: Let R = k [X, Y. Zl, k field. Then X, Y (1 - X), and Z (1 - X) is an R-sequence but Y (1 - X), Z (1 - X). and X is not an R-sequence. However the following Theorem shows that any permutation of M-sequence contained in the Jacobson radical is an M-sequence.

"A'\ve first show that if aI' ••.• On is an M-sequence and N .' 1t~module annihilated by (01, ... , an) then _' Ext; (N, M) ""',HomR (N. (01'

.~~ an)M)-

'\0.

Consider the exact sequence 0 --»- M --»- M --»Theorem 2: Let R be a Noetherian ring, M a finitely generated R-module and a" ..., On E J(R) an M-sequence. Then any permutation of 0 1, "', On is also an M-sequence.

(

=- HomR (N, (a Ma )M) = 0 as .a a,.N = 0 and (In is not a zero divisor of (~""~_I)M' It follows then from (.) that llxt'R-l(N. a~) =- Ext~ (N, M). If M = :M'

1

Then alx E a.M, i.e.

~,

JI E aIM.

I

M. Hence 0IX = a.alz and x = a.z. as 0 1 is not a zero divisor of M. Hence x = O. If To prove (0, let N be the submodule of M annihilated by . . d" f M M x E N, 0aX =0 O. Since a. IS not a zero rvisor 0 -M' x E al , °t i.e. x = 0lJl. JI E M. Now a.x = 0 implies a.alJl =0 0, i.e. o.Y = 0 as QIis not a zero divisor of M. Hence any x EN' can be expressed as x = 1)1, JIE N. i.e. N ... o,N_ Since 0t E J(R>" by Nakayama Lemma. N = O. t.e.' JI

= alz,

Z

by induction

E

CoroDary: Let R be a Noetherian local ring and M a finitely generated R-modu1e. Any permutation of an M-sequence is an M-sequence. Proof: Since an M-sequence can not contain units they are contained in m = J(R).

1'''')

II-J

'

Ext~-I(N'a7M) =-Hom N'(a.,.~an)M) ""'HOmR(N·(al... ~n)M)

0..

°

(.)

where 71:1 is also mUltiplication by 71 0 1 (Exercise 7, 7.2). Since 0IN = O. 71;' = 0 and hence 8 is surjective. By induction. ,Ext'R-l(N, M)

0lX = o,.y, JI E M. Since o. is not a zero divisor of

71:.

Ext'R (N. M)

° is not a zero divisor of M/aiM. = 0, x e..M/o.M.

M)8

~ Ext;-l(N. M) ~ Ext;-l N, 0tM ~ Ext'R(N, M) --»-

(0 a. is not a zero divisor of M and To prove (ii), let alx

t

il 100lar" multiplication by at" This induces an exact sequence

Proof: It is sufficient to prove the result for a transposition of successive elements. In view of Proposition 1, it is sufficient to prove this for the transposition (1.2). i.e. "

(ii)

M M --»- 0 where 710 •

0

R(

and the proof is complete. If we take N = RII. with ai' a., an E I, we have e."

Ext; (Rlf, M) Now HomR(RII.

M

=- HomR( Rlf. (a...~o.)M)-

(al ... ·,a.)M

zero element' of M =

) #' 0 if and only if there is a non-

M )M annihilated by I, (al.···,a"

P E Ass(M). Hence Ext; (RII, M) = 0 f ¢, U p. P E Ass(M). " at , .... an is not a maximal M-sequencein I, as 1M =F M.

i.e, 1 c P,

218

DIMBNSION

219

COMMUTTAIVIl ALGEBRA

Hence the smallest integer n for which ExtR (RII. M) =I- 0 is the number of elements in any maximal M-sequenee contained in 1. Definition: Let R be a Noetherian ring and M a finitely generated R-module. I an ideal with 1M =F M. Then depthI(M) is the number of elements in a maximal M-sequence contained in I. Clearly i~.is the sma!lest i?teger n for which ExtR (RII, M) =I- O. If R IS a local nng with maximal ideal m. then depth",(M) is called the depth of M and is denoted by depth (M).

Theorem 4: Let M be an R module and aI' .... a. an M-sequence. The homology of the Koszul complex X = K(M, a" a., .... an) is given by _ ) M H,(X) = 0, i~ I and Ho(X 0=. -(all -a )M' ... ,. 1J

Example: Let R be a Noetherian local ring with maximal ideal m. Then depth (M) = 0 if and only if HomR(Rlm. M) =I- 0 mE Ass(M).

Proof: Let I = (all a., ...• an) so that it is sufficient to show that the augmented complex \ d. _ do M X: 0 --* X. --* XII-l ...,;. X, --* X'-l --* ... Xl --* Xo --* 1M --* 0

Remark: Proposition 3 shows that if M is a [g. module over a Noetherian local ring. then depth (M) ~ dim M. Modules for 'which depth and dimension coincide form a special class and we will study them in the next section. • If I is an ideal in a Noetherian ring with depthI(R) ~n. then ExtR(RII. R)¢:O. so that pdR(R/l»n. We show that if all .... a. is an R-scquence and I = (a•• .. . . a.). then pdR RII = n. We do this by constructing an explicit free resolution of RII as an R-module. This is known as the Koszul resolution and we construct it more generally for an R-module M. Let M - be an R-module. a,• .... a. non-units of R and 1= (al • a•• ... . a.). Let Fbe a free module of rank n over R with

is acyclic. where do: X'o

a,:

r

y(k):

r-l

i (- I)J+l ail ei J~'

X(k-"):

(1 ~ i l

< i. < ... < i, ~ n)

where ei1 ,\ ... ;" J.... 1\ ei! is the term obtained by omitting e,IJ _ from e,. 1\ e'l ... I\. ec . Since a'-10, = OCr ~ 1). we have a complex of R-modules. Tensoring this complex with an R-module M. we have the Koszul complex K(M, a l • a•• .. .. a.) ~

o --* X n -+ XII-' --*

0 -e-

0 --*

~

.,. X, --* X'-l --* ... --* Xl --* Xo--* 0

.n

k

)

--* ~1 --*... --*

xL:"il ) --* xL":.-,})

A ... ;i/'" A ei, •

1

.

d"

xf') --* I"M

--* 0

where Ik = (a l .... , ak)' Clearlythe acyclicity of X(k) is equivalent to the acyclicity of y(k). Since by induction yek-l) is acyclic. it is sufficient to show that the quotient complex yek)! y(k-ll is acyclic (Exercise 2. 7.1). Consider the following commutative diagram

1\ F --*- t.. F. by B,(ei,1\. ei.· .. A ei) =

is the natural projection. We

ld

r

e..

i:t

complex 0 --* M --* M --* 0 and is clearly acyclic. Assume now that the complex X(1 dePt~:(M")' (iii). depthr(M'):> depthI( M) If depthI( M) = depthI(M ).

Let R be a Noetherian local ring with maximal ideal m and m-adic completion R. If M is a [g. R-module, show that any ma:aiD1al M·sequence of R is a maximal Sf-sequence of ft..

Let R be a Noetherian local ring with maximal ideal m, ai' a.,

. . .a. E m and M a finitely generated R-module. ~ its m-adic completion. Show that the following are equivalent: (i) ai' a•• '.' .• a. is a system of parameters (or M. (ii) a1• at.· , .• a. is a system of parameters for.~. (iii) a1• a••. _., a. is a system of parameters for RIAnn(M). Let R be a ring, S a multiplicatively closed set, and a1 .... a. an R-sequence. Show that (ai/I), ... (a.'I) is an Rs sequence provided they generate a proper ideal in R s. Let M be an R-module, a1,... , a. E R such that a/ = b/c/. Show that (i) If a1,... a'-I' bh aI+ 1-' "a. and a1,... a/-1• Ch a'+1' ..,a. are M-sequ~nces. so is al.. .. ah' .. a•. (ii) If a1 •••• a. isanM-sequenceand(a1 , •• • a/-1 , b" al+l ... a.)M =1= M then a1, · • _a,_1o b.; a1+1 ' _ ,a. is an M -sequence, (iii) Deduce that for arbitrary positive integers k 1, k t , .•• , k., a~" ... , 1/.' is an M-sequence if and only if a1 , a., .. . , a. is an M-sequence. Let M be a finitely generated module over a Noetherian local ring and a E m a non-zero divisor on M. Show that depth (M/aM) = depth (M) - 1. Let R be a Noetherian ring, M a finitely generated R-module, I an ideal of R. Show that depthr(M) = Inf depth/,Mp where

223

Deduce that depth(M) = depth(kt). 8.5. Coben-Macaulay modules We have seen that if R is a Noetherian ring and I an ideal generated by n·elements. then any minimal prime ideal of I has ht ~ n (Corollary 5, Theorem 1, 8 .2). The followingExample shows that the different minimal prime ideals of I can have different heights. Example: (X)

n (X -

R = k [X, y}, k field and I = (XY. X(X - 1» "'" I. Y). I has minimal prime ideals (X) and (X-I, Y)

of heights 1 and 2 respectively. However if ht 1= n, all minimal prime ideals of I will have height equal to n, but all prime ideals associated to I need not be of the sal11e height as the following Example Shows. Example: R = k[X. F], k field, S = k[Xi. }'11. XY. Xi}. The ScR is an inte$ral extension and dim S = dim R = 2. If I = (X8) in S, then by Krull's principal ideal Theorem ht I = I. However the ideal P = S n (X. Y) has height 2 and is associated to I as P is the annihilator of X' E S/I. For some special types of rings, an ideal I generated by ht I elements bas all its associated prime ideals of the same height. A classical Theorem of Macaulay says that this is true in the polynomial ring k[X1 , X 2•••• , X.]. k field. Rings with this property are called Coben Macaulay rings. We give an alternative description of these rings and show that it coincides with the classical definition. Let .R be a Noetherian local ring and M a finitely generated R-module. DefiDitiOll:

M is called a Cohen-Macaulay (C.M.) module if

224

COMMUTATl\'B ALGEBRA

DIMENSION

225

>.~ •

depth(M) = dim(M). R is called Cohen-Macaulay (C.M.) ring if the R-module R is C.M.

Examples: (i) If dim M = 0, then 'depth M = 0 as depth M:;:;; dim M. Hence M is C.M. In particular an Arjinian local ring is C.M. (ii) If dim R = 1, then R is C.M., if depth (R) = 1, i.e. m is not an associated prime of (0). Thus, any one-dimensional local domain is C.M. (iii) If dim R = 1 and all the elements of m are zero divisors, depth (R) = 0 and R is not a C.M. ring. In particular the localisation of

(;lJ,' ~)' k

{thowing that HomR (RII. M') = O. We now prove the Proposition

.;'by

induction on dim RIP. If dim RIP = 0, m = PE Ass(M) and '; 'tkpth (M) = O. Assume dim RIP> 0 and choose a E m. such ". that a is not 'a zero divisor of M. If N = MlaM, there exists , Q E Ass(N), with Q :J P Ra, By induction depth (N):;:;; dimRIQ -e dim RIP. By Exercise 4, 8.4, d,pth N = depth (M) -1 is C.M. Assume

an M-sequence and (ai' ... ,

Ass(M) so that P

'. dtm M p = ht Ann(M) =

is an M-sequence and that (_ ) M is C.M. at,_..M ot ak

(0 ,.;; k ,.;; n). This is clearly true for k

now that

mm a. e emen~ 0

If depth (M)

";

It h therefore sufficient to prove the result when I'

.

P . id IS,I\ pnme .. LotI . =. a prIme 1 cal and ht P=r. By CoroUary 1 above R,,: ISa a.M. of dim" i.e. depth Rp = depthp R =,. Hen~ ther~

Ideal

eXIst ) th en I' a.. C ..., M a, E P which is an R-sequence . If I =' (n. ... ..... Or, . R/ I S . . and has dimensions n - r where n = dim R S' P Is of •ht r containing an-R-sequence aI' .... ar, P I'S a mInima ..' I In~ "d . prune 1. eal of 1.= (01, ••• , ar). Hence by Oorollary 1 to Theorem 1 dim R;P = depthRII =dimRII = n -,. ht P+ dim RIP =n = dim R.

:~~tiOD:

A chain of prime ideals Po C PI C ... C p. is said to be '."'. -F -F 9= tura(O.~ed if there exists no primo ideal strictly lying between P, and . T 1. depthm R[X] ~ dbnR[Xl m= T+l and depthmR[Xj = r+ 1.

But

Corollary: (MacaulaY's Theorem). If K is a field. K[X1 • XI' ... , X.I is a C.M. ring.

231

EXERCISES

·.Show that if R is an integrally closed Noetherian local domain 'of dimension 2. then R is C.M. (Hint: If R is a Noetherian' integrally closed domain. principal ideals of R have no embedded components). Let M be an R-module over a Noetherian local ring with maxi. mal ideal m, Show that M is C.M. over R if and only if M is .C.M. over It 3. Letl: R -+ S be a homomorphism of Noetherian local rings with maximal ideals m and /I such that f(m) C n. Assume that . Sis a finitely generated R-module. Show that a finitely generated S-module Mis ,C.M. over S if and only if it is C.M. . over R. 4. Let M be a finitely generated module of dimension /I over a Noetherian local ring R with maximal ideal m. Suppose for every all ... ,o,E m withdim(

Ai

)M=/I-T

at.... , Q r

dim.RjP =

/I -

T, for every P E Ass (

we

have

M ). Show that (ab· .. ,a,)M

MisC.M. Let S be a C.M. local ring. [ an ideal in Sand R = Sll, Show that R is C.M. if [is generated by dint S-dim R elements. . Let R be a Noetherian local ring having a system of parameters a l ..... a, and [ = (ol' .... a,). Assume that the mapping ep:Rf1[XII X 2 , · · · . X,}-+ gTl(R) given by ep(X,)=a, [1(1 ~ i ~ r). ep jRjf = Id is an isomorphism. Show that R is C.M.

+

'.,

REGULAR LOCAL RINGS

233

DefiJiitioD: A generating sot of d elements for m is called a regular system of parameters of R. CHAPTER IX

REGULAR LOCAL RINGS

Examples: (i) If dim R = 0 then R is regular if and only if R is a field. (ii) If dim R = 1 then R is regular if and only if R is a DVR. (iii) R = k[[X1, X., ....X dll, k fleld is regular of dimension d and Xl' XI .... ,Xd is a regular system of parameters. (iv) Let R be the localisation of

The local ring of an affine variety at a point P reflects the local properties of the variety at P. The point f will be nonsingular if and only if the local ring at P is a regular local ring. We define the concept of regularity for any Noetherian local' ring R and obtain equivalent characterisations of regularity including the homological characterisation. viz. g/ dim R is finite. As a consequence. Rp will be a regular local ring when R is .regular local and P a prime ideal of R. We also show that a regular local ring has UPD property and is a Cohen-Macaulay ring. We then investigate two conditions concerning depth and regularity viz. (R k ) and (S,,) (k ~ 0) for a Noetherian ring R. The ring R will have no nilpotent elements if and' only if it satisfies the conditions Ro and SI' The conditions R 1 and S. are equivalent to the normality of the ring, t.e. the localisation at every prime ideal is an integrally closed domain. In the last section we study complete local rings. We prove the Cohen structure Theorem on the existence of a coefficient ring for any complete local ring and deduce the structure of complete regular local rings. 9.1. Regular local rings Let R be a Noetherian local ring with maximal ideal m and dimension d and let k = Rlm. From the dimension Theorem we. know that any generating set for m has at least d elements. Definition: Let R be a Noetherian local ring with maximal ideal m and dimension d. R is called regular if m has a generating set of d elements.

(;!~ ~)'

k field, at the

maximal ideal m = (X, V). Then R is not regular. We have dim R =ht m= 1 as ht(X, Y) = 2 and XI - ya is a non-zero divisor of k[X. Y]. But (X, Y) is a minimal generating for

m.

YJ ) (v) Let R = (X. _ YI) m' k field, where .

(

k[X,

m = (X -1. Y - 1). Then dim R = 1 and R is regular. In this case (X - 1, Y - 1) is not a minimal generating set for mas (Y + 1) (Y - 1) = (XI + X + 1) (X - 1) and Y + 1, XI + X + 1 are units in R. Theorem 1: Let R be a Noetherian local ring of dimension d, maximal ideal m and k = RIm. The following condition are equivalent. (i) R is regular. (ii) grm(R) is isomorphic to k[XI.X." .• ,Xd] as graded kalgebras. (iii) dim R = dim" (mImi). Proof: (i) => (ii). Let a1' al, ....ad be a regular system of para-' meters for R. The mapping if>: k[XI ••.•• .:r.l-+ grm(R) given by ",(X,) = a, + ml (1 .,;;i";; d) can be extended to a graded k-algebra homomorphism. By Theorem 3. 8.1, ", is an isomorphism as deg P",(R, n) = dim R = d. (ii) => (iii). Let : k[X1..... Xd] -+ grm(R) be a graded isomorphism as k-algebras. Then the first homogeneous components are isomorphic as k-spaces. In particular dim" (m/m·) = d. (iii) => (i). If dim" (mlm~ = dim R .,. d. then any basis of d-elements of mImI can be lifted to a generating set of m.. Hence

234

RBGULAR LOCAL RINGS

COMMUTATIVE ALGBBllA

rated by al." .a.; 01+1,· ••• a". Hence R is regular and al>" .a, is a part of a regular system of parameters.

R is regular,

..

Corollary 1: A regular local ring is a domain. Proof: Since gr..{R) .". k[Xu •..• X,,] is a domain. R is also a domain (Proposition 3.6.4). We will show in the next section that Ii regular local ring is a UFD.

Corollary 2:

R its m-adic completion. R is regular.

Let R be a local ring and

Then R is regular if and only if

Proof: By Proposition 1. 6.4. grm(R) 0< gr:;, (R) as graded rings. The result now follows from condition (ii) of Theorem 1. If R is a regular local ring and P a prime ideal of R. RfP need not be a regular local ring in general. as is shown by Example (iv). The following Proposition shows conditions under which RfP will be regular. PropositioD 1: Let R be a~regular local ring of diinensiond and a1> al, ...• a, E In (1 ~ 1 ~ d). The following conditions are equivalent. (i) al. a•• ..., a, is a part of a regular system of parameters for

R. (ii) The images al ....a, of au ... a/ modulo ml are linearly independent over k, (iii) Rf(al•... o,) is a regular local ring of dimension d - I.

Proof: (i) (ii). The elements al....a, is a part of a regular system of parameters 0 1,,, .0" if and only if al' iii•... a, is a part of al.... a". a k-basis of mfm' al•... a, are k-Iinearly independent._ (i) => (iii). let.R = Rf(al ....0,) and iii = mf(ill•... a,}. Let al• a., ... a" a'+l'... a" be a regular system of parameters for R. Their images generate By Theorem 1. 8.4. dim R as an R-module is d - I. This implies dim R = d - I. Since iii is generated by d - 1 elements. R is regular. (iii) => (i), Let al+l •••• a" E m be such that their images in

m.

m=

-( ..!!!.--) is a regular a1>" .a,

235

system of parameters. Then m is gene-

Corollary 1: let R be a Noetherian local ring with maximal ideal m. Then R is regular (i) follows from Proposition I. (i) => (ii) •. Let RfP be regular of dimension in = mlP then dim" (mimi) = d - I.

d - I.

If

-f- I C!. m m l +P •

mm

Since

from the exact sequence

o -+

(ml

+ P)fm l -i> mimi -i> m/(m l + P)· -+ O.

we have dim" Choose

(mlm~P) = dim" (:1) - dim" (ml:

01" .. a, E

P such that their

P )=d-(d"":'I)=1

images modulo ml span

236 (mZ

COMMUTATIVE ALGEBRA

REGULAR LOCAL RINGS

+ P)lniz over k,

Choose a,+1.... Od E m which together with span m modulo mi. Then a1.... ad is a regular system of parameters for R. ·By Corollary 1. it is an R-scquence. Hence P' = (a1.... 0/) is a prime ideal of height t contained in P. Since P' c P and dim.RIP = d - t we have P' = P. . The following Theorem is the algebraic analogue of the geometric result which says that P is a non-singular point of a variety if and only if the local ring at P is a regular local ring.

9.1. EXERCISES

~1'" .0,

Theorem2: Let R = k[X1.... X.]. k field. 1= u;,... .f,) an ideal of R and m = (X1-a1... :X.-a.) a maximal ideal of R containing 1. If S = (RI1)m/l. then S is a regular local ring if and only if dim S =

n-r. where r is the rank of the matrix (8f') . (a) .= "x} (.)

(0 1 , az....a.).

Proof: Consider the k·linear map A: R -+k· defined by

A(f)=((~J. ... (~.)J

fER.

Then 11. maps m onto k· and mZ onto zero so that it induces a k-isomorphism

r:;m =. k·,

If r

=

ronk(Of)

.'tben

!JXj (II)

-

r = dim" A(1)

=

(1-mr +m . Z

dim"

)

From the exact sequence 1+m'

0-+

we have dim" Thus

e

~tZ) =

mlin' =.

dim S = dim" ~ =

m'

m

dim" (:') -dim"

~im"(I';m z} = n-r.

of S so that

m

-mr- -+ mZ -+ I + m'

(I; ml)'

Let in be the unique maximal ideal

I':m' as k-spaces.

n-r.

-+ 0

Thus S is regular ( ~ )

237

1.

Give an example of a Cohen Macaulay local ring whichis not a regular local ring. . 2. Give examples of local rings ReS where either may be regular without the other being so. 3. Let R be a Noetherian ring. I an ideal of Rand S = R/I. S is said to be regularly imbedded in R if I is generated by an R-sequence. Show that if Rand S are regular local rings. then S is regularly imbedded in R. 4. Let R be a Noetherian local ring with maximal ideal m and a Em-mI. Show that R is regular implies RI(a) is regular and conversely if a does not belong to any minimal prime ideal of R. 5. Let R be a local ring. Show tbat R is regular