Commutative algebra

Let S be a multiplicatively closed set and I an idea!

LOCALISATION

of R with InS =
49

~ I such

Proof' IR • id . ' • so IS a proper I eal of Rsand hence itis contained in some maximal.Ideal ~f R s. The contraction of this maximal ideal givOl the required pnme ideal. Proposition 7: (Localisation and Quotients). Let I be an ideal of ~ andp : R -+ R/I the natural projection. For any multipli_ catively closed set S of R not intersecting I let T be 't . , I S Image under Th . !!. . eo.p defines a ring homomorphism R s -. (R/l)r which IS surjective and Rs/IRs = (R/I)r.

i:

Proof: Since S n 't = .I. h '. . closed Th . - '1'. we ave 0 ¢ T and T IS mu1tlphcatively a E R' s e m.apPlng p : R s -+, (R/I)r given by peals) = (p(a)/p(s», If ( /). E,} I~ a Well defined ring homomorphism and is surjective. a t~Aer p, then tp(a) = 0, t E T. P(SI)' then p(s,a) = 0 and hence s a E I Thi . I' l ' IS Imp ies th t ( /) IR a as E s- Hence, Ker p = IRs and Rs/IRs = (R/ l)r.

tf

Prop~si.t~""':8: ' (Localisation of Localisation). Let Sand S' be multlp1tcatlvely closed sets in R with S c S' and h: R . h nat a! defi . s-.Rs,(e . a ur map e ned by heals) = (a/s), s E S, a E R. If/: R -. R IS ,the natural map given by a -+ (a/I), then h maps/(S') ont~ uru,ts o~ RSI ~nd induces a ring homomorphism 1i: (R) , R which IS an Isomorphism. . s If (8 ) -+ S'.

;&~or: Clearly any element of/(S') is of the type (a/I) a E S' and :~~ Ima:e . un~er h is a ~ni* in Rs' . By Proposition 2 there is an uce nng

omomorphlsm 7i: (Rs)/(s') -+ R s', given by

7i (a/s)/(S'/I») = (a/ss'), s E S, s' E S', a E R. Clearly 1i is surjective. I

M?reov~r.'A =.(a/s)/(3'/I) E Ker Ii implies

s,'a = 0 for some

This ~P1tes x = 0, in (Rs)/CS') as (s//I) (a/s) = O. Hence h~.ISES: an ISomorphism.

~doronaryf : ~et S be a multiplicatively closed set and P a prime I ealo RWlthSnp=.p. Then Rp=(Rs)p ••

50

COMMUTAnVIl ALGIlBRA

Proof: Take 8' = R-P. A property is called a local property if its validity for a ring or a module is equivalent to its validity for each of'the localisations . under prime (maximal) ideals. We study some local properties. Proposition 9: Let M be an R-module. are equivalent, .

The following conditions

(i) M=O. (ii) M p = 0 for all prime ideals P. (iii) M .. = 0 for all maximal ideals m.

Proof: Clearly (i) implies (Ii) which in turn implies (iii). Assume now condition (iii) and suppose there exists x EM, x ¥= O. If 1= Ann(x), then I ¥= R, as I rt I. Choose a maximal ideal m with Ie m. By (iii), (xII) EM.. is zero, i.e. ax = 0 for some art m, a contradiction. CoroUary: Let M and N be an R-modules and j": M -+ N anRhomomorphism. The following conditions are equivalent .. (i) I is injective (surjective). (ii) fp: M p -+ N p is injective (surjective) for all prime ideals P. (iii) I",: M .. ~ N.. is injective (surjective) for all maximal ideals m. Proof: Proposition 5 shows that (i) implies (ii) which clearly implies (iii). Assume now (iii) and consider an R-homomorphism f: M ~ N. Let K (resp L) be Ker I (Coker f). Then there is an exact sequence of R-modules i f P O~K_M~N __ L~O. This implies the exactness of

t.

f. N.. __ P. .. _ M.. __ L..

O~K

~O

for-all maximal ideals m. Since/.. is injective (surjective), K-(resp L..) is zero for each m. This implies Kiresp L) is zero. Hence I is injective (surjective).

Proposition 10: Let M be an R-module. The following conditions are equivalent.

LOCALISAnON

51

(i) M is R-flat. (ii) M p is Rp-flat for every prime ideal P. (iii) M .. is R.. flat for every maximal ideal m.

Proof:

Since M is R-flat, its scalar extension M p """ Rp®M is Rp.

\:

..

,

R

flaj by Proposition 2, 1.4. This shows that (i) implies (ii) which in turn implies (iii). Assume now condition (iii) and consider an injective homomorphism 0 '-+0 N'

f~_1 N O~ N' ® M _

~ N.

It is sufficient to show that N ' ~ f. N", IS . ow O -+ N.. 0 -+ N~ ® M m -+ N m ® M m

® M'IS IDJec . . tive.

injective and since M m is flat over R m,

Rm

Rm

is injective. By Corollary 4 to Proposition 5, N.. ® M m """ (N ® M)",· Rm

R

under a natural isomorphism. It follows that 0 ~ (N' ® M)m-oR

(N

® M)m is injective. This implies from Proposition

!},

that

R

o ~ N' ® M~ N ® M

is injective. Hence M is R-flat. We show using localisation that for any ring homomorphism I : R ~ 8 with 8 faithfully flat over R, . prime ideals of R arc precisely obtained by contracting prime ideals of S and this property characterises faithfully flat extensions. Proposition 11: Let I : R ..... 8 be a horqomorphism ofrings and P a prime ideal of R. Then P is the contraction of a prime ideal of S if and .only if P = P«. Proof: If P is the contraction of a prime ideal, then P = P's, P' prime ideal in S. Then P« = P"" = r' = P. Conversely assume that Pr-P.. and let T=f(R-P). Then p. does not intersect Tand hence by Oorollary 4 to Proposition 6, there exists a prime ideal Q of S containing p. with Q n T =.p. Since Q :::> 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

1

and ljI(a ® b) = ab,

Since IjI4> = Id and {J,I\I are R-homomorphisms, Im(
linear maps

M®S R

~~ M®S®S~~M®S . R R R

Clearly 1M ® '" is injective. Since S is faithfully flat over R. the map M -+- M ® S given by x -+- x ® 1 is injective. In particular R

if M

=

Rjl,l any ideal of R. the map Rll >» Rjl ® S is injective. R

This implies I" = I for any ideal I of R. The result now follows from Proposition 11.' Conversely assume that S is R·flat and pee = P for all prime ideals P in R. Then for each maximal ideal m of R. me. = m. In partlcularer' -:F S. This Implies that S is a faithfully flat R-module by Corollary 2 to Theorem ~' 1.4. '

2.3. EXERCISES 1. If M is a finitely generated R-module; show that M s = 0 if and only if there exists some 8 E S with 8M = O. 2. Let I be an ideal of Rand S = 1 + I. Show that the extension of I under R -+- R s is contained in the Jacobson radical of Rs• 3. Let M be an R-module and I an ideal of R such that M m = 0 for all maximal ideals m :J I. Show that M = 1M. 4. Let S be the set of non-zero divisors of R. The 'ring R s is called the total ring of fractions of R. Show (i) S is the largest multiplicatively closed set for which R -+- R s is injective. (ii) Every element of R s is either a unit or a zero divisor. (iii) If R is such that every non-unit is a zero divisor, then R -+- R s is bijective. 5. Let SeT be multiplicatively closed sets, Show that natural map R s -+ Rr given by (ajs) -+ (aM is bijective if and only if every prime ideal of R which intersects T also intersects S. 6.. A multiplicatively closed set S is called saturated if ab E S implies a E S and b E S. Show that

\

1

,

LOCALISATION

7.

8. 9.

10.

S3

(i) S is saturated if and only if R-S is a union of prime ideals. (ii) Every multiplicatively closed setSis contained in a smallest saturated multiplicatively closed set S. (iii) For multiplicatively closed sets Sand T, we have SeT if and only if every R-module M withMs = 0, satisfies Mr=O. Show that for finitely presented R-modules M and N. (Homs(M, N»s = HomR.(Ms• Ns) for any multiplicatively closed set S. ' Show that an R-module (R domain) is torsion free if and only if M p is torsion free for all prime ideals P. Show that a ring R has no non- zero nilpotent elements' if and only if Rp has the same property for every prime ideal P. Is it true that Rp is a domain for every prime ideal P implies that R is a domain? Why? Give an example of an R-module M to show that {AnnR(M)}s may not be isomorphic to AnnRs (Ms ).

2.4. Applications This section is concerned with the applications of localisation in the study of projective modules as locally free modules. Proposition 1: Let u:M -+- N be a homomorphism of R-modules. (i) If N is finitely generated and up:Mp -+ N p is surjective for some prime ideal P, there exists some fER-P such that Uf: M N, is surjective. (ii) If M is finitely generated, N is finitely presented and Up is bijective for some prime ideal P, then there exists some gER-P such tbat u. bijective.

,-+-

Proof: Let K (resp L) be the Kernel (Cokernel) of u. Consider the exact sequence 0-+ K

.L; M ~N --!......... L -+ O.

The sequence O-l- Kp -+Mp ....!!!-l- N p -l- L p -+ 0 is also exact. Assume now that Up is surjective so that L p = O. Since N is finitely generated L is also finitely generated and hence (Exercised. 2.3) there exists some f E R-P with fL = O. This implies L, = O. Hence from the exactness of the sequence M I-+- N, -+ Lf -+ 0' it follows that uf: Mf-+N, is surjective.

S4

I

COMMUTATlVl! ALGEBRA

Assume now that Up is bijective. By (i). there exists some IE R-P with L,,,,. O. Consider the exact sequence 0-+

4. I

I

Kf-+MJ~Nf-+-O.

Since M is finitely generated and N is finitely presented, by Corollary to Proposition 6, 1.2, K, is finitely generated. By Corollary to Proposition 8, 2.3 we have (K')PR Of. Kp == O. Hence there. exis_some 11 E R,-PR, with 11K, == D. Write h.= aff. a E R-P. Since (III) is invertible in R" (all)K, == O. Hence . K'aOf. (K')Call) == O. Then g == fa E R-P is the required element.

,-I

i

Corollary: Let N be a finitely presented R-module such that N» is a free Rp-module of rank n. Then there exists some IE R - P with N, free of rank n over R,. Proof: Choose {x" ••• '"x.} in N such that {(X;! I)} is a basis of N p over R p • Consider the R-homomorphism u:R' -+ N, where R' is free with basis {e1, ••• , t.}. and u(e,) == Xi' 1 ~ i ~ n. Since by assumption, Up is bijective, there exists some lER-P with Uf bijective. Hence N, is free of rank n over R,. Proposition 2:

Let (Ji)'C/C. be elements of R such that ~ RJi==R.

.

'~l

. Then the ring S ... ; Rf, is a faithfully fiat R-module.

'-I

Proof: B! Corollary 2 to Proposition S, 2.3, each Rf, is R-ilat and hence S IS fiat over R. Let P be a prime ideal of R. Since

i Rf, = R, there exists some Ji ¢ P, say h. ¢ P. '-1

Then PJi = PRf is

,

,

a prime ideal of R~ and PS C PRf 1 x ~2 R~ =F S t since P Rfi1 oF Rf i' Jl JI Hence S is faithfully fiat over R by Corollary 2 to Theorem 2, 1.4. CoroUary: Let (f,h<;,S.' be elements of R such that IRJi = R. An R-module M is finitely generated (finitely presented) if and only if each Mit is finitely generated (finitely presented) over R jj' Proof:

Clearly M finitely generated (finitely presented) implies the

..

-

.

LOCALISATION

SS

same for eac? M", Conversely assume that each MIl is finitely generated (fiDitely presented) over R I ~ i < n. Then there exists an exact sequence

,I,

Rj. -+

RJ, -+ Mfi -+ 0,

where m and p can be chosen to M'==

be independent of i, If

~ M"thenM'=MlZlS whereS==nRjj • Since S'Is faith.R

1"1

I

'

fully fiat over R, the result follows from Proposition 4, 1.4. Definition: For any commutative ring R, X= Spec R denotes the set of prime ideals of R. For an ideal I of R, V(l) denotes the set of prime ideals of R containing I and D(I) = X - V(RI), IE R. (.1

Proposition 3: X== Spec R is a topological space in which the closed sets are precisely {V(I) I I an ideal of R}'and the collection {D(/)},cR is a base for open sets. Moreover, X is quasi-compact. Proof: The axioms for closed sets are satisfied in view of the followingproperties of V(l) whichfollow readily from the definition. (i)

VCR) = ' V(O)

(ii) V(IJU V(I,) (iii)

n V(Ia) = '"

=

=

X.

V(I,! J .

V(Il",) .

'"

Let U be an open set in X so that U = X - V(l), for some ideal I. By (iii). n V(RI) == V(I) so that U == X- V(I) = U D(!), lel lel i;e. {D(/)} is a base for open sets. To prove that X is quasi-compact, consider an open covering of X by basic open sets, viz. X = U D(f,,). If I is the ideal gene-

'" n V(RI",). Hence '" X-Vel) = UD(fJ = X. '"

rated by the {I,,}, then Vel) =

-

Hence V(l) == , i.e, 1= R. This implies = V(I)

=

V(

In particular I == 3: g""/"", ga, E R.

i RI..,) = nV(f",,),

'-1

showing that X is quasi-compact.

1

,-,

i.e, X = (j D(f",,). 1

56

COMMUTATIVE ALGEBRA

CorollllrY: . X Proof:

=U D(f"t) <:> R = 1

£Rj"j' 1

Follows from the last part of the above proof.

Theorem 1: Let M be an R-module. The following conditions are equivalent. (i) M is finitely generated projective over R. (ii) M is finitely presented and M", is a free R", module for every maximal ideal m. (iii) M is finitely generated and M p is free R p module for every prime ideal P and the map rauk: Spec R -+ Z given by P -+rankR,Mp is continuous. (iv) There exist elements (fillet",n in

R such that

R

=i

Rfi

1

and each M r, is a free R Ji module of finite rank. (v) For every maximal ideal m of R, there exists some fE R-m with M I free of finite rank over R I · Proof: (i) => (ii), Since M is finitely generated and projective, Mis finitely presented. Moreover, M is finitely generated and projective implies M m = M ® R", is also finitely generated and R",-projective R

and since R m is a local ring, M", is R",-free. (ii) => (v). Let m be a maximal ideal of R. Since M", = R",®M. R

.

is Rm-free, choose a basis {I ® x,} of M m over R", (I ~ i ~ n). Consider the map u: Rn -+ M given by u(e,) = x" I ~ i ~ n, where {e,} is the standard base for R», By assumption, u'" is an isomorphism. Hence by Proposition 2, ul is an isomorphism for some fE R-m. This implies (v). (v) => (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 whole ring R. In particular 1 =

i o,fi for some fi E '-1

Y, a, E R.

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

LOCALISATION

57

only if P E U D(f/), where the union is taken over all j, for which J

MIl is of rank n. Hence rank: P -+ rankR,Mp is continuous. (iii) => (v). Let m be a maximal ideal and choose {x ..... , xn} in M such that {(x,/I)}(I" i ~ n) is a basis of M", over R",. Define

an R-linear map u: Rn -+ M, by ute,) = xi, I ~ i ~ n, where (e,) is the standard basis of R". The mapping u",: R::' -+ M", is surjective and since M is finitely generated, by Proposition I, there exists some g E R - m such that ull : R~ -+ Mil is surjective. Since rank M", =n and the rank function is continuous there exists some se: R-m such that for all P E D(h}, rank M p = n. Let f = gh. ThenfE R-m and the map u/:Rj-+ MJ is surjective as UII is so and rankMp =nforaiIPED(f}. Thus up:R~.-+Mp is surjective for all P E D(f). Since R'j. and M p are bdth free modules of the same rank, Up is an isomorphism for all P E D(f). Let m' be a maximal ideal of RI and m" its contraction to R so that m" E D(f). Consider the commutative diagram (Rj}m' 1

!

R:'~,

-

(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

,

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

'

I

I

HenceM' is finitely generated projective over S. Since M' c:>!. M I8l s 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, Proposition 4: X = Spec R is disconnected .. 1 = el + e. where e1, e. are mutually orthogonal, non-trivial idempotents in R. 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.

'*

»

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. 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,.

T

'1

i T

LOCAUSATION

Proof:

59

Follows from Theorem 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).

=

I

1'\

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,.

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

M

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

COMMUTATIVE ALGEBRA

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. •

R

Moreover if (li) holds, N is isomorphic to the dual module M* = 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

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 R

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

,

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

...... 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

® N "'" (M ® M*) ® N R

R

R

"",'(M® N)® M* "",R ®M* cy-M*. R

R

R

Corollary: If M and N are projective R-modules of rank one, M® Nand M* are projective R-modules of rank one. R

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

tive, The relations

R

+

LOCALISATION

61

(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 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. 2.4. EXERCISES 1.

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

= ; 1-1

2.

R I and P

=;

PI> where

1-1

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. .

R

3.

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.

4.

Let (Ji)l
5.

R-module. Suppose XI E 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

i Rfj = Rand M an '-1 Mft are such that XI and XJ have the

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 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 deR

•.' 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

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.

1

1

CHAPTER In

NOETHERIAN

RINGS

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. 3.1.

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

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 of M. Choose any M 1 E li. If M 1 is not maximal. choose M E I.

M 1 C M..

IrM. is not maximal continue this process. After a

=1= finite number of' steps, there exists someM", E I which is maximal for otherwise condition (ii) will be violated. • DefiDition: An R-module M is called Noetherian if it satisfies any one of the above equivalent conditions. DefiDition: A ring R is called a Noetherian ring if the R.module R 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. 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. Proof: Assume M is Noetherian.

Since every submodule of M'

1

NOETHERIAN RINGS

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. Corol1ary 1: Any homomorphic image of a Noetherian ring is Noetherian. 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. Corollary 3: Let R be a Noetherian ring and M a finitely generated R-module. Then M 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 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: 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.

I

I

r

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/

Proof:

t

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 .,~

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[ . .

InK.

:.:'ir/,-.k.

.

Corollary 1: If R is Noetherian.

Noetherian, the rmg R[

• 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

~

1

NOIlTFlllRIAN RINGS

2.

67

(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 _

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

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

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

R but p2 is not P:primary. However any power of a maximal ideal is primary. More generally we have the following Proposition.

Proposition 1: Let I be an ideal of R with ideal. Then I is m-primary.

VI = m a

maximal

Proof: Let ab E I, and b¢ m, Then m + Rb == R so that c + Ab = 1, for some c E m, AE R. Since = 'm, ck E I for some k;> 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

'./

r

,

lI

I

NOETHBRIAN RINGS 69

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 <; i ~ r). From any primary decomposition we can always obtain a reduced primary decomposition in view of the following Proposition. Proposition 2: Let HI' N., .... Nk be P-primary submodules of M. Then N1nN.n ... nNk is P-primary. Proof: It is sufficient to prove the result for k = 2. Let N=N, n N., and suppose aXEN,aER,xEM. Ifa¢P.thenaxENlnN. and N N 2 are P-primary implies x E NI n N 2 = N. If a E P, then " a"'M eN, and a"'l'>1 C N 2 for some n, andn•. Hence a"'+"M eN, t.e. a E rM(N). Hence N is P-primary. We now prove the existence of primary decomposition for finitely generated modules over a Noetherian ring Definition: A submodule N of M is called irreducible if N *- M. and N cannot be expressed as an intersection of two submodules properly containing N. Proposition 3: Let M be a Noetherian module. Then any irreducible submodule N of M is primary. Proof: Let N be an irreducible submodule and suppose N is not primary. Then there exists some a E R, such that the homothety Ao : MIN - MIN is neither injective nor nilpotent. Consider the increasing chain of submodules of MIN Ker~ocKer~c..... cKerA:C ......

70 COMMUTA'IlVB ALGBBRA

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 (2 = Ker rP and x = rP(y) = O. Since Aa is neither injective. nor nilpotent. Ker =I- 0 and Im.p =F O. If p:M -+ MIN is the natural projection and N, =p-' (Ker rP). N. = p-I (1m rP). then N = N, nN2 is a decomposition of N as an intersection of two submodules properly containing N. 'This contradicts the irreducibility of N. Hence N is primary. Theorem 1: Let M be a Noetherian R-module. Then any proper submodule N of M admits a reduced primary decomposition N = nNh N I being PI-primary.

~I

;' '.

~

,

Proof: We first show thatN can be expressed as a finite intersection of irreducible submodules, Let 1: be the- collection of alI submodules of M which cannot be expressed as a finite intersection of irreducible submodule s, We claim that I =-rP. for if 2: =I- rP. it has a maximal element N. Clearly N is not irreducible so that N=N1nN•• NeN, and NeN•. The maximality of N implies

=F

Ii

=F

that N, and Nt can be expressed as a finite intersection of irreducible submodules and N also has such' a decomposition. This is a contradiction as N E~. Hence ~ = . i.e. every submodule of M can be expressed as a finite intersection of irreducible submodules. Since an irreducible submodule is primary. we have N = nN I • I

N I being primary. By grouping together all P,-primary submodules, we can assume the PI are distinct. By deleting the superfluous N" i.e. those N, which satisfy NI"::) n N, we get a reduced primary

decomposition for N.

i

j"l

Remark: The primary decomposition is not unique. Let R=k[X. YJ. kfield and I=(XI.XY). If II = (X). I.=(Y-Ax,X2), AEk, then I = II n I. is a reduced decomposition of I as an intersection of primary ideals. 12 is not unique as AE k is arbitrary. We now investigate the uniqueness of the primary decomposition. Definition: Let M be an R-module. A prime ideal P of R is said

~

+

I

I

-I

I

+

I

NOETHERIAN RINGS

71

to be associated to M if P is the annihilator of some non-zero xEM. . Clearly P is associated to M if and only if there exists an . injective R-homomorphism RIP -+ M. The set Ass(M) denotes the set of prime ideals associated toM.

Theorem 2: Let R be a Noetherian ring and M a finitely generated R-module. Let (0) = N, n N 2 n ... n N, be a reduced primary decomposition of 0 in M. N, being Prprimary (1 ..;; i ~ r). Then Ass (M)= {Pi' p., ..... , PrJ. Proof: Let P E Ass(M) so that P = Ann(x) for some non-zero x EM. Since x#O, by rearranging the N" we can find aj such that xr/iN,UN.U .... UNj and xENJ+ln ... nN,,(I~j the homothety Aa : MINI ..... MIN, is nilpotent so that P';'M eNI• for some n,;> 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). Corollary 1: (First Uniqueness). Let M bea finitelygenerated module over a Noetherian ring R. If'N = ;; N I is a reduced primary de1

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

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 peAss(M)

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

N , be a reduced

1-1

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=

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< r). A component which is not minimal is called an embedded component.

Example: Let R=k[X, Yj, k field, I =(X·, XY). Let 11 = (X) and I. = (Y-AX, XI). A E k. Then 1= 11 n I. is a reduced primary decomposition of 1 and I. is P.-primary where p. = (X. Y). Since p."::) PI = h.l. is an embedded component and it is not unique. We now show that the minimal primary components of N are uniquely determined. Proposition 5: Let M be an R-module and S a multiplicatively closed set in R. Let C be the set of prime ideals of R which do not intersect S. Then the mapping P -+ Psis an injection of AssR(M) nC into AssR.(Ms) and the.map is a bijection if R is Noetherian.

.

,

NOBTHIlRIAN RINGS

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).

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

= M' if P ¢ P' N = f-I(N') if Per.

(i) N' (ii)

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.

.. Corollary: (Second Uniqueness Theorem).

Let N

=

nN, be a

'-1 reduced primary decomposition of N. N, being P,-primary.

If P, is

minimal. then N, is uniquely determined by N. Proof: Suppose PI is minimal. Then P, ¢ PI for i > 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

CQMMUTATlVB ALGEBRA

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

-. .,

• PI .... P,}. I 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.

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

n

P,. Conversely if aE

nPh

<

)..:M/N,-+MIN,

1_1

' ..I

is nilpotent for each i, so that a"IM eN" n, ~ l. then a·M = o and a E y'A"nR(M).

If n

= Max n" '

Corollary: If R is a Noetherian ring then the . nil radical N(R)= n P. PeAss(R)

Proof: Take M =R. Definition: For an R-module M. the set {P I P prime ideal of R. M» O} is called the support of M and is denoted by SUPP(M). Clearly M = 0 if and only if Supp(M) = .,..

*'

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).

~

I

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

-.n 1

P, =

V AnnR(M).

(i) =>- (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

(ii) =>- (iii).

. no'

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

i

t

NOETHERIAN RINGS

75

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). Corollary: Ass(M)eSupp(M) and the minimal elements of Ass(M) and Supp(M) are the same. Proposition 8: Let 04 M' -.. M -.. M" -.. 0 be an exact sequence of R-modules. Then (i) Supp(M) = Supp(M')U Supp(M"). (ii) For finitely generated R-modules L and K, Supp (L@K)= Supp(L) n Supp(K). R Proof: (i) For each prime ideal- P ofR the sequence 0 -.. M~-.. M~ -+ M;. -+ 0 is exact. Now P E Supp(M) if and only if M 1'-=1:0. <> Either M~ -=I: 0 or M'; 0 <:> P E Supp(M') or P E Supp(M"). (ii) (L ® K)p 0< L p @ K p. Since R p is a local ring

*

*' 0 <> R

L p ® Kp Rr

Rr

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

3.2.

I.

2. 3.

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

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

I

I

-+

\

I

T

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. 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 M'+l eM" Ml+lnM'

= M,nM' and

M i +1

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

+ M' = M,+M' Ml+l'

Hence M is

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

78

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. 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. 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

NOETHERIAN RINGS

79

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 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

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<,. Since R

= ~R 1

"

by the Chinese Remai~der Theorem, the / ,(1 .e;;; t «; r) are

comaximal and

n I,

'=1

= (0).

Let P, be the unique prime ideal of

"'t {P,) (I .e;;; i.e;;; r). Then m, is a prime ideal orR and hence maximal. Since P, is nilpotent, it follows that 1, is

R, and m, =

1

80

COMMUTATIVB ALGBBRA

m,-primary.

Hence (0) =

n is a reduced primary decomposition I,

1

of (0) in R. Since 1,(1 ~ i ~ r) are pairwise comaximal, so are m,(1 ~ i ~ r), Hence I,(l ~ i ~ r) are isolated components of the primary decomposition of O. Hence 1,(1 ~ i ~ r) are uniquely determined by R. i.e. R, 0< Ril, are unique (I ~ i ~ r). 3.3. I.

2. 3.

4. 5. 6.

7.

EXERCISES

s,

Let N I and be submodules of Msuch that both MIN~ and MINa are Artinian modules. Show that MIN1nN. IS an Artinian module. Let M be an Artinian R-module and I: M ~ M an injective homomorphism. Show that/is an isomorphism. Let M be an R-module which is both Artinian and Noetherian and I: M ~ M. R-homomorphism. Show that. for some n ~ I. M = In(M) EEl Ker (P). Deduce that if I: V ~ V is a linear transformation of a finite dimensional vector space V. then V = V1 EB V•• where I I VI is non-singular and II Va is nilpotent. Let M be an Artinian R-module. Is the ring RIAnn(M) Artinian? Why? Show that a finitely generated algebra A over a field k is Artinian if and only if A is finite dimensional over k. A ring R is called a primary ring if it has just one prime ideal. Show that (i) For any m-primary ideal I, with m maximal. RII is a primary ring. (ii) A primary ring has no idempotents other than 0 or 1. Give an example ofa primary ring which is not Noetherian. Let R be an Artinian local ring with maximal ideal m, Show that the following conditions are equivalent. (i) Every ideal of R is principal. (ii) m is principal. (iii) dimk (m/m") ~ I. where k = Rim. ,

8.

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

9.

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

'-1

Show that R has

\

NOErHaRJAN RINGS

81

3.4. Length of a modUle DefinItion: An R.module M;j= 0 is called a simple module if it has 110 submodules except 0 and M. Examples: (i) Anyone dimensional vector space over a field k is a simple k-module. (ii) Any group of prime order is a simple Z·module. DeflDition: A composition series of an R-module M is a finite decreasing sequence of submodules M =

u, ::::> M I ::::> u, ::::> ... ::J M n = {O}

such that each Mt/M'+1 (0 ~ i ~ n - I) is simple. Examples: (i) Let V be a vector space of dimension it with basis {el' e...... en} and Vn~ the subspace generated by {ell ea•...• e,}. Then J' = Vo ::::> 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. 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. 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

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=

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

=1=

\

~

1('1 "", •

If m

+ I <: 1(M), we

=1=

=1=

=1= =1='

=l=

=l=

repeat the process. After a finite number of steps, we get a composition series of M,

a

composition, the length of Definition: If an R-module M has any composition series of M is called the length of M and is denoted by IR(M). If M has no composition 'series, IR(M) is defined to be co, Proposition 1: An R-module M has a composition series if and only if it is both a Noetherian and an Artinian module.

:~"-

, NOETHERIAN RINGS, 83 Proof: Assume that M has a co .'. any strictly increasing or decrea;POSltlOn 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;)

=/(M'}::::>f(M~)::::> ... f(M;)={O} By CorolIary to Proposition 1, 1. I g-:I(M;') Mt,. , --I M" ~ -,-;- 18 sim, pie for each i. Hence the above s . . g ('+1) M/+I M of length k + I, i.e. I(M) = I(M:;I~ ;;.:;,,}~omposition series of PrOPosition 3: Let R be a N - heri . \generated R-moduJe Then t~et erl~n ring and M ~ 0 a finitely , . ere eXIsts a decreasing chll.in M M.::::>Mil-I::::> '" ::::> M o = 0 such that MtIM/-1 """ RIP P prune. . '

=

to,

Proof: Let I be the COllection of submod I . the required properly G.. ", ~. ~,"'~ u es of M whleh have . C.~ "',lI.S 0 E:E. ~t N be a maximal

'1

84

COMMlITATIVB ALGIlBRA

element of};. If N oF M, M/N -1= O. Hence Ass(M/N) oF.p. and choose P E Ass(M/N), so that R/Pl~NJN. Then NI also has the required property and N I ~ N, contradicting the maximality of N. Hence N

= M.

=1=

Corollary: With the same notation as above, we have Ass(M) C {PI. P I •• _, P n} C Supp(M) and the minimal elements of all the three sets are the same. Proof: Consider the chain M = M n ~ M'-l ~ ... ~Mo = {O} with M,/M'_I ~R/P" P, prime (I EO; i <; n). Then, by Exercise 6, 3.2 Ass(l\f)

c

Ass(M/M._1)UAss(M.-I/Mlt'-s)U", UAss(MI) C

{PI' Pl... ·' P n} as M,/M H ~ RIP,·

Since P, E Supp(R/P,) = Supp(M,/M,-I) and Supp(M,/M,-J SupP(M,) c Supp(M), we have {PI,... P.} c Supp(M).

c

The minimal elements of all the three sets are the same by , Proposition 7, 3.2. Theorem 2: Let M be a/g. module over a Noetherian ring R. Then the folloWing are equivalent (i) IR(M).is finite. (ii) Every P E Ass (M) is a maximal ideal. (iii) Every P E Supp (M) is a maximal ideal. Proof: (i) =:. (ii). Choose a decreasing chain as in Propositi~n: 3, M = M n ~ M_I~"" ~Mo = {O} with M,/M,-I ~ R/P,. Prpnme, (I EO; t e; n). Since IR(M) < 00 and RIP, is isomor~hi~ to. a subquotient of M, we have IR(R/P,) < 00 for each i. ThIS implies that R/P. is an Artinian R-module and hence Artinia? R/p.~module,.i.e. an Artinian ring. Hence, it is a field, i.e. P, IS maximal. SIDce Ass M c {PI' ...• P n},every P E Ass (M) is maximal. (ii) => (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

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)

<

00

*P E Ass(M)andP is minimal.

Proposition 4: Let R be a Noetherian ring. The following conditions are equivalent. 0) R is Artinian. (ii) Every prime ideal of R is maximal. (iii) All elements of Ass (R) are.maximal. Proof: (i) =>(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

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) <: co.

aa

Corollary 1: R is an Artinian ring if and only if R is a Noetherian ring and every prime ideal of R is maximal. Proof: If R is Noetherian and every prime ideal of R is maximal, R is Artinian by Proposition 4. Conversely assume R is Artinian. Then lR(R) < co and hence R is Noetherian. From Proposition 3, 3.3 every prime ideal of R is maximal. I ~.

I

Corollary 2: Let R be an Artinian ring and M a finitely generated R-module. Then lR(M) <: ee. Proof: Since R is also a Noetherian ring and M is finitely generated, M is both an Artinian and a Noethcrilin module. Hence . lR(M) < co.

3.4. EXERCISES 1.

Let 0 -+ M 1 -+ M.

-+ ..•• -+ M. -+

0 be an exact sequence of

R-modules of finite length. Show that 2.

=

E

IRp(Mp).

PeAss(M)

-

Let P 1> P 2, • _ ., p. be the pri..me ideals associated to the primary decomposition-of (0) in a Noetherian ring R. Show that if all

UP"

4.

(-lY IR(M,) = O.

Show that if M is an R-module _of finite length over a Noetherian ring R, then . lR(M)

3.

:Z

'=1

then R s is an Artinian the P, are minimal, and S = R 1 ring. . Let R be a Noetherian ring and M a .finitely generated R-module with decreasing chain as in Proposition 3. Let P E Ass (M) be a minimal element. Show that IRp(Mp) is equal to the number ofi for WhichP = P,.

>0_

.

NOI!THI!RIAN lUNas

5.

6.

87

Let M be a finitely generated module over a Noetherian R. Consider a decreasing chain as in Proposition 3. Illustrate by an example that {PH Pa' ... , p.} may not be unique. Let I be an ideal in a Noetherian ring R having primary decomposition without embedded components and S the complement of the union of prime ideals associated to I. Show that A(I) = lRs(RsIIRs) is finite. If R is Artinian show that A(I) = lR(R)-IR(I).

CHAPTER IV

INTEGRAL EXTENSIONS

The study of solutions of diophantine equations in number theory led to the notion of algebraic integers. A parallel development in algebraic geometry arising out of a study algebraic curves led to the notion or algebraic functions of one- variable and integral functions. Both these developments are unified under the study of integral extensions. Integral extensions are well behaved with . respect to extension and contraction of prime and maximal ideals. A basic Theorem due to Noether on normalisation shows that any finitely generated k-algebra is integral over a suitablY chosen polynomial SUb-algebra. From this Theorem, we obtain Hilbert's NUllstellensatz, which says that if I is a proper ideal in the polynomial ring k[X" ... , X.], k algebraically closed field, there is at least one commonsolutlon for 'the polynomials in I. Varieties Whose local rings are integrally closed are called normal varieties. The normalisation operation, i.e. taking the integral closure is a standard tool in geometry. 4.1. Integral elements We assume throughout this chapter that R is a commutative ring with unit element and S is an e»tension ring of R with the same unit clement. DetlDition: An element «E S is said to be integral over R if it satisfies a monic polynomial relation of the type a.0

+a

Examples: (i)

1 l Gl. ...

+....+ ao = 0, af E

R (I E;; i E;; n).

Any II E R is integral over R.

INTBGRAL EXTENSIONS

89

(ii) Let K be a field extension of k and « E K algebraic over k, Then « is integral over k, (iii) Let R = Z and S = Z + iZ, the ring of Gaussian integers. Then « = I + i is integral over Z. _ (iv) Let R = Z, and S = Q. Then at E Q-Z cannot be -integral over Z, for if at = p/q, (p, q) = I, satisfies a relation of the type GIn + alGI· -1 + _..+ a. =: 0, a, E Z, then pn + a1p.- lq +

'" + a.q. = O.

This implies that q divides]J" and hence a prime factor of q divides This contradicts the assumption that (p, q) == I. We have the following equivalent characterisations for integral elements.

p..

Theorem 1: The following conditions are equivalent. (i) The element at E S is integral over R. (ii) R[at] is a finitely generated R-module. (iii) R[«] c R' c S, where R' is a subring of S which is finitely generated as an R-module. (iv) There exists a faithful R[at]-module M which is finitely generated as an R-module. Proof:

(i) => ~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

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

on E I. Consider the

)

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

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+ Corollary 1: Let

Clt10

.•••

+

0,

= 0,

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

«s, ... , «, E S be integral over R.

Then

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

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. Corollary 2: The set of elements of S integral over R is a subring of S containing R. 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. Definition: The subring of all elements of S integral over R is called the integral closure of R in S. Examples: (i) The integral closure of Z in Q is Z. (ii) The integral closure of Z in Q(i) is Z + tz. (iii) The integral closure of Z in C the· field of complex numbers is the ring of algebraic integers. •

Dt:fini~on: If the integral closure of R in S coincides with R. then R IS said to be integrally closed in S.

.

INTEGRAL BXTENSIONS

o

91

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. Corollary: If closed in S.

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. 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•

o

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 (<
4. 1. EXERCISES

1

1. 2.

Show that a unique faetorisation domain is integrally closed in its quotient field. Show tl\at the integral closure of Z in Q(v'2) is the ring of elements of the type a + bv''i., a. bE Z.

92

COMMUTATIVB ALOBBRA

Let ReS be an integral extension and n a maximal ideal of Sand m = R n n, Illustrate by an example that Sn need not be integral over R m • 4. Let R be a ring and II an invertible element of S ~ R. Show that R[oe] n R[rl ] is an integral extension of R. S. Let K b&a field and S = K[X, Y]. Let R be the subalgebra of S generated by the monomials of the type XIH Y' (i ~ 0). Show that oe "" XY is such that R[oe] is contained in a finitely _generated R-module, but a is not integral over R. (Theorem I is not true, if R' is not a subring). 6. An R-algebra A is said to be integral over R if it is integral over R' the image of R under the natural homomorphism given by a ~ a.IA. a E R. Show that if A,(i ~ i ~ n) are

3.

integral over R, then ~ A f is integral over R. f~1

1.

Let A be an R-algebra which is integral over Rand j: R -+ S a ring homomorphism. Show that the scalar extension S ® A R

8.

is integral over S Show that if A and 'B areR-algebras both integral over R, then A ® B is integral over R. R

9.

4.2.

Let ReS be such that S-R is closed under multiplication. _ Show that R is integrally closed in S. Integral extensions

Proposition 1: Let ReS be domains and S integral over R. Then R is a field if and only if S is a field.

*"

Proof: Assume' that R is a field and let II E S, oe O. Then oe satisfies a relation of the type I1.n + a lll....1 +an = 0, af E R and n can be chosen to be minimum. This implies that an 0, as S is a domain. Hence (-lla.)(oen+a,Il·-I+ +a 1ll)='1, so that II has an inverse ~=(-I{an)(oen-l+(hll ·+ +an_J). Hence S is a field. Conversely assume that S is a field and let a E R with O. Then II = a-I E S is integral over R. It satisfies a relation of the type I1.n + a1..· -1 + an = 0, a, E R. Multiplying by an on both sides we have I + ala + .. : + anO" = 0, so that a E R has an inverse b E R given by b =-(a1 a.o+ ...+ a.a-1 ) .

+...

a*"

+ ...

+

"

-

*"

INTEGRAL EXTENSIONS

93

Coronary 1: Let ReS with S is integral over R, P' a prime ideal of S and P = P' n R. Then P is a maximal ideal of R if and only if P' is a maximal ideal of S. Proof:. R{P is a subring of S{P' and SIP' is integral over RIP. HenceR{P is a field if and only if S{P' is a field. • Corollary 2: Let R C S with S is integral over R, PI' r, prime ideals of S with PI C p. and p,nR = p,nR = P. Then

p 1=p.· Proof: Sp is integral over R p by Proposition 2,4.1 and P l5pcP.5p are both prime ideals of 51'.' Since P1Sp Rp = p.SpnRp = PRp a maximal ideal of Rp , both P15p and PaSP arc maximal ideals of .51'.' Hence P1S p "" p.Sp ; This implies that PI = P••

n

Proposition 2: Let R c 5 with S is integral over R. For any prime ideal P of R, there exists a prime ideal P' of 5, such that p'nR=p. Proof: Since S is integral over R, 51'. is integral over Rp • Consider the commutative diagram. i

R _5

p

1 iii I R p _ Sp

Where the horizontal map below is induced by the inclusion map and the v.ertical maps are-natural homomorphisms. Let m be any maximal ideal of Sp. Then mnRp' is a maximal ideal of R p • Hence it must be equal to the unique maximal ideal of R p , i,e. mnRp = PR p • Let P' be the contraction of m under the map Then P' is a prime idoal of Sand P' nR = P.

i.

Definition: Let R c 5 be an extension of'rings. A prime ideal P' of S is said to lie above the prime ideal P of R if P = P' nR. Let ReS be an integral extension. Proposition 2, shows that for any prime ideal P of R there exists a prime ideal P' of S lying

94

COMMUTATIVB ALGEBRA

above P.

Moreover, two such prime ideals P t and PI satisfying

P t C PI must coincide by Corollary 2,_Proposition l.

Example: It is possible for two distinct prime ideals P~ and P; of S to lie over the same prime ideal P of R. Consider R = Z, S = Z[i] the ring of Gaussian integers. Then P~ = (2 + i) and P~ = (2-i) are prime ideals of S lying over the same prime ideal P = (5) of R. Theorem 1: (Going Up Theorem). Let S be an integral extension of R. Let P 1 C PIC: ... C p. be a chain of prime ideals of R and p~ c P; c ... c P;"(m < n) a chain of prime ideals of S with P; n R = P, (I .;;,; i .;;,; m) Then there exist prime ideals P;"+h ... , P: of S such that P;" c P;"+l c. ..c P: and P; n R = P, (1 .;;,; i .;;,; n). Proof: It is sufficient to prove the result for n = 2. Let P t C p. be prime ideals of R and P~ a prime ideal of S with P~ nR = Pt. Then SIP; is an integral extension of RIPt and PI/Pt is a prime ideal of R/Pt. Hence there exists a prime ideal of S/P~. say p;IP;, P; prime in S, lying above PI/Pt. 'Clearly P; c P; and P; n.R = PI'

r:

Proposition 3: Let S be an integral extension of Rand R -+ n a ring homomorphism of R into an algebraically dosed field n. Then / can be extended to a ring homomorphism g: S -+ n. Proof: Let P = Ker (f) so that P is a prime ideal of R. By Proposition 2, there exists a prime ideal P' of S lying above P. Now /induces an injective ring homomorphism R/P -+ n which can be extended to an injection 1: K -+ n where K is the quotient field of RIP. Let L be the quotient field of SIP'. Then L is an algebraic extension of K as SIP' is an integra! extension of RIP. Since n is algebraically closed1 can be extended to an isomorphism ,g of L into n. Let g = gp where p: S -+ SIP' is the natural projection. Then g is the required ring homomorphism. Proposition 4: Let S be a ring extension of R which is finitely generated as an R-module. Then for any prime ideal P of R, the . set of prime ideals of S lying above P is a finite set. Proof: By considering the extension Sp over Rp we assume without

INTEGRAL EXTBNSION

95

loss of generality that R is a local ring with unique maximal ideal P. By considering extension SIPS over R/P, we may also assume that P = (0), i.e. R is a field. Since the extension Sis finitely generated R-module, where R is afield, Sis Artinian by exercise 5, 3.3. Hence S has only finitely many prime ideals.

4.2. 1.

2. 3. 4.

5.

6.

7.

EXERCISES

Let ReS, where S is integral over R. Show that a E R is a unit in S if and only if it is a unit in R. Show further that the Jacobson radical J(R) = RnJ(S). Let ReS be domains, where S is integral over R. Show that if / is a ring homomorphism of S into a domain" such that the restriction of'fto R is i~ective, then/is injective. Let G be a finite group of automorphisms of a ring S and R = sa = {a E S I"(0) = a, for all "E G}. Show that S is an integral extension of R. Let G be a finite group of automorphism of Sand SG= R the subring of invariants as in exercise 3. Let T be a multiplicatively closed set in S with ,,(T) c T, for all "E G. Show that G acts on ST and (ST)a is isomorphic to RTo where To= TnR. Let G be a finite group of automorphisms of S and Sa = R the subring of invariants. Show that G acts transitively on the set of prime ideals of S lying above a given prime ideal P of R and deduce that this set is finite. Let ReS be an ring extension for which the Going up Theorem holds. Show that the contraction of a maximal ideal of S is a maximal ideal in R. Let ReS be a ring extension. Show that the following conditions are equivalent. ' (i) Going up Theorem holds for the extension ReS. (ii) For every prime ideal P of R, and T = R-P, the ideal P' of S maximal with respect to the property P' n T = '" lies aboveP.

4.3. Integrally closed domains Definition: A domain R is said to be integrally closed if it is

96

1

COMMUTATlVs ALGEBRA

integrally closed in its quotient field. Examples: (I) The argument in Example (iv), 4.1, shows that any unique factorisation domain is integrally closed. In particular k[X1 , XI' ... , X,J. k field, is integrally closed. (li) The domain R = k [X, YJ/(X"- Y8) is not integrally closed as ", tV is integral over R but does not belong to R.

=:x

Proposition 1: Let ReS be an ~xtension of rings and T c R multiplicatively closed set in R. If R is the integral closure of R in S, then RT is the integral closure of R .. in ST' Proof: Clearly Rr is integral over RT by Proposition 2, 4.1. Let (",/t) E ST be integral over RT. Then it satisfies a relation of the type (",/t)"+ (a1/ll)(",/ t )"- 1 + ... +(o./t.) =0, a,ER, I" tE T. Ift' =nt" multiplying both sides by (It')", we have an integral

,

.

dependence of I'", over R. Hence r'", E

R, i.e. (",/I)"";(/'«/t'/) E RT.

Corollary: If R is an integrally closed domain, RT is integrally closed for any multiplicatively closed set ,T cR. In particular Rp is integrally closed for every prime ideal P of R.

Proof: Since

R=

R,

RT =

RT •

Example: If R is an integrally closed domain and P a prime ideal of R, then R/P need not be integrally closed. Let R = k[X, YJ and P = (XI- f1'). Then R/P is not integrally closed. . Proposition 2: Let R be a domain. The following conditions are equivalent. (i) R is integrally closed. (ii) R» is integrally closed for every prime ideal Pof R. (iii) R m is integrally closed for every maximal ideal m of R. Proof: (i) => (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

1

INTEGRAL BXTENSIONS

i m: Rm-+R.. is surjective for every maximal ideal (iii). Hence i is surjective. This implies (i),

97

m, by assumption

Proposition 3: Let R be a ring andftX) and g(X) be monic polynomial over R] ::> R such that the product heX) = f(X) g(X) has coefficients integral over R. Then f(X) and g(X) have coefficients integral over R. Proof: We observe that (as in the case of a field) for any monic polynomialf(X) E R1[Xl, there exists an extension ring R'::> R1 in which/(X) is a product of linear factors. Choose a ring R' ::>R1 such that bothf(x) and g(X) split into linear factors over R'. -t.e. /(X) = 7t(X-",,) and g(X) 0= 7t(X-M, "'" ~J E R'. Let S be the

,

/

integral closure of R in R'. Since heX) = f(X)g(X) has coefficle(lts integral over R, heX) E S[X]. Now h(",,) = 0 and h(M = 0 are integral relations for the "" and the ~/ over S. Since S is the integral closure of R in R', all the e and the ~J belong to S. Since the coefficients off(X) and g(X) are sums and products of the "" and the ('>J respectively the coefficients of f(X) and g(X) belong to S, i.e. they are integral over R. ' Corollary 1: Let R be a domain with quotient field Kand K' an extension field of K. If", E K' is integral. over R, the minimum polynomial of", over K has all its coefficients integral over R. Proof: Let h(rr.) = 0, heX)E R[X] be an integral dependence of ", over Rand let/(X) E K[X] be the minimum polynomial of", over K. Then /(X) divides heX) in K[X] and heX) = f(X) K(X), g(X) E K[X]. By Proposition 3, f(X) has coefficients integral over R. Corollary 2: Let R be an integrally closed domain with quotient field K and K' an extension field of K. If", E K' is integral over R, the minimum polynomial of ", over K has all its coefficient in R. Proof: Follows from Corollary 1. 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

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 n i.e. h + 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):

INTEGRAL EXTENSIONS

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..) <:(L~, a~) if and only if L" c L~ and a~ = a~ I L,.. By Zorn's Lemma, :E has a maximal element (L*, a*). It is-sufficient to show that L* = L. Then a*(P') = Q', a* E G(L/K}. If L* -=F L, choose a E L* with a ¢.L. Since the number of conjugates of a is finite,

100

COMMUTATlVB ALGBBRA

there exists a smallest normal extension L I containing L*(a). and finite over L *. Then <1* can be extended to an automorphism of the algebraic closure of L and since L is normal. a* can be considered as an element of G = G(L/K). Then a*(P' n~) and Q' nL1 ~ lying over Q' L*. Since G(LI/L*) are prime ideals of S is finite by the first part of the proof there exists T E G(L1/L*) with Ta~(P' n L I ) = Q' n L I • Hence (Lt. Ta*) E I. contradicting the maximality of (L*. <1*).

n

J

n

Theorem 1: (Going Down Theorem). Let R be an integrally closed domain and S a domain which is an integral extension of R. Let PI:::> p. :::> ••• :::> P n be a descending chain of prime ideals of Rand a prime ideal of S lying above PI' Then there exists a chain P~ :::> P~ :::> .••:::> P~ of prime ideals of S with Pi = P; n R (I ~ i ~ n).

P;

Proof: Assume first that S is a normal extension of R. By going up theorem,there exists is a chainP~ C P::- I C ... C P;' of prime ideals of S. with P;' n R = P" (1 ~ i Ill;; n). By Proposition 5. there exists some a E G(S/R)with<1P; = P;. If P; =ap;'(2 ~ i ~n), then P; :::> P~ :::> ... :::> p~ is the required chain.· . In the general case let S* be a normal extension of R containing Sand pr, a prime ideai of S* lying above P~. Then by the first part of the proof there exists a chain pr:::> p::::> ... :::> P: of prime ideals lying above PI:::> p.:::> '" :::> r; If P; = P'! n S, (2 ~ i « n). then P; :::> P~ :::> ... :::> P: is the required chain.

4.3. EXERCISES I.

Let R be the ring of Gaussian integers with even imaginary part. Show that R is not integrally closed. What is its integral closure? 2. Let R be ring of all formal power series in X with coefficients in a field K containing no term' 'in X. Show that R is not integrally closed. What is its integral closure? 3. A domain R is-called a g.c.d. domain if any two elementSOt R have ~ greatest common divisor. Show that any g.c.d. domain is integrally closed. 4. Let R = k[XC. Y4] and S = k[X', X3Y. X~. ~] :::> R, where k

1

J

INTEGRAL BXTBNSIONS

5.

6.

7.

1

4.4.

101

is a field. Let P = (Y') be the prime ideal of R. Show that the ideal (XC, X3Y. XY·. Y') of S is a prime ideal associated to PS in S but does not lie above P. Let ReS be an extension and R the integral closure of R in S. Let I be an ideal of R. Show that if I« is the extension of I in R, the integral closure of I in S is 'liTe. Let R be a Noetherian ring. K its total quotient ring and at E K. Show that at is integral over R in each of the following cases (i) There is a non-zero divisor ~ E R with ~ocn E R for each n ;:. 1. (li) There is an ideal I of R containing a non-zero divisor such that «r C 1. Let ReS be an integral extension. Rand S domains and R integrally closed. If P be a prime ideal of R, show that the set of prime ideals of S lying above P are minimal elements of the set of prime ideals of S containing PS. Finiteness of integral closure

Proposition 1: Let R be an integrally closed domain with quotient field K and L a finite separable extension field of K. Let S be the integral closure of R in L. Then there exists a basis {el' e...... en} of Lover K, such that S

c

:i Rei.

'-1

Proof: We first show that there exists a basis of Lover K which is contained in S. If at E L satisfies a minimum polynomial equation etn + aletn- I +...+ an = O. a, E K. choose bE R. b #. 0 such that ba, E R. (I ~ i";;; n). Then (bet)n + alb(boc"y>-l +...+ anb" = O. so that b« is integral over R. i.e. bet E S. for some bE R. b =,6 O. Given any basis of Lover K. by suitably multiplying by a non-zero element of R. we can assume that the basis lies in S. Let such a basis be {e~. e~, .... e~}. Consider the bilinear form Lx L -+ K. given by (x. y) -+ Tr(xy). It is non-degenerate as L/K is separable. Henc e there exists a dual basis {eJ • e...... en} of L/K such that Tr(e;e) = 5,). If et E Sand '" = }'; a,e" then ate; E S. for all ], I

( 1 ,;;:: j ~ n) so that Tr(ate;) = Tr(I a,e,e;) = a). Now by Corollary "" 2 to Proposition 3. 4.3. the trace of any element of S belongs to R. Hence Q) E R, (1 ~ j ~ n).Thus ScIRe).

,

)

102

COMMUTATIVE ALGEBRA

Corollary 1: Let R be a Noetherian integrally closed domain with quotient field K and L a finite separable extension of L. Then the integral closure S of R in L is Noetherian. Proof: Choose a basis {el , .... e.} of Lover K such that S

C

1

,

:E Re,.

Since R is a Noetherian ring, Sis a NoetherianR-module and hence it is also a Noetherian s-mcdule, i.e. it is a Noetherian ring. Corollary 2: Let R be a principal ideal domain with quotient field K and L a finite separable extension of K. If S is the integral closure of R in L. there exists a basis {flJfl, ...• f.} of Lover Ksuch that S = Rf,. .

i

'-I

Proof: By Proposition I. we can choose a basis {elJe...... e.} of L/ K such that S C ~ Rei' Since R is a principal ideal domain and S

,

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. is also a basis of Lover K. Hence m = PI. This implies S where {fi.

t..... ,f.} is a basis of L/K.

=

i

'-I

Rfi.

Theorem 1: (Noether's Normalisation Theorem). Let k be a field and R a finitely generated k-a1gebra. Then there exist )11' )It..... )I, E R algebraically independent over k such that R'is integral over k[YlJ )I.,.... )I,]. Proof: Let R be generated by {xlJ x••' ...x.} over k, The proof is by induction on n. Arrange the generating set such that Xl. x.,.... X, are algebraically independent over k and XT+l. X'H• . .. . x. are algebraic over k[Xh x., ....x,]. If r = PI. there is nothing to prove. Assume that r
,a"-

x;

4

I

.l

I,

.I

I

1

I

INTEGRAL EXTENSIONS

103

+ (terms of degree less than d in x.) = O. This shows that x. is integral over R' = kIx;" x;,....x:_ I ] . By induction, there exist{)lI.·.. •)I,} algebraically independent over k such that R' is integral over k[)I, • ...• y,]. Since each x,(1 <;; i ,.;; n) is integral over R', R isintegral over R' and hence over k[Yl"" y,].

F(alJ a..... , a'-I' I)~

Case (ii): ' Let k be any field. Consider the polynomial f(XlJ X ...... X.) such that f(x l , x., ... , x.) = O. Choose t > total deg f, Attach weights m, = t·-l , m. = t·- 2 ..... m.=1 to the variables Xl' X., X. respectively and define the weight w(M) of a monomial Xf· occurring infasw(M) =

~ m,oc,.

x:·

If M' =

1

Xf". ..x:"

is

another

monomial occurring in/, then it is easy to verify that w{M»w(M') if and only if (ocl' ocf... ·.oc.) > (<<;. oc:) in the lexicographic ordering defined by (ocl,,,.oc.) > (oc;; oc:) when there exists some m with oc/ = "'; (1 ,.;;i .,,;:; m) and 0Cm+1 > OC~I' Thus the distinct monomials occurring infhave different weights. Choose M =' U1'...X:' A =1= O. A E k occurring in f with maximum weight. Make the substitutions x; = X, + x::"(1 .,,;:; i .,,;:; PI-I) in the relationf(x1, Xl ....' x.) = o. By the choice ofthe monomial M, we have a relation of the type h:(M) + (terms of lower degree in x.) = 0, with coefficients in R' = k[x;, ... x:_ 1] . Hence x. is integral over R' = k[x;, ... X:_I)' The proof is completed by induction on PI as in.case (i),

«;

Corollary 1: Let R be a finitely generated k-algebra, k field. and m a maximal ideal R. Then R/m is a finite extension of k, Proof: Since R' ~ RIm is afg. k-algebra, by Theorem 1. it is integral over k [Yl' " .• Y,], Yl' ..• Y, algebraically independent over k. Since R' is a field, so is k [Yi' ...• y.]. Hence r = 0 and R' is a finite extension of k, Corollary 2: Let R be af,g. k-algebra. k field and I an ideal of R.· Then ';T is the intersection of alI maximal ideals of R containing I. Proof: We may assume that 1=0. Clearly ';0 is contained in every maximal ideal or R. Conversely suppose fER - V'O. Then Rf=f:O. Let PI be a maximal ideal of R, and m =nnR. Now R,ln is af.g. k-algebra and it is a field. Hence it is algebraic over k, Bot R,/Plt::¥(R/m),. Hence R/m is integral over k. 1'Iiis

104 COMMUTA'TIVE ALGBBRA

implies that Rim is a field. i.e. m is a maximal ideal of R. Since [If; m, the intersection of all maximal 'ideals of R is contained ~~

.

Corollary 3: (Weak Nullstellensatz). Any maximal ideal m of R = k[Xlo ••• , X.]. k algebraically closed. is of the type m = (Xl - al •

• • ••

XD-O D). a, E k,

(I";; i";; n).

'.~

Proof: Since k is algebraically closed by Corollary I. Rim =- k, Let a, be the image of X, + m under this isomorphism. Then (Xl-a, • . . . • X.-a,,) c m and since it is maximal. it is equal to m. Corollary 4: (Hilbert's Nullstellensatz). Let R algebraical1yclosed and I an ideal of R. Let V(l)

= {(an· ..• aD) E

=

k [X, ..... X,,]. k

k" [leal' ... , a.) = O. for all [ E l}.

Then J = {h E R [h(al> "', 0.) is an ideal equal to V t.

= O.

for all (aI' ... , a.) E V(l)}

Proof: Clearly J is an ideal containing I and hence J~ 'III. To show that J C 'III. consider g E J. Now for every maximal m ~ I we have m = (Xl-a,• . . . • XD-aD) ~ I. This implies a=(a" . . . • aD)E V(I)

and hence g(a) = O. i.e. gEm. Hence g E

n

m:JI

m

=

'Ill.

Theorem 2: Let R be a domain which;is a finitely generated algebra over a field k, K the quotient field of Rand Lafioitefield extension of K. If S is the integral closure of R in L. then S is a module of finite type over R. Proof: By Tbeorem I. there exist {Yl"'" Yr} algebraically independent over k such that R is integral over R' = k [y]. YI' ••.• Yr}' Since the integral closure of R' in L is also S, we can assume with. out loss of generality that R itself is a polynomial algebra i.e. R = k[X" Xl..... X.]. Since L is a finite extension of K, tbere exists a finite normal extension L' of K such that L c L'. If S' is the integral closure of R in L' and if we show that S' is a finite R-module. then it will follow that S is also a finite R-module al

•«-

J

INTEGRAL EXTENSIONS

105

S c S' and R is Noetherian. Hence we assume without loss of generality that L is a normal extension of K. Then there exists an extension FIK.with KeF c L such that L is separable over F and F is purely inseparable over K. If S, is the integral closure of R in F and S is the integral closure of S, in L then S is the integral closure of R in L. It is sufficient to show that Sl is a finite Rmodule. for by Proposition 1. S is a finite SJ-module as LIF is separable. Hence we assume without loss of generality that LIK is purely inseparable. say L = K(<
4.4.

I.

2.

EXERCISES

Let ReS be domains such that S is a finitely generated Ralgebra. Show that there exists [E R; [=F O. and elements {Y],·... Yr} E S algebraically independent over R such that SI is integral over sf where S' = R[Yl' YI..... Yr]' Let ReS be domains. S a finitely generated R-algebra. Show that there exists a E R, :f. 0 such that any homomorphism

a

106

COMMtl'rATlVE ALGEBRA

f: R -+ n, n algebraically closed withf(a):I= O.can be extended n~ to a homomorphism g: S 3. Let k be a field. S=k[X. YJ{(X·. XY). Let x and y be the cosets of X and y. Show that x and y"(n ~ 0), form a basis of S . over k and S is not integral over R = k[xJ. Show further that

-+

4. S.

6.

7.

there is no element of S which is algebraically independent over R. Let ReS be domains. S a finitely generated R-algebra. If R has Jacobson radical zero, show that the Jacobson radical of S is also zero. A. ring R is called a Jacobson ring trevery prime ideal of R is an intersection of maximal ideals. Show that R is a Jacobson ring if and only if for every ideal I of R, R{I has both nil and Jacobson radical equal. Show that a finitely generated algebra over a Jacobson ring is a Jacobson ring. Prove the following version of Noether's Normalisation Theorem. Let R = k[X1 ..... X.]. k field. h E R- k. Then there exist t., .... t.-1 E R algebraically independent over k, such that R is integral over k[t.. t•• .. .. rn-l' h].

a" •....a,.

+

}; Xf' ... X~·· In the relation - h ~ah •.. a,. X~l ... 14.. = 0, make the substitutions t, = X, X::" (I ~ i ~ n-I) for suitable m,as in Theorem I and argue as in case (ii) by obtaining an integral relation for X. over k[t•• .... tn-I> hJ. (Hint: Write h =

+

1

1

CHAPTER V

DEDEKIND DOMAINS

While studying the ring of integers in an algebraic number field we come across rings which do not have the unique factorisation property for elements. .The uniqueness of factorisation can be restored in some cases by replacing elements by ideals; i e. every proper ideal in such a-ring can be expressed uniquely as a product of powers of prime ideals. Rings with this property are called Dedekind domains. They can be characterised equivalently as domains which are Noetherian, integrally closed and having the property that every non-zero prime ideal is maximal. This characterisation is useful in geometry as co-ordinate rings of nonsingular curves are of this type. We show that the integral closure of a Dedekind domain in a finite .extension is a Dedekind domain. The ring Op of merom orphic functions defined at a nonsingular point. P of a .curve is a special kind of ring which is a local ring as well as a Dedekind domain. Every element hE 0, can be assigned an integer Ordph which is nothing but the order of contact of h with the curve at P. The map h -+ Ordph is generalised to the concept of a valuation and the ring Op is an example of a valuation ring. specifiqally a discrete valuation ring. The generalised concept of valuation is used to develop the concept of order of contact of two algebraic varieties along generalised branches. 5.1. Valuation rings Let ReS be an extension of rings and n an algebraically closed field. The following extension problem arises quite often in geo~etry. . Given a ring homcmorphism h: R -+ n. does it have

108 COMMUTATIVE ALGEBRA

an extension ii as a ring homomorphism 'Ii:S....".. o.? We have seen in Proposition 3, 4 2, that such an extension is possible if S is integral over R. If R is a subring of a field X, we look for a subring S of X containing R for which h has a maximal extension. Such maximal extensions turn out to be valuation rings. Proposition 1: Let ReX, X field, and h: R -+ 0. a non-trivial ring homomorphism of R into 0., an algebraically closed field. If GC E X, lit #- 0, then h can be extended to a ring homomorphism li of either R[IIt]"""" n'or R(IIt-I ] _ n. Proof: We may assume without loss of generality that R is a local ring and heR) is a subfield of 0., for if P = Ker h, P is a prime ideal of Rand h can always be extended to E: Rp """" 0. by defining 1i(alb) =

~~:;, h(b) #- 0:

. Since R p is local and 1i(Rp) is isomorphic

to Rp/PRp , a subfield of 0. we can replace (R, h) by (Rp , ]i). Assume therefore that R is a local ring with maximal ideal m and heR) =F, a subfield of n. Now h can be extended to a homomorphism h: R(X] ... F(XJ by defining

,

= l:, ii,X',

h('Z 0,1")

where iit = h(o,). Let I = {IE R[J:'] Iftilt) = O} and J =h(I). Then J'is an ideal inF[XJ and henceJ= (6(X» for some 6(X) EF[X]. If 6(X) is a non-unit, it has a root ~ E 0. and 'Ii: R(1It] ... 0. can be defined by setting1i(IIt) =~, and ]i I R = h. If 6(X) is a unit, we have a relation of the type

.

,

l: a,GC'= 0, with

'-0

iio = I, ii, = 0, i t» I

(*)

Choose r minimum with this property. Similarly if h has no extension Ii: R(GC-I ] -+ 0., I;1t-I satisfies a relation similar to (*), i.e.

i

'-0

b,GC-'

= 0, lio =

1,

li, = 0, t » I

(s minimum). Wemay assume without loss of generality that r ~ s.

Since b. = I, b. - I E Kerh em and bo ¢ m, i.e, bo is a unit. So the second relation may be rewritten after multiplying by lit', as fl.'

+ Cilltl-I + .,. + c. = 0, c, = 0, I <: i .s;; s.

l

I

I

l

DEDEKIND DOMAINS

109

By subtracting O,lltf'-8 multiple of this from (*) we obtain a relation of type (*) of degree less than r satisfied by lit, contradicting the minimality of r. Hence h has an extension 7i: R(IIt]'" 0. or h: R[IIt-I ] ... n. Definition: A subring R of a field X is called a valuation riog of K if for every lit E X, CI. 01= 0, either lit E R or GC-I E R. Examples: (i) Any field K is a valuation ring of X. (ii) Let K be a field and R

= {~~ri lfl:X), g(X)EK(X), degf e; degg}.

R is a subring of X(X) and is a valuation ring of X(X). (iii) Let K be a field and R=

{i ,-0 a,XI. I a, E

K}

the ring of formal power series in X over X. L =

{~i ,_N o,X'1

If

a,EK,NEZ}

is the field of all Laurent series ill X over K, then R is a valuation ring of L. (iv) Letp be a fixed prime. Let R c Q, the field of rationals, be defined by .

R={P'~ I r>O, (m,p) = I, (n,p) = I ~\J 1.0

'5

Then R is a valuation ring of Q.. Proposition 2: nontrivial ring field n. The valuation ring

Let R btl a subring of a field K and h: R ... 0. a homomorphism of R into an algebraically closed' h has a maximal extension E: V -+ 0., where V is a of X.

ProoT: Consider the setl: of all pairs(R", h",), R" ring, R c R" c K and ha.: R." ...... 0., a ring homomorphism which is an extension of h. Define (R", h".) .s;; (R" h~) if and only if R", c R~ and h", = h~ I.R.. By Zorn's lemma, l: has a maximal element (V, Ji). Then clearly V is a valuation ring of X for if GC E K, lit 01= by Proposition I,

°

110

COMMUTATIVE ALGEBRA

h has an extension hi: V[ctj-+- n or 'hI: V[at-1j-+V[ctj = Vor V[ct-1j = V i.e. ct E Vor ",-1 E V.

n, By

maximality

Proposition 3: Let V he a valuation ring of K. Then (i) K is the quotient field of V. (ii) Any subring of K containing V is a valuation ring of K. (iii) V is a local ring. (iv) V is integrally closed. Proof: (i) Clearly I E K implies I = 1-1 E V. For ctE K. IX =F- 0, either IX E Vor IX-I E V so that or. can be expressed as or. = or.j1 or or. = l/or.-1 , i.e. K is the quotient field of V. (if) If V c V' c K, and IX E K, or. =F- 0, then or. E V or ",-1 E V and hence or. E V' or IX-I E V'. (iii) Let m be the set of non-units of V. If a, b Em, a =F- 0, b '# 0, then either afb E V or blo E V. If ajb E V, then a + b = (I + alb) bE m and if bjaE Y then a + b = (I + bla; a E m. Moreover, for any r E R we have ra E m. Thus m is an ideal and V is a local ring (iv) Let e E-K satisfy an integral relation of the type ct' + a1or.·-1 +

... + a. =

0,

0,

E V.

If or. ¢ V, then or.- E V. This implies ct = - at-ataat-1 ... _a.or.-(n-l) E V a contradiction. 'Hence or. E V, i.e. V is integrally closed inK. 1

Proposition 4: The ideals of a valuation ring are totally ordered by inclusion. Conversely if the ideals of a domain V with quotient field K are totally ordered by inclusion then V is a valuation ring ofK. Proof: Assume that V is a valuation ring of K and I, J ideals of V with j ¢ J. Choose a E I with a ¢ J. Then for any b E J, we must have bla E V for otherwise alb E V and then a = b(a/b) E J, which is impossible. Now bla E V implies b = a(bja) E I. Hence J c I. Conversely assume that the ideals of V are totally ordered by inclusion and consider or. E K, or. '# 0, Write or. = alb, a, bE V. Then either (a) c (b) or (b) c (a) so that alb E V or bja E V, i.e. atE V or or.-1 Eo V. Hence V is a valuation ring of K.

I

DEDEKIND DOMAINS

111

Corollary 1: If V is a valuation ring of K and P is a prime ideal of V, then Vp and V/P are valuation rings. Proof: If the ideals of V are totaIly ordered for inclusion, so are the ideals of Vp and V/P.

.

'

Corollary 2: domain.

Any Noetherian valuation ring is a principal ideal

Proof: Let V be a Noetherian valuation ring and I an ideal of V. If 1= {o..... , a.}. then there is a largest ideal among (a l ) , (a.) ..... (a.) which must be I. Hence I is principal. Corollary 3: Let V be a Noetherian valuation ring. Then there exists an irreducible element p E V such that every ideal Iof V is of the type 1= (p"'), m

~ I and

n(pm)

=

m-l

o.

Proof: Si~ce V is a principal ideal domain, the maximal ideal m of V is principal. i.e. m = (p), p irreducible. Now every non-unit a E V is contained in m and hence divisible by p, so that by succesI, u unit. Hence (a) = sive division by p, we have a = up", m

(pm). Clearly

n(pm)

m ....l

>

= (0) as V is a UFD. .

-

Theorem 1: Let R be a subring of a field K. Then the integral closure R of R in K is the intersection of all valuation rings Vof K containing R. Proof: Let V be a valuation ring of K containing R. Since V is integrally closed, V :::l R.. Conversely let 0 ¢ R. We construct a valuation ring V of K such that V:::l R and a ¢ V. Since a ¢ R. we haveo¢R[a-1]=R'. Hence a- 1 is a non-unit in R' and hence contained in a maximal ideal m' of R'. Let n be an algebraic closure of k = R'[m' and h : R' -+ n the composition of the natural projection and inclusion k -+- n. Now h can be extended to a maximal homomorphism h : V -+ n, where V is a valuation ring of K, containing R' by Proposition 2. Then h(a- 1) = h(tr 1) = 0 as 0-1 Em'. This implies a ¢ V for, if a E V, then 1= h(l) = -1I(atr1 ) = h(a)h(tr1 ) = o·

112

COMMlfTATIVE ALGEBRA

Corollary 1: A domain R with quotient field K is integrally closed if and only if it is an intersection of a family {V..} of valuation rings of K. Proof: If R is integrally closed, then R = R =

n V , the

intersec-

V~R

tion of all valuation rings of K containing R. Conversely if R = n V.., V.. valuation ring in K, R is integrally closed as each V. is integrally closed. We now give another characterisation of valuation rings.

..

.

Definition: Let Rand S be local rings with unique maximal ideals and ms respectively. S is said to dominate R if R is a subring of Sand m» = R n ms.

mil

Theorem 2: Let K be a field and I = {(R, mR) IRe K} where R is a local subring of K with maximal ideal mR. Define a relation .,.;; in I by (R, M R) ,.;; (S, ms) if (S, ms) dominates (R, mR)' Then (V, my) E l: is maximal if and only if V is a valuation ring of K. Assume (V, my) is a maximal element of I. Let k = V/my, and h : V -+ n the composite of natural projection V -+ Vlm» = k and the inclusion k ...... n. By Proposition 2, it is sufficient to show that (V, h) is a maximal extension. Let (R', h') be an extension of (V, h) where we may assume R' to be a local ring with Ker h' = mS' the unique maximal ideal of R'. Since h'is an extension of h, mR' n V = my so that (R', mR') dominates (V, my). Since (V, my) is maximal in l:,(V, my) = (R', mR') and the proof is complete. Conversely assume that (V, my) E l:, where V is a valuation ring of K and suppose (V, my)";; (R, mR) E:E. If IX E R, then either IX E V or IX-I E V. If IX ¢ V, then oc-I EVe R, i.e. oe l E my = mR n V, i.e. a.~1 E mR, a contradiction. Hence for every IX E R we have IX E V, i.e. V = R. Hence (V, my) is a maximal element Proof:

n an algebraic closure of k

on.

Corollary: Every subring of K which is a local ring is dominated by at least one valuation ring of K. Theorem 3: Let R be a principal ideal domain with quotient

1

1

DEDEKIND DOMAINS 113

field K. Then the valuation rings of K containing R and different from K are of the type R pR, for some irreducible elementp E R. Proof: If pER is an irreducible element, clearly R~R is a valuation ring of K containing R (Example (iv), 5. 1). Conversely assumethat V is a valuation ring of K Ci=K) containing R. Since V =F K. my contains a non-zero element IX E K. Let CIt = a/b. a, b E R. Then my n R contains the non-zero element a and since it is a prime ideal of R, my n R = (p) for some irreducible element pER. Then (R pR, pR pR)";; (V, my) and by maximality R pR = V. Corollary 1: The valuation rings of Q different from Q are of the type Zl'z, p prime. Corollary 2: Let K(X) be the field of rational functions in X over a field Kand V a valuation ring of K(X) containing K different from K(X). Then either V = K[X!J(x) for some /(X) E K[X] irreducible or V= K[X-l]IX-')' If X E V, then K[X] c V and by the previous Proposition V= K[X](rIX» for some irreducible /(X) E K[X]. If X ¢ V. then X-I E V so that K[X-I] C V. Then by Theorem 3, V = K[X-I]p for some prime ideal P. Since X-I is not invertible in V. X-I E P, i.e. (X-I) C P. But (X-I) is a maximal ideal so that (X-I) = P. Hence V = K[X-Ib_.) (Example (it), 5. I). Proof:

We now invOlltigate relations between valuation rings and valuations. Definition: A group r is called an ordered group if it has a total ordering EO; which is compatible with the group structure. i.e, ar: < ~, implies "(iX ,.;; y~, IX"(";; ~"(, for aU IX, ~, y E r and IS-I,.;; iX-I. Definition: Let r be an ordered group. A valuation I I on a field K with' ValUCIO in r is a mapping I I: K* -+ r satisfying the conditions: (i) Iab I = I a II b I

-c

(ii) la + bl Max ( Ia 1.lb/) It is sometimes convenient to extend r to a larger set r u{O} by introducing an element 0 with the properties, 0·0 = 0.0·« =ot·O

114

COMMUTATIVE ALGEBRA

=O,"'ErandO<",forall"'Ef. By defining 1°1=0,/ I is extended to all the elements of K and the two conditions (i) and (ii) on I I are still satisfied. Example: Every valuation ring defines a valuation. Let V be a valuation ring of K. Denne an order ~ in K* by stipulating a ~ b if and only if abr) E V, a, bE K. * Then ~ satisfies all the conditions of total ordering except the anti symmetry. because a ~ b and b ~ a implies abr! E V and ba-1 E V, t.e. ab-1 is a unit of V. Let G denote the multiplicative group of units of Vand r = K*jG. Then the ordering on K* induces an ordering on rwhich makes it a totally ordered group. The mapping I I: K* -+ r given by Ia I = aG is a valuation on K. Conversely every valuation on K defines a valuation ring of K. Let I I:K~rbeavaluationonK. Then V={aEKI lal~l} is a valuation ring of K. There is an alternative notation for valuation which is particularly convenient in geometry. Here the value group is written additively and the order is reversed. Let A be a totally ordered abelian group written additively. Definition: A valuation v on K with values in A is a mapping v : K* ... A satisfying: (l) v(ab) = veal + v(b) (ii) v(a

+ b);;;;' Min {veal.

v(b)}.

It is sometimes convenient to adjoin an element 00 to A and extend the operations by setting '" + 00 = 00 + '" = 00, lX E A, 00 + 00 = 00 and «<: 00 for all « E 11. Then v can be extended to a mapping v:K ... A by defining '1'(0) = 00 and the two conditions on the valuation are still satisfied. Examples: (i) Let K = k(X) be the field of rational functions in X over k and p(X) an irreducible polynomial in k[X]. Any nonzero element of K can be uniquely expressed as ,

I(X)

O(X) = p(x)r. g(X) ,

Z and p(X) does not divide I(X) or g(X). Then the map v: K -+ Z given by v(6(X» = r is a valuation on k(X) called the p(X)-adic valuation. rE

<, DBDEKIND DOMAINS

115

(ii) More generally let R be a PID with quotient field K and pER an irreducible element. If « E K, write '" = p'. bjc, (p, b) = 1, (p, c) = I, r EZ. Define v:K ... Z by '1'(<<) = r. Then ~ is a valuation on. K called. the p-adic valuation on K. In p.ticular If R = Z, the p-adlc valuatIon on Q is defined. (iii) Let K = k(X), the field of rational functions in X over k, For any non-zero element/(X)jg(X) E k(X),define v

(~i~) = degg(X)-degl(X).

Then v is a valuation on K. (iv) Let K be the field of Lauent series in X over k, and for « E K, lX = ~ al XI, Q, E k, NEZ define '1'("') = N, the order of «. Then v is a valuation of K. Each of the valuations in the above Examples have the value group Z. Such valuations are called discrete valuations and the corresponding ~alua!ion rings arc called discrete valuation rings. . The valuatIon nng corresponding to the valuation v is clearly grven by v = {a E

KI v(a);;;;, O}

and the maximal ideal m

= {a E KI '1'(0) > O} 5.1. EXERCISES

1. 2. 3.

Show that an intersection of a totally ordered collection of valuation rings of K is a valuation ring of K. Prove that in a valuation ring any radical ideal is prime. Let R be a domain with quotient field K (;6 R). Prove that the following conditions are equivalent. (i)

~ is a ~aluation ring IS

maximal.

in which every non-zero prime ideal

(ii) There are no rings properly between Rand K.

4.

~et K b~ a field, R integrally closed subdomain of K with K as Its q.uotlent field. Let {R x } be a family of valuation rings of Ro: = R. Show that the integral closure of R in L,

K With

Q

116 COMMUTATIVE ALGEBRA

an extension field of K is the intersection of all valuation rings of L which dominate one of the R",. 00

5.

Let R be a local ring with maximal ideal m

= (0), show that the only ideals

= (p).

If

n

(pn)

0=1

of Rare (0) or (p"') for some

m ;;. I. Show further that either p is nilpotent or R is a 6.

7.

valuation ring. Let R be a local domain with quotient field K. Show that R is a valuation ring of K if and only if any ring Ii.' with R c R' c K contains the inverse of some non unit of R. If v is a valuation on K with values in t1, show that (i)

v(

i

at)

1-1

:>

Min {v (ai)}' 10000 .. n

(ii) Equality holds in (I) if there exists some j such that v (aJ) =

Min {v (al)}' 111;III;n

8.

Let v: K -+ t1 be a valuation on K. VII = {a E K

.9. 10.

I veal >. Il}

For GI E t1 define

and

V; ={a E K I v(a) :> a} Show that· VII and V; are ideals of the valuation ring V and every non-zero ideal of V contains some Show that a field which is algebraic over a finite field has no nontrivial valuation. Let K = k(X), the rational function field in X over k and Il E K. Show that 1: v(GI)=O, where the summation is taken

v:..

.

over all the p (X)-adic valuations, each p (X).adic valuation counted deg p (X) times. and the valuation given in Example (iii), 5.1. 5.2. Discrete valuation riDgs DefiDitioo: Let K be a field. A discrete valuaticn on K is a valuation v: K· -+ Z which is surjective. The corresponding valuation ring is called a discrete valuation ring. Examples: All the Examples given in 5.1 are discrete valuations. The corresponding valuation rings are discrete valuation rings.

l

DEDEKIND DOMAINS

117

Definition: Let v be a discrete valuation on K. An element with t E K is called a uniformizing parameter for v if vet) = I. Examples: In the Examples (i), (ii), (iii) and (iv) given in 5. I the uniformizing parameters are respectively (i) p(X) (ii) p (iii) IjX and (iv) X. Theorem 1: Let v be a discrete valuation on K with discrete valuation ring R and the corresponding maximal ideal m. Then (i) m is principal and t E R is a uniformizing parameter for v if and only if m is generated by t. (ii) If t is a uniformizing parameter, every a E K can be uniquely expressed as a = ut», u unit in R, n E Z and K = R,. (iii) Every non-zero ideal of R is uniquely of the type m" (n ;;. I). In particular R is a principal ideal domain. Proof: (i) We first show that m is principal. Since v is surjective, there exists some t E R with vet) = I. Clearly, (t) em and let a Em. Then v(a):> I and v(at-1) = v(a) - vet) :> O. Hence at-I E R so that a E (t) showing m = {r). Thus any uniformizing parameter generates m. Conversely assume that m = (t'). Then vet') :> I. Since t E (t') = m we have t = at', a E R, so that I = vet) :> vet') 1, i.e. vet') = I. (ii) If n = veal, then v(at ......) = 0, so that at-n = u, u unit. This shows that a = ut" and n is unique as n = v(a). Clearly this also implies that K = RI • (iii) LetI be a non-zero ideal of R. Choose a E 1 with veal = n, n least non-negative integer. If t is a uniformizing parameter, then v(ar O) = 0 so that atr" = u, u unit, i.e. a = ut«. Hence (t n) C 1. If bE 1, with v(b) = k :> n, then v(bt-k ) = 0, i.e. b = u'tk, u' unit. and b E (t n) . Hence 1 = (t n ) = m" and n is unique.

>

Definition: Let 1 be a non-zero ideal of a discrete valuation ring R. If 1 = m", then we de6ne v (1) = n. Proposition 1 : Let R be a discrete valuation ring with maximal ideal m and 1 a non-zero ideal of R. Then (i)

v(l) = lR(RjI).

(ii) If Rfm contains a field k, then dimk (Rj1)

= [Rjm:

k]v(l).

118

COMMUTATIVS ALGEBRA

Proof: (i) Let m = (t) and I = (In) with n = vel). Consider the sequence of ideals of R R ::J m ::J ml ::J ... ::J mn = I. Then

IR(RIl) = ~~: IR (m~1 ) by repeated USe of Proposition 2, 3.4

IR(m~:I) =

Now

as m' =1= ml+1 for if m' =

mi+l,

dimRlm

(;~1) =

1.

then m = 0 by Nakayama Lemma.

Hence l,,(Rfl) = n = vel). (ii)

If Rim contains a field k, then

. (m') = [Rim: k] dlmR/m . (m') m

d,mk m'+l

+! •

'

Hence

dimk(RII) = [

.

=

m' ) ~ dimk ( I+i m

n-l

'-0

n-l •

(

m' )

Rim: k ] ,~o d,mR/m m'+1

( m' ) = IRlm(mTil'l +1 )

But d,mRlm m'+1

= IR ( mm'+!)

' Hence dimk (RII) = [Rim: k]/R (Rfl) = [Rim: k]v(l}.

Theorem 2: Let R be a Noetherian local domain with unique maximal ideal m =1= 0 and K the quotient field of R. The following conditions are equivalent. ' (i) R isa discrete valuation ring. (ii) R is a principal ideal domain. (iii) m is principal. . (iv) R is integrally closed and every non-zero prime ideal of R is maximal. (v) Every non-zero ideal of R is a power of m. Proof:

(i)

~

(ii) follows from Theorem 1.

(ii) ~ (iv) is clear from definition. (iv) ~ (iii). Choose t'E m, t oF O. Since m is the only non-zero prime ideal, (r) is m-primary. Choose n such that mnC(I)

[

DEDJ!KIND DOMAINS

119

and m n- 1 ¢ (t). Choose a E m n- 1, a ¢ (t) and let «= tla E K. Then 11- 1 ¢ R and since R is integrally closed, at- l is not integral over R. This implies by Theorem 1,4.1 that «-1 m ¢ m. But by construction at_I m C R so that at-I m = R. Hence m = RIX is principal.

~ (ii). Let m=(t). We claim that

n

n(t

=0

mn=(O). If aEmn for '-0 n, all n ;) 0, then a = bnl b. E R. Consider the increasing chain of ideals (bol C (b I ) c ... C(bn) c ...· Since R is Noetherian th~r~ eXi~ts some n such that (bn) = (bn+I ) , i.e. bn+l = eb., e E R. ThIS implies that b.(1 - er) = 0, so that b. = 0 as I is a non-unit, i.e, a = O. Let I be any non-zero ideal of R. Then there exists an integer n such that I C m' and I ¢ mn+I • Choose a E I, a ¢ mn+l. Then a = ut n, u ¢ m and hence u is a unit. Hence In E I, t.e. I = (In). (ii) ~ [i). Since R is a principal ideal domain, m is principal i,e. (iii)

m = (I) for some t

E

R.

Clearly

n=O

n)

as shown in

the

previous implication. Hence every a E R can be expressed uniquely as a = ut", u unit. Consequently if at E K, Cl' 0, CIt can _ be written uniquely as " = ut», mE Z, u unit in R. If we define V(I1) = m, v is a discrete valuation on K whose valuation ring i.s R. Thus the conditions (i) to (iv) are equivalent. Clearly (I) ~ (v) by Theorem I. We show that (v) ~ (iii). By Nakayama Lemma, m mi. Choose t E m - mI. Then (/) = mn for some n ;;;. I and clearly n = 1 by the choice of t. Hence all the conditions (i) to (v) are equivalent.

'*

'*

Corollary: If R is a discrete valuation ring and IE m - m 2, then t is a uniformizing parameter.

5.2. EXERCISES 1. 2. 3.

Show that a valuation ring other than a field is Noetherian if . and only if it is a discrete valuation ring. Let R be a domain and P a prime ideal in R[X] which contracts to (0) in R. Show that R[X]p is a discrete valuation ring. Let R be a domain with quotient field K such that R contains an element p 0 with (p) prime ideal and R[p-I] = K. Show

'*

120

4. 5.

COMMUTATIVE ALGBBRA

that R is a discrete valuation ring. Give an example of a local domain whose maximal ideal is principal and which is not a valuation ring. Let R be a local domain with quotient field K and maximal ideal m. Let v be a valuation on K whose valuation ring dominates R. Suppose m is finitely generated or v is a discrete valuation, show that there exists a E m with v(a) = In! v(b). bem

6.

7.

5.3.

Let R be a valuation ring which is not a field. Show that R is a discrete valuation ring if and only if every prime ideal of R is principal. Let R be a valuation domain with quotient field K and L a subfield of K. Let R' = L n R. Show that R' is a valuation domain with quotient field L and that if R is a discrete valuation ring so is R'. Dedekind domain

Definition: A Dedekind domain is a Noetherian integrally closed domain in which every non-zero prime ideal is maximal. Examples: (i) Any principal ideal domain is a Dedekind domain, (ii) Let K be an algebraic number field, i.e. a finite extension field of the rational number field Q and let R be the ring of integers in K, i.e. the integral closure of Z in K. Then R is a Dedekind domain as shown in Proposition 1. In particular the ring Z[y-5] = {a + by -5 I a, bE Z} is a Dedekind domain. (iii) The co-ordinate ring of a non-singular curve is a Dedekind domain. Proposition 1: The ring of integers in an algebraic number field is a Dedekind domain. Proof: Let K be an extension field of Q of degree n and let R be the integral closure of Z in K. By Proposition I, 4.4, there exists. a basis eel' e2 , ... , en} of K/Q such that

Rei Zel. 1_1

Thus R is a

finitely generated Z-module and hence Noetherian. R is integrally closed by definition. Let P be a non-zero prime ideal of R. Now

DBOBKIND DOMAINS

121

n

p Z -:f- 0 as (0) is the unique prime ideal of R lying above (0) (Corollary 2, Proposition 1, 4.2). Hence P n Z is a maximal ideal of Z. This implies that P is a maximal ideal of R showing that R is a Dedekind domain.

Theorem 1: Let R be a Noetherian domain in which every nonzero prime ideal is maximal. The following conditions are equivalent. (i) R is a Dedekind domain. (ii) Rp is a DVR (discrete valuation ring) for every non-zero prime ideal P of R. (iii) Every primary ideal of R is a power of a prime ideal. Proof: R is integrally closed if and only if Rr is integrally closed for every prime ideal P of R. Hence condition (i) implies (ii) by Theorem 2,5.2. Conversely condition (ii) implies that R is integrally closed and then R is a Dedekind domain. Now we show that (ii) => (iii). Let Q be a P-primary ideal of R and we may assume P -:f- O. Then R p is a DVR. Hence by Theorem 2,5.2 QR p = pnR p for I. Since P is maximal, P" is P-primary and hence some n

>

Q = QRpn R =pnRpnR =pn. To show (iii) => (ii) let Q' he a non-zero ideal of Rp Q = Q' n R. Since every non-zero prime ideal of R p maximal,Q' is PR p primary. Hence Q is P-primaryand Q = P» (n 1). This implies that Q' = pnR p and hence DVR by Theorem 2,5.2.

>

and let is also by (iii) Rp is a

Corollary: Let R be a Noetherian domain in which every nonzero prime ideal is maximal and let K be the quotient field of R. The following conditions are equivalent. (i) R is a Dedekind domain. [ii) R is an intersection of a collection of DVR's of K. (iii) R is the intersection of al1 the DVR's of K containing R. Proof: (i) => (ii) follows from the relation R =

nR

p•

p

(ii) => (iii) is clear. (iii) => (i), Since eachDVR is integrally closed, R is integrally closed and hence R is a Dedekind domain.

122

COMMUTATIVE ALGEBRA

Theorem 2: Let R be a Dedekind domain with quotient field K and I a non-zero ideal of R. Then I can be uniquely expressed as I = P~' p;' . . . P:' = P~'nP:'n ... nP;',

where PloP" ... ,P, are prime ideals of R containing I. The P?sitive intege~s"l are given by = vp,(IRp,) where vr, is the discrete valuation of K corresponding to the valuation ring

n,

Rp,(l ~ i ~ r).

Proof: Since every primary ideal of R is a power of a prime ideal, I has a primary decomposition of the type I = P~' n ... n p;', where PI; P 2 '" P, are non-zero prime ideals of R and hence maximal. This implies PI' ... , P, are pairwise comaximal so that I = P~'

n... ,nPl' = P:'. p;' . . . . P;'.

Since IRp, = P;/Rp" we have vp,(IRp,) """ n, (I ~ i ~ r). The uniqueness part follows from the above relation and the uniqueness of primary decomposition. A Dedekind domain R can also be characterised by the property stated in Theorem 2. (See O. Zariski and P. Samuel: Commutative Algebra-Vol. I). We now give another characterisation of Dedekind domain which is useful in number theory. Let R be a domain with quotient field K. 0

Definition: A fractionary ideal of R is an R-submodule M of K such that aM C R, for some a E R, a::f= 0. Examples: (i) Letl be an ideal of Rand 01: E K, 01:*0. Then oM = «l is a fractionary ideal of R. In particular if 01: = 1, then any ideal of R is also a fractionary ideal. These are calJed integral ideals. A fractionary ideal of the type Rot, 01: E K is called a principal fractionary ideal. (ii) If M is a finitely generated R·submodule of K, then clearly M is a fractionary ideal. Conversely every fractionary ideal of R is a finitely generated R·module if R is Noetherian, for if M is a fractionary ideal, aM is an integral ideal for some a E R, a i' O. By choosing a generating set {ai' ... , a.} for aM, we have {a- I a1' ... , a- 1 a.} as a generating set of Mover R.

123

DEDEKIND DOMAINS

Definition: A fractionary ideal M is called invertible if there exists another fractionary ideal N such that MN'= R.

1

Example: The principal fractionary ideal M is invertible with inverse N = ROI:-I.

=

ROI:, 01: E K, 01: =I: 0

Remark: (i) If M is invertible, its inverse is unique and is equal to (R: M) = {OI: E K, OI:M c R} because MN = R, implies Nc(R:M)= (R:M)MNcN

and hence N = (R: M). (li) The invertible ideals of R from a group for multiplication with R as unit element and the inverse of Mis (R: M). °

Proposition :z: Let M be a fractionary ideal of R. The following conditions are equivalent. . (i) M is invertible. (ii) M is finitely generated and M p is invertible in Rp for every prime ideal P of R. (iii) M is finitely generated and M m is invertible in R m for . every maximal ideal m of R. . Proof: (i) ~ (ii). Let M be an invertible ideal so that M(R: M) = R. This implies that 1 = 1: a,b" 0, E M, b, E (R: M). If x E M, then x =

~ (b,x)

,

a" where bix E R.

Hence M is generated over R

/

by the {a,}. Moreover

M(R:M) = R implies Mp(R: M)p = R p i.e. Mp(Rp: M p ) = R p for every prime ideal P. Hence M» is invertible. (ii) ;> (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. mality m- 1l is a power of contradiction.

111,

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.

125

DEDEKIND DOMAINS

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

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.).

This implies

, a1 b~, with b1 EKe L.

1=I

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

DEDEKIND DOMAINS

127

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)

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

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

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

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

,

have a welldefinedhomomorphism S ~ SIPS"

SIQ~I

and therefore

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

Chinese Remainder Theorem. Its Kernel is

n Qi' =

PS.

I

Hence SIPS ~ ~ SIQi'.

This also is an isomorphism as k-spaces.

'-I It is therefore sufficient to show that dim" SIQ!' = But

SIQi'

~

SOI_ ~ Sal. Qj'SQI PSQI

e,f,.

But by Proposition I. 5.2

. = e(QIIP) [SIQI: k]

=e,fj

and the proof is complete. 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». veE

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

m.n

DBDBKIND DOMAINS

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

5.3. 1. 2.

EXERCISES

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 .

I'

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. (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. 6.

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.

CHAPTER VI

COMPLETIONS

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. 6.1.

Filtered riogs and modules

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",

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

f 1

f

COMPLBTIONS

131

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

l'v) I~ ~ l~

fl'VI,

e:t

q~~ .

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. 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.

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. .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 .;00

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' --<0\ ~tL<.M 1' ... IS. 0 , Examples: (i) For any ring R, Ro ... R. R. = O. 71 ;> 1 defines a gradation on R called the trivial gradation. (ii) .Let R = K [Xh .... X.J, K field and ~ the set of homo• geneous polynomials over K in ·XI .... ! X. of degree d. Then R ,,;,

EB }; .

d~

~. and Rtl~, c ~+tI' 80 that {~}, d ~ 0 is

a gradation-

132

COMMUTATIVB ALGBBRA

Definition: Let R be a graded ring. An R-module M is caned a graded R-module if M can be expressed as a direct sum of subgroups {Mnh t.e,M = EEl 1: M n, such that RmMn C M m+., m,n,> O. n>O

t

,

~~k " &>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..<.. ~V" e'" ..!~ 'Y'-""J.,.~~ ,-: v-y ) (ii) Let M be an R-module. The tensor algebra T(M) of M defined by T(M) = Ee I T/ (M) where T/(M) is the tensor product /~

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..

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";;;'0

module M/N is also graded. 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. 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{

(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

1>0.

Ii

1

COMPLBTIONS

133

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
Definition: Let M be a filtered R-module over a filtered ring R, with filtrations {Mn}.>o and {Rn}n>o respectively. Let grn(M) = Mn/MO+ l and gT(M) = EEl I sr, (M;. ·>0

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. 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 h:N-+ K.

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). ~_:>

.

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

COMMUTATlVI! ALGIIBRA

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.

?

COMPLETIONS

135

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*· 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

1. Let' R = EEl

~

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.

jJ~

2.

3.

4.

5.

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. Let M = EEl ~ M. be a graded R-module with trivial gradation .>0

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

M

gr({)

o -+

6.

7.

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

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. 6.2.

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).

137

COMPLBTIONS

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). 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/ /

and the submodule MeN defined by M

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

= xu i

:> O}.

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). lim

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

,

.'~

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

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

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) .

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. Coronary: The topology defined by the filtration is Hausdorff if and only if

nu, o .

=

(0).

Proof: The topology is Hausdorff if and-only if

{O} = {OJ =

nu; 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

Proof: Let

~

JY.I

= _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",

x:

COMPLETIONS

139

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)

+

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

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

S:t1l

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

COMMUTATIVE ALGEBRA

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

•

= -rrM;; and

•

it is easy to check that the follow-

If

ing diagram is commutative f 0 -+ M' -+

M"

-+

-+

0

-+

0

dNl dM"!

dM'l 0 -+

g

M

f

M'

g

M

-+

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. 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

o -+ M'

-+

M -+ Mil

-+ 0 is exact.

Proof: For each n, we have an exact sequence of R-modules M' M Mil 0-+ M' nf-l(MIf} -+ M. -+ g(M,,) -+ 0

such that the following diagram is commutative

o -+

M'

M'

nf-l(M.+ 1)

6~+1

1

~

-e-

....

M.+t

~ -+ 0

g(M.+ 1)

1 1

6.+1

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

-+

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

"

.... Mil -+ 0

0: is surjective for each n.

I I

I

I

4

COMPLETIONS

Corollary 2:

141

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

completion M. Then the completion

M. of M. with respect

induced filtration on M. is a submodule of

= M.

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

~ o -+ M.

Sf and ~

" (M) M.

-+ M -+

M.

to the

"" MM for all n. If

and Mil = M/M. we

-+ 0

t.e.

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

Since n is fixed .M'' :

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

Hence Mil ""'"

M" and this implies ~

"" M .

• M.,

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

induces a filtration

This

'

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

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

M.

By Corollary 2, we have

Z.,

ot

it under a natural isomorit.

phism making the following diagram commutative

M

if

- - -+-.

Mn+l

Hence

lim

M

lim

Sf

M:

--"""'~-If

M.

If

M"

8+1

~

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= <:B 'J- In be the abelian "

where

p~z' R being a countable"direct

sum of Z/pZ. Show that

the completion of R with respect to the p-adic filtration is isomorphic to R but the completion of R with respect to the filtration induced on R by the p-adic filtration on S is the

· z direct product 7C taken countable number of times. 4.

pz Use Exercise 3 to show that there exists filtered modules M', M, M" such that 0 -+ M.' -+ M o -+M." -+ 0 is exact for all n,

• but S.

6.

(J!.:

Mo) -+

(J!.:

Mo") -+ 0 is not exact.

Let R = k [[X, Y]) the power series ring in X ~nd Y over a field k of ch 2, m = (X, 1') and P = (XI- }'lI). For the m-adic filtration on R, show that P is a prime ideal in R closed in the m-adic topology but gr(P) is not a prime ideal in

*

gr(R). Let M be a filtered module,which is Hausdorff and complete

with respect to the topology defined by the filtration. Show that a series Io X O, X o E M converges in M if and only if x 0 7.

tends to zero. Let M be a filtered R-module with filtration {M o} such that M is complete with respect to the filtration and N asubmodule

J

COMPLETIONS

143

of M. Show that MIN is complete with respect to the induced filtration on M/N. 6.3. I-adic filtration We consider in this section some special properties of com- ' pletions with respect to I-adic filtrations. Let M be an R-module and I an ideal in R. Definltlon: The topology defined on M by the I-adic filtration is called the I-adic topology and the completion is called the I-adic completion M. EDDlples: . (i) R = k[X'J, k field and I = (X). Then the compte• II. lim R k UX]], 'the rmg . 0 f ~rorma I' power senes . In . .K " : " ; : (X")"': non X over k. , (ii) Let R ~

R

=

lim Z ~ -Z

Z and 1= (p), p prime. The completion .. is called the ring of p-adlc integers. =

n pO (iii) The completion of R = k[Xwo., X.] with respect to the I-adic topology, where I = (Xl' Xl •..., X o) is the power series ring k fIXI , XI" .., X o]].

Let M be an R-module 'and I ~ ideal in R. Let 1:1 and R denote respectivelythe I-adic completions of M and R. Define the scalar operation of RonM by {an} {Xo} = {anXoh where {a.} is a Cauchy sequence in Rand {X o} is a Cauchy sequence in M. This operation is well defined and for this operation Mis 'an R-module. Let M and N be R-modules and I:M -)oN an R-linear map.

,

'

N

Since 1(1"M)c t-», n ~ O,finducesan R-Hnear map!.: I~-+ PN such that the diagram M [n+1M I

! -

M I'M

is commutative.

"

-

N lo+lN

-

N I"M

l.tI "

f.

1

.-

144

COMMUTATIVB ALGEBRA

On passage to the inverse limit we have an R-homomorphism

".,.,

A

,f:: M _ N. Clearly Id mapg:N-K.

=

A

AA

fd and gf = gf for

another R·IiDear

f f/ Proposition 1: Let 0 .... M', ~ M. _ M" .... 0 be an exact sequence of finitely generated R-modules over a Noetherian ring R and I an ideal in R. Then the I-adic completions

-L g .L £1"

_

O-+:tl' modules.

0 is an exact sequence of

D-

Proof: By Artin-Rees Lemma, the f-adic completion M' is the same as the completion of M' with respect to the filtration on M' induced by the f·adic filtration on M. (Exercise 2, 6.2). Similarly the f-adic completion £1" is the same as the completion of M" under the filtration which is the image of the f-adic rotrationonM. The result now follows from Oorollary 1 to Proposition 3, 6.2. Let M be an R-module and I an ideal in R. Consider the natural map M _

Mgiven

by x -+ {(x

+ M.)}

and

a

corres-

It These maps induce natural R linear R® M -+ R® MO!. M giving rise to a natural

ponding map from R to maps

R® M

-+

R

f

R

R-linear map ¢M:

R® M R

-+

M.

The following Proposition gives

conditions under which ¢M is an isomorphism. Proposition 2: Let R be a ring and M a finitely generated R· module.

Then the map' ¢M: ~ ® M -+ ,

R

f:J is surjective.

If fur-

ther R is Noetherian it is an isomorphism. Proof: Clearly 'R:

R® R -+ R is R

an

isomorphism aad since

'M

tensor product commutes with direct sums, is an isomorphism when M is free of finite rank. If M is a finitely generated Rmodule, there exists an exaot sequence of R-modules

COMPLETIONS

.L;

145

.s.;

O-+N F M -+ 0 where F is a free module of finite rank. Consider the commutative diagram ~

~

-- J

~

0

A

If

~

0

with the first row exact. By Corollary I, to Proposition 3, 6.2,

a~d the second part of the proof of Proposition I, g is surjective. SlDce ¢F is an isomorphism, it follows that "'M is surjective. , Assume. DOW that R is Noetherian. Then by Proposition 1 the ~ot~om row .IS .exac.t. Since N is also finitely generated, ¢N is 'surjective. This Implies that is injective and the proof is complete.

'M

\ CorOIIar~ 1: Let R be.a Noetherian ring, I an ideal in R. T~ the l-adlc completion R is a fiat R-algebra. ' Proof: Since

Rflj M

"'"

M, the

result follows from Proposition 1.

CorOUa? 2: Let R be a Noetherian ring, 1 an ideal in R and I;; the ]. adic I t'. ..../'li comp e Ion. If a E R is a non-zero divisor of R, Iit ]IS also a non-zero divisor of R. Proof: The homothety given by A.: R -+ R A (b) = ab I' . . ti '" ~ A. ' • s lDJec Ive k .. . IS mjec-

and hence A.: R -+ R which is also a homothety by a E '~ tive, Thus a E l{ is a non-zero divisor.

Remark' If R is d '" . domah a omam, It IS not 10 general true that a omam (Sec Exercise 6, 6.3).

ii

is also

Proposition 3: Let. R be a N oe therian etlan ei ring, 1 an Ideal in Rand th e ].-a dirc completion. Then (i)

i """ ~ ® t CO! RI R

I;; l{

146

COMMUTATIVE ALGEBRA

(ii) (f}8 c><(F) (iii) 1"//8+1 =.

P./fn+ 1

(iv) iis contained in the Jacobson radical of Proof: (i)

R.

Since R is,Noetherian, I is finitely generated and l:

A ® I -+ J R

is an isomorphism. Since

R is fiat over R,

the natural map

R(l)I-+ R®R R

R

is injective with image isomorphic to Ri. .(ii) Apply (i) to the ideal I". Then (10) c« RI" = (Rl)" =

(If.

(iii) By Corollary. 2 to Proposition 3, 6.2 we have for each

n an isomorphism at": R/[" c>< R/j"making the following diagram with exact rows commutative o -+ /"/[0+1 -+ R/I"+l -+ R/I" -+ 0

1 o -+

j"/[O+I

1

lotn+

1

-+

Rlllt+l

at"

-+

A/i" -+

0

In particular /"//0+1 c:=.i"/P"+l. (iv)

By Corollary 3 to Proposition 3, 6.2, Ais complete with the series 1 + a + al

the j.adic topology. Hence for any a

E t,

+... a"+ ... converges in R say to s.

Hence (1 - a) is a unit in

with inverse s.

This implies 'i c J (R) the Jacobson radical of

~

it

Corollary: Let ~ be a local ring with unique maximal ideal m and A its m-adic completion. Then maximal ideal Proof:

m.

R/m =.

R

is a local ring with unique

RI';', a field, so tIrat ,;,

is a maximal i?eal in A. By

COMPLETIONS

147

_ (iv) Proposition 3,';' is contained in the Jacobson radical of Rand hence it is the unique maximal of

R.

Theorem 1: (Krull's Intersection Theorem). Let M be a finitely generated R-module over a Noetherian ring R and I an ideal in R,

M the I·adic completion of

M. The natural map M -+ M has Kernel consisting of those elements of M which are annihilated by some element of 1 + I.

Clearly the Kernel of the map M -+

Proof:

M is ii I"M. o

If (1+a)x

=.O;,'Cor some aEI, then x=-ax=a1x=-a1x= ... E/"M for all fl. Conversely let N = ,

n1

8M.

By Artin·Rees Lemma, there

0

exists some n such that (I"+JtM)nN = J1'(I"MnN), k ~ O. This implies for k = I, Nc IN and hence N = IN. SinceR is Noetherian N is finitely generated and by Proposition 2, 2.2 there exists some a E I with (1 + a) N = 0 and the proof is complete. .Example:" Krull's theorem may be false when R is not Noetherian. Let R be the ring of Coo-functions on the real line and I the ideal .of functions vanishing at 0, the origin. Then, Ker R -+ A consists of functions whose derivatives vanish at 0 but elements annihilated by 1 + I consist of functions vanishing identically in a neighbourhood ofO. Corollary 1: Let R be a Noetherian domain, I =f. R an ideal in R. oo Then n I" = (0). o

Proof: No non-zero element of R can be annihilated by an element of 1 + I. CwoJIary 2: Let M be a finitely generated module over a Noetherian ring Rand IcJ(R). ThennI"M= (0), l.e, the I-adic topology on M is Hausdorff.

o

148

COMMUTATIVB ALGEBRA

Proof: Every element of 1 + I is a unit. Coronary 3:

Let M be a finitely generated module over a Noethe-

rian local ring R with unique maximal ideal m. Then

nmnM

=

O.

i.e. the m-adic topology on M is Hausdorff. In pa~ticular the m-adic topology on R is Hausdorff. Proof: Follows from Corollary 2, as m = J (R).

6.3. EXERCISES

1. Let R be Noetherian ring. I an ideal in Rand M a finitely

.

generated R-module.

nI'M= 1

Show that

n

m:Jl

Ker(M-+oMm ) .

Let R be a Noetherian ring. M a finitely generated R-module. I and J ideals in R. Show that the J-adic completion of the I-adic completion of M is isomorphic to the (I + J)-adic completion of M. Let m" tn ...... m, be distinct maximal ideals of a N~etheria~

2.

3. .

ring R and 1=

nm,. IT R is the I-adic completion of Rand R""" 1

is the completion of R mi under the m,Rmi-adic topology. show that

R""< ~ R~" 1

4.

Let I be an ideal in R. Show that if R is complete for the I-adic topology. then R [[X]J is complete for the (I. X)-adic topology. 5 ~ Let R be a Noetherian ring. I an ideal in Rand S = 1 + I.

6.

Show that R s can be identified with a subring of it Let R = C[X. Y]. C the complex number field and f(X. Y) = X(X'

+ YO) + (X'I- YI).

Show that the ideal (f) is a prime ideal in R.

Let S = R/(f)

and m = (X. Y)/(f) the maximal ideal of S. Show that S the m-adic completion of S is not a domain. (Hint: In th~ ring C [[X. YJJ. f decomposes into a product of two power series).

CONPLBTlONS

6.4.

149

Associated graded rings

Let R be a ring. M an R-module and I an ideal in R. Consider the I-adic filtrations on M and R and the corresponding associated graded module sr, (M) over 8rl (R). We consider in this section relations between properties of the 8rl (R)-module grl (M).and the R-module M. Proposition 1: Let R be a Noetherian ring. M a finitely generated - R-module. and I an ideal in R. Let {M,} be an I-stable filtration on M. Then (i) grl (R) == E9 :E /'//0+1 is Noetherian. ,<;10

(ii) grl(R) "'" g'1 (k) as graded rings. I (iii) grl (M) is a finitely generated grl (R)-module .

Proof: (i) Let I be generated by {ab a an}and let ii, (1 ~ i :0:;; n) be the image of a, in 1/1·. Then [ii t • ii ii.} generate g'l (R) as an algebra over R/I. Since RII is Noetherian. grl (R) is Noetherian. (i}) By Proposition 3.6.2, I'/lnH """ I'llnH • n ~ O. under a natural isomorphism which can be extended to an isomorphism of

(.k) as graded rings. (iii) Since {M.} is an I-stable filtration. there exists an m such that 1M, '"= M n+1 for all n ;;;. m. i.e. IrM m = Mm+r> for a.llr > 1.

gr, (R) onto 8'7

Hence 8', (M) is generated by

i

n=O

M.IM.i-J. over 8I',(R).

Since

M./M.+1is Noetherian and is annihilated by I for all n it is Noetherian as an Rl1-module and hence finitely generated over RII.

Hence Et> ~ M,IM.+l is finitely generated OVer RfI. This implies a

that sr, (M) is finitely generated over 8rl (R). <:IW'

Proposition 2: Let M and N be ftltered R·modules with I-filtrations and t/>: M -+ N a homomorphism of filtered modules. Let

gr(t/»: gr (M)-+gr (N) and $:1:1 -+oN be. the homomorphisms induced respectively on the associated graded modules and tho completions. Then sr (t/» is injective (surjective) implies that ~ is injective (surjective).

150

COMMUTATIVE ALGBBRA

Proof: Since r/>: M -+ N is a homomorphism of filtered modules, r/>(Mn) c Nn (n;e; 0). Now r/> induces natural R-linear maps r/>n:MIMn--l>-NIN. and In:MnIM.+l-+N.INn+l (n~O) making the following diagram commutative 0-+ M n -+ M -+ M -+ 0 u, . M n+l M n+l I

.N.If,

o -+ - -

-+

. Nn+l

1··+1 1·· N Nn+l

-+

N -+ 0 Nn

By Proposition 6, 2.2 we have an exact sequence

o -+

Ker In -+ Ker r/>n+t -+ Ker r/>n -+ Coker In -+ Coker

rPn+l

--I>- Coker

if>n

--I>-

O.

Assume now that gr (if» is injective (surjective). Then Ker In = 0 (Coker fn = 0). Hence Ker if>n

= 0 (Coker if>n = 0) implies Ker if>-+1 = 0 (Coker rPn+1 = 0). Hence by induction, if>n is injective (surjective) for all n. This implies from Proposition 3, 6.2 that ~ is injective (surjective). Theorem 1: Let R be a ring, I an ideal in R, M a filtered R-module with I-filtration {M n}. Assume that R is complete in the I-adic topology and

n

M.

1

= (0).

If gr/(M) is a finitely gene-

rated gr/(R) module, then M is a finitely generated R-module. Proof: Choose a finite generating set of gr,(M) over gr,(R) consisting of homogeneous elements{Yl ... , y,} where deg y, = n, (1 ~ j.:s;;; t) and choose x, E Mnl with y, = X, + Mnl+J.' Let F = ~ ED ... R, t times and F. = {(al) I 01 E r-«, 1 ~ j~ r}, where P = R if k ~ O. This defines a filtration on F and the map if>: F -+ M, given by ." [(a,)] = 1:a,x, is a homomorphism of filtered R-modules. 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

COMPLETIONS

151

Since R is complete and F is a free R-module of finite rank. I is an isomorphism. Since

n o

M.

= (0). g is injective.

.This implies that

~ is surjective as ~ is surjective. Hence M is finitely generated overR.

Corollary 1: With the same assumptions as above, gr, (M) is a Noetherian gr, (R)-module,implies.M is II:. Noetherian R-module•.

Pr~f: Let Nbc a submodule of M and N.=NnM•. Then {N.} is an I-filtration on N and the inclusion N. -+ M. induces an N•

injection H

11+1

-+ MM. n+l

'";> 0) and hence an injection

gri (N) -+

Since 81'1 (M) is a Noetherian grc(R)-module, grl (N) is finitely generated over gr, (R) and hence by Theorem ,1, N is finitely generated over R. This implies that M is a Noetherian R-module.

grr{M).

Corollary 2: Let R be a Noetherian ring, I an ideal in R. Then the I-adic completion R is also Noetherian. Proof: Since R is a Noetherian ring, gr,(R) is also Noetherian by Proposition 1. Since grr (R) "'" grI(R), it follows that srt (R)

n

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. . 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.

152

COMMUTATIVB ALGEBRA

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.

Hence there exists m, n such that

1

aElm, arf:lm+l, bE I·, brf:l·+l. Letli=a+lm+tEgrl(R) and Z; = b + /0+1 E sr, (R). Then Ii =I- 0, Ii =I- 0, and hence ab-= (l·b:t= O. This implies ab-=I- 0, i.e. R is a domain.

Theorem 2: (Hensel's Lemma).. Let R be a local rlng with maximal ideal m such that R is complete with respect to the m-adic 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 {g.(Xn and {h.(X)}

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

= «(X), ]i.(X) =

~(X)

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

and Let n

=

I.

Since 1(X) = lX(X) ~(X), there exist gt(X), ht(X) E R[X]

such that ~(X)

= «(X), Jit(X) = ~(X),

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

and

deg gt(X)

=

r, deg ht(X)

=d-

r.

COMPLBTIONS

153

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

d

}; 1.,X',)., Em·. o

Since «(X), ~(X) ~re relatively prime, there exist polynomials iIi,(X), ~,(X) E k[X]

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

= q;,(X) lX(X) + fJ<.X)

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

_

= ~(X)

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

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

are or degre~;utmost d-r and r resPectively, we have XI - ",,(X) g.(X) - ojI,(X) h.(X) Em R[X'J. gn+l(X)

Let

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

=[ -

h.+1 =[ - (g.

Ji. = ~.

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

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

_ h. };"Iq., I

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

mn+ t R(X].

Since gn+l _ g. E m· R(X], the coefficients of {g.(X)} is a Ca;Ch! sequence and since R is complete, {g.(X)} ~ g(X) E R[Xj. Im~. larIy {h.(X)} ... heX) E R[X] '. Since [~~.h.• e m·R[X] and R IS complete, we have, by passmg to the limit [(X) = g(x) heX), i(X) = «(X) and 1i(x) =~(X).

154

COMMUTATIVll ALGEBRA

The highest degree coefficients of g(X) and heX) are respectively of the type (1 + a)Xr and (1 + a)"::l Xd-r, a E m, as (X(X) and f(X) are monic, By replacing g and h respectively by (l + a)-lg and (1 + a)h, we can make both g(X) and heX) monic and the proof is complete. Corollary 1: Let R be a local ring with maximal ideal m, complete with respect to the m-adic topology. LetJtX) E R[X] be a monic polynomial such that/(X) E R/m[X] has a simple root ~ E Rlm, Thenf(X) has a simple root a E R such that Ii = ~. Proof: Let l[xJ = (X - ~) ~(X), where X - ~ and ~(X') E R/m[X] .are relatively prime. Then there exists a monic polynomial X - a € R[XJ with Ii = ~, which divides/eX). Hence a is a root of I(X) and it has to be a simple root ofJtX') as ~ is a simple root of leX). Corollary 2: (Implicit Function Theorem). Let R= k [[Xl' ... , X,]], the power series ring in Xl' XI" .. , X, over a field k, Let

/(Z) =

Z"

+ aleX) zo-l + ... + a.(X) a,eX') E R such that Z" + a1(0) ZO-l+.,o,

be a monic polynomial, +a.(O) admits a simple root ~ E k. Then g(X) E R with g(0) = ~ and/(g(X» = O.

there exists some

Proof: Apply Corollary I to the local riog R [XI>' , ., X,]] which is m adically complete for m = (Xl' X••. . . , X,). Corollary 3: Let R = k [[Xl' XI•...') X,]] the power series ring in Xl> XI" .. , X, over a field k and d ao integer relatively prime to ch. k, If a(X) E R has constant term 0(0) 0 and is 'a doth power in k, then a(X) itself is a doth power In R.

*

Proof: Apply Corollary 2 to the polynomial/(z)

= zd -

a(X).

6.4. EXERCISES I.

Let R be a ring, I a finitely generated ideal of R, such that R/I is Noetherian and R is complete for the I-adie topology. Show th~t R is Noetherian.

COMPLBTIONS

lS~

LetR be a NoetheriaD semi-local ring, I. its nil.radieal aDd.J the Jacobsoo radical. Show that ~f R With J-adlc topology IS such that R!I is complete, then R IS complete. L t R be a Noetherian ring, I an ideal of R, J(R). the Jacoblo~ 3. r:diealof R. Show that I c J(R) if and only If every ma.xlmal ideal of R is closed for the I-adie topology. (Such a nDg • • . is called a Zariski ring). Let R be a ring with I-adie completion R. .Sh.o~ that R IS 4. faithfully flat over R if and only if ~ is a Zariski ring. Let R be a Zariaki ring with I-adlc topology, I~ J(R). L~t S. M be a finitely generated,R-module. Show ~at If the l-adlc

2.

completion it is a free A-module. then M is a free R:mo~ul~. Deduce that if A is a domain, J an ideal of R with JR pnncipal, then J is principal.

CHAPTER VII

HOMOLOGY

In the first chapter we observed that an exact sequence of R-modules may. fail to be exact on the left when tensoredwith another R-module. Evaluating the Kernel of the tensored sequence at the first as well as successive stages' leads to the notion of the Tor functor. 'Similarly, an exact sequence of R-modules may fail to be exact on the right after applying the Hom functor and tho evaluation of the Cokernel at the first as well as successive stages leads to the notion of the Ext.functor. It is possible to develop a unified theory of derived functors and deduce the Tor and Ext functor as their special cases. We outline this development and derive some elementary properties of the Tor and Ext functors needed in the subsequent chapters.

7.1. Complexes Let R be a commutative ring with 1. A (positive) complex X of R-modules is a sequence {Xn}n.O of R-modules and a sequence {d.}••o of R-linear maps. dn: X n -+- X n- 1 such that d.dn+l = 0 (n:> 0). We denote such a complex by the symbol

n.

d1

X: ... -+- X. --+- 'Xn-l -+- •.• Xl ---. Xo -+. O. Let X = (Xn• d.).~o and X' .;; (X'•• d'.)•• 0 be two complexes. A homomorphism F: X -+ X' of complexes is a collection {Fn}•• o of R-Hnear maps, F.: X. -+ X'. such that the diagram ' . F'+1 vJ X.+l ___ .4.+1

d·+,l

x.

is commutative for all n.

F.

1~~+1 ,

---+ .X.

~"

HOMOLOGY

157

Let x be a complex of R-modules. TheR-submodules Z.(X) = Ker d. and B.(X) = Imdn+l are called respectively the il-cycles and n-boundaries of X. Since d.dn+l = 0, B.(X) c Z.(X) (n ;>0) and the n-th homology module H.(X) is defined to be ~.

JiJ..Jl)' PropositioD 1: Let F:X--+ X' be a mapping of complexes. Then Finduces an R-linear map H.(X) -+B.(X') for all n. Moreover (i) fd· = Id and '(ii) If G: X' -+ X" is another mapping of complexes, (GF):=G:F: for all n.

r::

--

Proof: Since F isa ma~ping of complexes. we have the commutative diagram: '

Since d~+l F.+ , = F.dn+I' we have F.+1(Z.+I(X}) c Zn+I(X'} and F. (B.(X» c B.(X') for all n. Hence Fn defines a map

* Z.(A:') Z.(X') P• : B.(X) -+- B.(X') given by F: (IX +B.(X)} = F.(IX} + B.(X'). IX E Z.(X). Then well defined and satisfies (i) and (ii],

F:

is

Definition: Two maps F, G: X -+- X' are said to be homotopic f there exists a collection {s.}.~o of R-linear maps, . Sn : X. -+ x,;+1 such that (+1 S. + Sn-I d. = F.-G. for.all n. F :;,., G denotes that mapping of complexes F and G are homotopic underthe homotopy s. Proposition :z: Let F. G: X -+ X' be mapping of complexes such that F"", G. Then~. = for all n.

G:

158

COMMUTATIVE ALGIlBRA

Proof:

Let 09

=

{s.}...o be a homotopy between F and G. Then F.II - G.. If IX E Z.(X), then d.IX = 0, so that

09 + 09n-l d.n = d'.11+1"

a.+l S.(IX) B.(X'). Hence ~ (IX + B.(X» = G: (IX + B.{X» for all n. F.(IX)-G.(IX)

=

E

Definition: Two complexes X and X' are said to be homotopically equivalent if there exist mappings F:.X -+ X' and G:X' -+ X such that GF 0< l x and FG 0< lx«.

Coroll~ry:

If F: X -+ X' is a homotopy equivalence, F:: H.{X) -+ H.{X')

is an isomorphism for all n. Proof:

Inverse of

JiZ is G::Hn{X') _

H.(X).

Definition: A complex Xis said to be acyclic if Hn{X) = 0 for all n. Definition: A projective resolution of an R-module M is a comp!ex

X together 'with anR-linear map X o ~_ M such that the sequence

... - x. -.'!.__ Xn-l -+ ... -+ Xo -~ M -+ 0 is exact and all the X, are R-projective. If X is a projective resolution of M we denote it by the symbol X ~_ M -+ O. Clearly Hn{X) = 0, n > 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

•

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

r:

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

HOMOLOGY

159

Proof: Given an R-module M, choose a free R-module Xo and a surjection s: X o ..... M with Kernel K o giving rise to an' exact .

/

go

fo

sequence 0 -+ Ko --+ Xo --+ M -+ O. Now choose a free R-module Xl together with a surjection [,: Xl -+ Ko with Kernel Kl • Continuing process, we have a collection of exact sequences g. f. o -+ K.--+ X. --jo- K n- l -+ 0 {n;;" 0kro = • and K-l = M. If d.:X. -+ 1"'-1 is the map d. = g'-l/.. then the complex ~

~.

... -+ X. - - X'- l - ... Xl --+ Xo - _ M -+ 0 is a projective resolution of M. The following Theorem shows that if X and X' are two projective resolutions of M they are homotopically equivalent.

. Theorem 1: Let - I: M-+ M' be a homomorphism of R-modules, & I' , X - - M and X' - - M' projective resolutions of M and M' respectively. Then there exists a mapping of complexes F:X _ X' such that Ie = e'Fo' Moreover F is unique up to homotopy.

Proof:

We first prove the existence of F. X.

•.....

-

- , f· "

M

-+

Consider the diagram 0

1\

"'\.

x: -;?

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. -.-,

X~+!

"

__

cr.+!

l

,

'\.~ x:

_-+

• cr.

~

X;;-l

160

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

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

a1

£

~

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

---->-

X.

---+

X'.

/\'._0. t

t,/ X'

.+2

--+ d'"H

X'

.+1

d',,+1

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•

HOMOLOGY

161

Since X.+ 1 ts projective, there exists an R-Hnear map S.+l:Xn+l -+ X~H such that d~+J8.+1

= F.+1

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

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

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 X". -+ 0

tla"•

x:.:"l

-+

0

with exact rows. l'his induces a commutative diagram

162

CoMMUTATiVB ALGBBRA

X'.

B·F)

X.

-+

Xli..

B.(X')

!d"n

Id

d'.

0\

-+

B.~~')

-+

n

~

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

1""

H,,(X') .... H,,(X) -.. H,,(X")

i.e.

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

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

> O.

Remark: The exact homology sequence derived above satisfies the naturality condition (see Exercise 4, 7.1). Let 0-.. M' .... M....!..... M" .....!-7 0 be an exact sequ:rIteorem 3: . ." I Let X· •.... M' -+ 0 and X· --+ M" .... 0 be d ence of R -mo uies, . I Th th re

. projective resolutions of M' and M" respecttve y. . . .reso Iution ' X exists a projective sequence of complexes 0 .... the diagram

0

....

X' 0 e'

I

,j..

0 -+ M' Is commutative.

F x, ~

F

o --7

f

--7

en

e

• M"" Oof Mand an exact --7

Xo

£1 M

X

--!!-7

Go --7 X"0 a" g --+

X" -+ 0 such that

-+

0 (*)

I

1f"

-+

0

HOMOLOGY

163

Proof: Let X" =" ~ EB ~'. F,,:~ -+ X"' the natural inclusion x' -+ (x', 0) and G,,: X,,"" the natural projection (X', x") -+x". Clearly the sequence 0 -+ -+ X" -+ -+ 0 is exact and X" is . projective for all n. To define e : Xo -+ M and d,,: X" -+ X"-l satisfying the required conditions, we first construct a family of R-Hnear maps {«"}""I, «,,: -+ X~_I', «0: x'a' -+ M satisfying suitable conditions. We then define

x:.'

x:.

x,;

x,;

- respectively by

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

e("'o, x'J=fr.' x'o+Ofcri'., d"(x',,, x'")=(d'"x'"+IIl,,x",,, d""J(',,). Thoco~dltion.. 4-~d" =0, 11 ;> 1 and e d" = 0 'i~pose the following

. ;' C9nditiOlil '. (I)

-

.(ii) -

Oil

tho GI".

.

£'_1«. = GIn-Ia.. 11 > 2, f

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

X' o

-fe' -~

Since sOd"l =glllod''t = 0, [m(< [m(-fa') and -.the projectivity of .1"'t gives an R-linear map IXI: X/' -+ X'o with - fs'ar,,= otod"l' Using the same type of argument, {otn}1IdlI can be dCftnod inductively satisfying conditions [i), Hence e and d. (11

;> 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.

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

J

X'0

0 -+ c'

c

~

M'

0 -+

Xo

-+

I -+

~

M

X"0

-+

I

-+ 0

E" r

...~

-+

M',' -+ O r

is commutative.

Remark: The above Theorem shows the existence of such a projective resolution; 7.1. 1. 2.

3.

4.

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'

fl.' 0 -+

F --+

I

~ Y'

G

X --->-

--+

'"

X" -+ 0

I fl."

fl.1

t

Y --+

i-

Y"

-+ 0

oIJ

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

,

165

HOMOLOGY

F*

•.• -+

G*

0

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

1

~«

II

1'.

."I IX,.".

I •"

«'·

tX.._ 1

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

"':

oIJ:

(J'~

"\

. -+ '"

(Y')-+ n-)

...

is commutative. 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» satisfYing (i)

T(ld) =

(ii) T(gf)

u,

= T(g)T(f)

(iii) T(I + g)

(T(gf)

= T(f) T(g»

= T(f) + T(g).

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

The functor

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

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'

.... -+

X

we have the complex T(X): ...

-+

T(Xn)

n

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

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

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

;> 0)

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

t.e.

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' •

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

U

Theorem 1:

Let T be a covariant additive functor.

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

Then

HOMOLOGY

167

(ii) LoT=- Tif T is right exact. (iii) For any exact sequence of R-modules,

0_ M' ~ M .

... _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

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

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

T(X)

~

T(X") -+ 0

This induces an exact homology sequence 8•

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

168

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

Applying T on the dl --c>-

T(d J)

Jf ----+ 0

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.

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

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» i.e.

(n ~ 0)

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

~

0)

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

~

0, of R-modules there is an

exact sequence

1 R"T(g)

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

R"T(/) --c>-

which satisfies the naturality condition.

/

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

HOMOLOGY

169

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

then TorR (M, N) = Ho(X ® N). •

.

IfI: M

-+

M' is an R-Hnear map,

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 F=F®N:X®N .... X'®N. R

R

R

The induced map on the homology

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

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

on the homology

G:: H" (X® N) ..... H" (X®N'), i.e. R

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

f

Proposition 3: Let 0 -+ M'

--+

+

g

M --+ M" <jI

4

0

.

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

Tor. (f, N)

I

. R

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

.R

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

and

Tor. (M.

R

+)

a.

If

TOr;.'-l(M', N)-+... Il

... -+Tor;;(M,N') - - - - . - + Torn (M, N) Tor. (M. <\I)

R

"a.

II

I

------+ Tor. (M, N ) _ Tor._ 1 (M, N) which have the naturality property.

-~ ...

Proof: The first exact sequence is a consequence of Theorem 1 7.2, as Tor: (M, N) = LnT (M), for the functor T = ® N. R . To establish the second exact sequence, consider a projective resolution X of M and the exact sequence of complexes 0-+ X® N'

lz®+)o X®N I.. ®<\I) X®N" -+0.

R

R

B

induced by sequence 0 -+ N' -+_ N --! N" -+ O. By Theorem 2,7.1, there is-a natural exact homology sequence

... -+ H.(X®N')_ Hn(X®N) _ Ho(X®N")..!~ H._1(X®N')_ ... R

R

R

B

i.e:.-_ To~(M,N') -+ To~(M,N) -+ Tor=(M,N") -+ TOr:_l (M, N') -+ ... is exact.

We now consider the right derived functors of the Hom functor. Let N be a fixed R-module. Consider the functor T defined by M - T(M) = HomR (M, N) and for an R-linear map !: M -+ M', T(I) is given by TU) =Homll (f, IN) : Homll (M', N) - HomB (M, N).

Since T is a contravariant additive functor, ROT(n .;) 0) are defined and we denote RoT by Extii ( ,N). From the definition of RoT

HOMOLOGY

171

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

-+ ...

then Ext'R(M, N) = H; (HomR (X, N». If!: M -+ M ' is an Rlinear map, X, X' projective resolutions of M and M' respectively, ! can be lifted to a mapping of complexes F: X -+ X' which induces a map F: Homs. (X', N) -+ HomB (X, N) which in turn defines a map on homology i:: Ext'R(M', 'AT) _ Ext; (M, N). T~is map is denoted by Extii(f, N). The correspondence! -+ Ext; (f, N) satisfies conditions (i), (ii) and (iii) of a contravariant functor. Clearly Ext; (M, N) = 0 for n ;) 1 if M is projective and Ext~ (M. N) "" HomB(M, N) for all M and N (Theorem Z; 7.2). We will now show that the Ext functor has the property, that for any R-linear map"': N -+ N', there is an induced R-Iinear map Ext~(M,.p):

Ext;(M, N) _

Ext~(M,

N').

We have only to take a projective resolution X of M and consider the mapping F induced by .p on the complexes HomR (X, N)

-!:- HomB (X, N ')

which in turn induces a mapping on the homology

1 ..\ t

Ext;(M,
The correspondence '" -+ Ext; (M,.p) satisfies the conditions (i), (ii) and (iii) of a covariant functor• The following Proposition on Ext can be proved on lines similar to Proposition 3. Proposition 4: Let

o -+M' .L; M ....!_ M" _

and

O-+N' _ + N ._

<jI

0

N"_O

be exact sequences orR-modules. Then there exist exact sequences

172

COMMUTATIVB ALGBBRA

.,. -+

n (M" N) EXI'(g, N) 0 E xt R ' ----..ExtR(M,N)

Ext' (f, N)

8.

----+ Ext; (M', N) ~ Ext;+l (M", N) an

d

•..

~

E

0 (

xt R M,

-+ •••

N' Ext'(M• .,,) n ) ---~ ExtR(M, N)

Ext' (M.
----+ Ext~ (M, N") _

8.

Ext;+l (M, N') -+ •.•

which have the naturality property.

7.2. EXERCISES 1. 2. 3. 4. 5.

Show that functor given by T(M) ~ M ® M and T(f) = j®j is covariant but not additive. R Show that an additive functor carries homotopically equivalent complexes to homotopicalIy equivalent complexes. Show that LoT and RoT (n ~ 0) ate additive functors. Show that for any covariant functor T, the functor LoT is right exact. . Prove the naturality condition for long exact sequences viz., given a commutative diagram

I

--+

HOMOLOGY

173

As*: Ext; (M, N) -+ Ext; (M, N)

8.

which is also scalar multiplication by a. State and prove a corresponding result for the scalar multiplication on' the second variable and also for the Tor functor. Show that an additive functor carries a split exact sequence to a split exact sequence.

7.3. Homological dimension We give some applications of the Ext and Tor functor in the study of homological dimension of rings and modules. Homological dimension is closely related to the dimension of rings and modules to be studied in the next chapter.. Let M be an R·module. Theprojective dimension of Mover R, denoted as pdRM is defined as follows. Definition:

pdRM = n if X has a projective resolution of 'length n Xj

i.e,

=

0, i t» n, X o

*' 0,

and n is the least with this property. Otherwise pdRM is defined to be 00. Examples: (i) ]f M is R-projective, the sequence

o -+

/

Xu --+ M -+

1>, Xu = M,

is a projective resolution of M of length' O. Hence pdRM = O. The converse is also true, i.e. if ,0 ~ Xu -+ M -+ 0 is a projective resolution of' M of length 0, then M ~ Xo is projective. (ii) Let R = Z ..Since R is a PID, every projective R-module, being a submodule of a free module is free. Let M be an abelian group which has non-zero torsion. Express M as the quotient of a free abelian group X o with Kernel Xl' The sequence

o -+

i

•

Xl --+ Xo --+ M -+ 0

is exact, with X o, Xl free so that pdRM..o;; I. If pdRM = 0, then Hence pdRM = 1 if M has torsion elements. (iii) Let R ~ Z, and M = 2Z,. Consider the map

M is projective and hence free over Z.

p:

Z,""

Z,

174

COMMUTATIVE ALGBBRA

given by p(lX) = 2a., IX E Z.. gives an exact sequence -

Clearly Im p = 22'. = Ker p.

i

This

p

-~ Z. ---->- 2Z. -+ 0

o -.. 2Z. Consider the complex A

A

A

A

p_

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

where A = tp, Clearly it is a projective resolution of M =!Z over R = Z. of infinite length. It can be shown that pdRM = 00.' The global dimension of a ring R denoted by gr. dim R is defined as follows. Definition:

g/.dim R

= Sup pdRM,

where the supremum is taken

M

over all R-modules M. (i) If R is a field, pdRM =0, for all M, as. every R-module is free. Hence gl. dim R = O. (ii) If R = Z, we have seen.in Example (ii) above thatpdRM :>;; 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.

EXlUIlpJes:

(iii)

If R = Z. and M

= 2z., pdRM =

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

=

00

and gr. dim Z.

=

00.

Then Mis R-projective

0 for all R-modules N.

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

all N.

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) 0-+ Hom.(M, N')

L

Hom.(M, N)

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

HOMOL~Y

175

-+ 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.

The following conditions

(i) pdRM:>;; n (ii)

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:>;;

/I

~ 1). then Kil-l is projective.

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 io

•

i

d

O_Ko --+ X o --+ M_O 1 1 O-+Kl --+ X, --+ Ko _ 0

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

d._

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

where

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

176

COMMUTATIVE ALGEBRA

"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

we have we have

o -+ KI -+ Xl -+ 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 <: n - 1) projective. By assumption (iv), K"- l is projective so that we have a projective resolution of M of length n. Hence pdRM .r;; n, Corollary 1: pdR M .r;; n if and only if ExtR+1 (M, N) = 0, for all R-modules N. Corollary 2: gl idim R.r;; n if and only if ExtR*1 (M, N)=O for all Mand N. Proof: gl.dim R.r;; n, if and only if pdRM <: n, for all M, i.e. Ext.;+l (M, N) = 0, for all M and for all N. . The Ext functor, can also be developed by using injective

."','"

,,';\.'-";,

HOMOLOGY

177

resolutions instead of projective resolutions. The notion of an inj~ve module is dual to the notion of a projective module.

DeflDition: An R-module N is injective if and only if for any exact sequence 0 -+ M' sequence

~

M

of R-modules the

induced

Hom. (f, N) M' N) O' t BomR (M, N) - - - - - BomR ( , -+ IS exac

i.e, any R-Iinear map M' -+ N can be extended to an R-linear mapM-+N. Examples: (i) If R is field, any R-module N is injective as any R-Iinear map of a subspace into N can be extended to the whole space. (it) Z is not Z-injective as the homomorphism 0 -+ Z .--.!....... Q, induces f*: Homz (Q, Z) -+ Homz (Z, Z). If I" is surjective,Iz has a preimage under f*, t.e. Z is a direct summand of Q, a contradiction. We will show that an abelian group G is an injective Z» module if and only if it is divisible, i.e. for every x E G, nEZ, n "# 0, there exists some Y E G with ny = x, More generally,. we havo tho following._

.Theorem 2: An R-module N is injective if and only if for any. - ideal Iof R, anyR-homomorphismf:I -+ N. can be extended to an R~homomorphism f*: R -+ N. Proof: Clearly if N is injective, any homomorphismf: 1-+ N can be extended to f*:R -+ N. Conversely assume this condition for all ideals I of R and consider a submodule M' of M and an Rhomomorphismf: M' ~ N. We construct a maximal extension' for f and show that its domain is the whole of M. Consider :E = {(M", IJ} where M" is a submodule of M containing M' and f. is an extension of f. Define(M",f~) <; (M/I,f/l) if and only if M .. is a submodule of M~ and I" =h I M«. }; has a maximal element (M*,f*) by Zorn's lemma. We claim thatM=M*. If not, chooso Z E M, z If; M* and consider I = {a E R I az EM"'}. I is an ideal of R and the map h: 1--+ N, given by heal =/* (az) is R-linear. By assumption, h has an extension h*: R -+ N. Extend

178

COMMUTATIVE ALGEBRA

now f" to fi :M" + Rz = M,: -+ N, by defining f,eX

+ bz) = I" (x) + h"(b),

x E M, bE R.

Clearly J. is a well defined R-Iinear map and (M",f") < (MI,fl)' M" eM,. This contradicts the maximality of (M",f").

=l= Corollary: An abelian group G is an injective Z-module if and only if it is divisible. Proof: Let G be Z.iojective, x E G, and nEZ, n oF O. Consider 1= nZ and f: 1-+ G, defined by f(An) = Ax, AE Z. Thenf has an extensionj*; Z -+ G. If y = f(I), then ny = j*(n) = 1(1i) = x. . Thus G is divisible. Conversely assume that G is divisible. Let I = nZ be an ideal of Z and j": I -+ G any Z-homomorphism. If/(n) = x, there exists some y E G with ny = x. Define/": Z -+ G, by f0) = Ay, A E Z. Then I"(n) = nj*(I) = ny <= x = I(n) so that f* is an extension off Hence G is Z-injective. PropclsitioD 2: R-module.

Any R-module M can be embedded in an injective

Proof: Consider first the case where R = Z and M a free abelian group. Now Z c Q the additive group of rationals. Q is Zinjective as it is divisible. Hence M, being a direct sum of copies of Z can be imbedded in a direct sum of copies of Q, which is divi~ible. i.e. injective. Now any abelian group M is isomorphic to F/K, with Ffree. Choose an injectiveZ-module N with FeN so that M = F/K c NIX which i~ injective as the quotient of a divisible group is divisible. . We now consider the general case. Let M be an R-modulo and consider M" = Homz (M, Q/Z). Then M" is an R-module for the operation (af) (x) = f(ax), a E R, x E M,f EM". The natural map t«: M -+ M"", given by iM(X) (f) = I(x), x E M, IE M" is R-Iinear. We claim that iM is a (1,1) map. If x#O, there exists a Z-homomorphism h : Rx-+ Q/Z, with hex) i= 0, by assigning any' non-zero value to x if the order of x is infinite or the value I/n Z, if the order of x is n. Since Q/Z is Z-injective, h has an extension h": M -+ Q/Z, an abelian group homomorphism, with h"(x)= h(x) i= 0. Hence

+

HOMOLOGY

iu (x) (h) oF 0,

179

i.e, i M is (I, 1).

Express M" as a quotient of a free R-module F, i.e. a: F -+ M* is surjective R-Iinear map. Then a*: M** -+ F* is a (I, 1) Rhomomorphism and hence M is isomorphic to an R-submodule of F*. We will show that F* is R-injective. More generally we will show that P" is injective for any projective R-module P. Consider a diagram

where g is R-linear. Since Q/Z is Z-injective, i*: M* -+ M'· is Ilirjoctive. Conaiderthe diagram

P

i,

~

p**

Since P is R-projective, there exists an R-Iinear map h: P -+ M* with i*h =g*ip • If 1= h"iM: M -+ P*, then .f is an extension of g and p* is injective. Corollary: Any R-module M has an injective resolution, i.e. an exac1l sequence do

O-+ M -+ Qo - - + QI -+ ... -+ Q. Q, jnjective for all i :> 0.

d.

~

... of R-modules with

Proof: Embed M in an injective module Qo so that we have an exact sequence of R-modules 0-+ M -+ Qo -+ Qo/M -+ O.

Embed now Qo/M in an injective modules QI and continue the process as in Proposition 3, 7. I to yield an injective resolution forM. -

Proposition 3: An R-module N is injective if and only if Exli (M. N) = 0, for all R-modules M.

180

COMMUTATIVE ALGEBRA

Proof: Let N be injective. Express M as the quotient of a II.:.: module F with Kernel K so that we have an exact sequence of R· modules ' j p O-+K~F~

M-+O

By Proposition 4, 7.2 we have an exact sequence 8

~

-+ HomR(F, N)

~ HomR(K. N) ~ Ext1(M,N)

-..!.-- Ext1 (F, N) = 0, asF is free. Since N is injective. j* is surjective and hence a = O. Now Ext1 (M, N) = 0 follows as, Ker p= 1m a= O. Conversely assume that Ext1 (M, N) der an exact sequence .

O-+M'

i

~

=

0, for all M and consi-

M-+M/M'-+O

This gives an exact sequence

,"

-

-+HomR(M.N) ~ HomR(M',N)-+Ext1(M/M'.N)=0. This implies that i* is surjective and hence N is injective.

Proposition 4: Let 0 -+ M' -+ Q -s- M" -+ 0 be an exact sequence of R-modules with Q injective. Then .Ext~+l (M. M') ~ Ext; (M, M") (n;;;. 1) for al\ M.

Proof: We first show by induction on n that if N is R-injective Ext;(M, N) = 0 for all n;;;' 1 and for all M. This is true for n = 1 by Proposition 3. Assume the result for n - 1 (n ~ 2) and consider an exact sequence 0 -+ K -+ F -+ M -+ 0 with F free. This induces an exact sequence

Ext;-l (K, N) -+ Ext; (M, N) -+ Ext; (F, N) = O. By induction Ext;-t (K, N) = 0 and hence Ext; (M ~ N) = O. Now consider the given exact sequence

o -+ M'

-+ Q -+ M" -+ O.

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181

This Uiducesan oxact sequence ~ Ext; (M,Q) -+ Extli (M, M") -s- Ext;+l (M, M') -+ Ext;+l (M, Q) -+

. .

. SiDce 'the terms at the extremes are zero, the middle terms are .isomorphic for n ;> 1. We now define the injective dimension of an R-module M. DefiDition: Let N be an R-module. The injective dimension of N over R. denoted by idRNisnif Nhas an injective resolution of length nand n is the least with this propertY. Otherwise idR(M) = co. Theorem 3: Let N be an R-module. The following conditions are equivalent. (i) IdR(N) E;; n. (ii) Ext;+' (M, N) 1

(iii) Extii+ (M, N)

= 0, i;> =

1, for all M.

0 for all M.

(iv) If 0 -+ N -+ Qo-+ Ql --)- ... -+ Q'-l --)- Til-l -+ 0 is exact with Q, (0 E;; i E;; n - 1) injective, then T"- l is injective. do

d'- 1

Proof: (i) => (ii), Let 0 -+ N -e- Qo-+ ... Q.-l --+ Q. --)- 0 be an injective resolution of N oflength n. It splits into short exact sequences do

o -+ N --)- Qo-+ Ko --)-0 41

o -+ Ko -+ Ql -+ K1 -+ 0 d'- l

0--)-K..-. -+ Q..-l ~ Q.-+ 0 whereK, = 1m dr(O E;; 1 ~ n - 2). Repeated use of Proposition 4 gives ' Ext;+1 {M, N) ~ Extii+1-l(M, Ko) <:>< Ext;+H (M, K1) ... "'" Ext~+\M,K._'> "'" Ext~ (M, Q.) = 0 (I;;;' 1) as Q. is injective. (ii) ~ (iii) is clear.

182

COMMUTATIVe ALGBBRA

(iii) =>(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:

gl. dim R " n ~pdR (M) "n for all M ~ Ext;+l (M, N) ~ IdR (N)

Corollary 3: gl. dim R

= 0 for all M, N,

,.;;;; n for all N.

= Sup pdRRII, where the supremum is 1

taken over all ideals I of R. Proof:

If Sup pdRRII = 1

00,

then gl. dim R= 00. Assume therefore

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

< n for

any R-module N.

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

Consider an exact with Q, injective

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

T.-J

= 0

as pdRRII "n. This implies that T 11--1 is injective by Theorem 2. Hence IdRN" n. . We now study g]. dim R, when R is a local ring and show that the Tor functor can be used to compute the global dimension of local rings.

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183

Proposition 5: Fo~ R-modul~s M and N, there exists an isomorphism Tor: (M, N) c:<.'Tor: (N, M), which is functorial in M andN. 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 [

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

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

d,.

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

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.

P~O

isexaet.

R

We now prove the Proposition by induction on n. For n = 0, the result is true as M®NO!.N®M. R

R

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 O~K~F~M~O.

This gives rise to the following exact sequences

o=

Tor~

8.

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

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

By induction, there exist isomorphisms following diagram commutative

~

and

Tor=_l (K, N) ~ Tor:_ 1 (F, N)

1~

1 m

TO':_l (N, K) ~ Tor:_l (N, F)

IX

F). making the

184 ~

COMMUTATIVE ALGEBRA

induces therefore an isomorphism y: 1m an ~ 1m8n, i.e. y: Tor~ (M, N) ~ Tor~ (N, M) is an isomorphism.

ProIHlsition 6: An R-module M is flat if and only if Tor: (M, N) = 0 for all R-modules N. Proof:

.

Assume Torf (M, N) = 0 for all N and consider an exact •

ojJ

-

sequence 0 ~ N' -e- N ~ N" ~ O. This gives an exact sequence by Proposition 3, 7.2. 0= Torf (M, N") ~ M

~

® N' ~ M ® N ~ M ® N"~ O.

Hence Ker q; = 0, and M is R-llat. Conversely assume that M is R-llat, and let N be any R-module. Consider an exact sequence .,. o ~ K ~ F ~ N ~ 0, with F-free. This gives an exact sequence _ .~ 0=0 Torf(M, F)~ Torf(M, N)~K®N~F®N~M®N~O Since- N is flat, Ker ?> = O. Hence

R

Tor: (M, N) "" Im Torf (M, N)

R

R

= Ker q; = O.

Corollary: Let R be a Noetherian local ring and M a finitely generated R-module. Then M is free if and only if Torf(M, N) = 0 for all N. Proof:

Follows from Corollary to Theorem 2,2.2.

Theorem 4: Let R be a Noetherian local ring with maximal ideal = Rim. If M is a finitely generated R-module. then M is free if and only if Tor: (M, k) == O.

m and k

Proof: If M is free, Torf(M, k) = 0 by Proposition 6. Conversely assume that To': (M. k) = O. Let {Xl' .... xn} be a minimal generating set of M and let u: F ~ M be the surjective R-linear map given by u(e,) = x" where F is free with basis {e;} (1 <; i"';; n). u

If -K = Ker u, we have an exact sequence 0 -lo K ~ F ~ M ~ O. This gives an exact sequence -

HOMOLOGY

185

u O=Torf(M.k)~ K® k~F® k ~M®k~O

R

R

_

R

Since {Xl• .. .. x n} is a minimal generating set for M. a is an isomorphism. Hence KeF Q "" K ® k = O. i.e. K/mK = O. Since R is R

Noetherian, Kis finitelygenerated and by Nakayama Lemma, K=O and M is free. Proposition 7: Let R be a Noetherian local ring with maximal ideal m and k = Rim. Let M be a finitely generated R-module. Then pdRM ".;; n if and only if Tor:+l (M, N) = 0, for all R-modules N.

Proof: If pdRM ~ n, M has a projective resolution of length utmost n which can be used to compute Tor:+ l (M, N). This implies Tor:+l (M. N) = O. Conversely assume that TOr::+1 (M. N) = 0 for all N and consider an exae't sequence O~Kn-l ~Xn-l~ ... ~Xo~M ~O. (*)

with X, (1 E;; t « n - 1) projective. It is sufficient to show that Kn-l is projective. i.e. Tor: (K n- 1 • N) = O. The exact sequence (*) splits into short exact sequences. Using the argument as in Theorem I, on the Tor functor instead or the Ext functor. we have Tor~ (Kn-ll N) Of. Tor:+! (M. N) = O•. Corollary: With the same assumptions as above pdRM".;; n if and only if Tor~l (M. k) = O. Proof: If pdR M ".;; n. then Tor~+1 (M, k) = O. Conversely assume that Tor:+! (M, k) = O. Then Tor~ (Kn- h k) "" Tor:+1 (M. k) = 0 and this implies that Kn-l is projective, by Theorem 4. Hence pdRM"';;n.

Theorem 5: Let R be a Noetherian local ring with maximal ideal m and k = Rim. The following conditions are equivalent, (i) gl. dim R ".;; n (iO Tor:. 1 (M. N) = 0 for all M, N (iii) TO~l(k. k)

= O.

186

COMMUTATIVE ALGEBRA

Proof: (i) => (ii), If gr. dim R ,,:;;; n, pdRM ,,:;;; n for any R-module M and this implies (ii) by Proposition 7. (ii) => (iii) is clear. (iii) => (i). Assume that Tor:+! (k, k) = O. This implies by CorolIary to Proposition 7 that pd R k,,:;;; n. Hence Tor:+ 1 (k, M)=O for all M. By Theorem 2, Tor~l (M, k) 0,: Tor:+ 1 (k, M) = O. Hence by Corollary to Proposition 7, pdR M,,:;;; n, i.e, gl. dim R ~ n, Corollary: Let R be a Noetherian local ring with maximal ideal m and k = RIm. Then gr. dim R = pdR k. Proof: If pdR k = 00, Tor:+ 1 (k, k) =1= O. for each nand gl. dim R = 00. If pdR k < 00, then pd R k ,,:;;; n if and only if Tor:+dk, k)=O . if and only if gl, dim R,,,:;;; n. Remark: If gl.dim R = n, k has a projective resolution of length n. An explicit resolution of k over R is given by the Koszul complex which will be described in the next chapter. As an application of the Tor functor we derive "Local criterion for flatness". Proposition 8: Let f : R -+ S be a homomorphism of rings and Man R-module. The following conditions arc equivalent. (i)

Torf (M, N)

(ii) Tor: (M, S)

=

0 for any S-module N.

= 0 and

M

® S is a fiat S-module. R

*

Proof: (i) (ti), If N = S in (i), we have Torf (M, S) = O. If 0 -+ N' -+ N -+ N" -+ 0 is an exact sequence of S-modules, we have an exact sequence 0= Torf (M, N")-+M®N' -+M®N-+M®N"-+O R

i.e.

R

R

O-+M®N'-+M®N is exact. R

R

Since M ® S ® N =. M ® N, R

S

R

this implies M ® S is S-fiat. R

(ii) => (i) Since Torf (M, S) = 0, Torf (M, F)=O for any free S-module F. Express N as a quotient of a free S-module, so that i

we have an exact sequence 0 -+ K -+ F -+ N -+ O. This induces an exact sequence

HOMOLOGY

187

iO

0= TorR (M, F) -+Torf (M, .N)-+M® K-?M ®F . 1 . B R -?M®N-+O. II

Since M ® S is S·fiat, M ® S ® K -+ M ® S® F is injective, II

R

S

R

S

t.e. i* is injective. This implies ToiR (M, N) =0. Theorem 6: (Local criterion for flatness) Let f: R -+ S be a homomorphism of Noetherian rings, I an ideal of R such that its extension I' under f is contained in the Jacobson radical of S. Let M be a finitely generated S-module. The following conditions are equivalent. (i) M is R-flat. (ii) M

® RjI is RjI-fiat and Torf (M, Rjl) = O. R

(iii) Tor: (M, N) = 0 for alI R-modules N annihilated by I.

(iv) Torf (M, N) = 0 for alI R-modules N annihilated by P for some s » 1. (v) M ® Rjl' is RIl' flat for an s ;:: 1. R

.

Proof: (i) => (ii) folIows from Proposition 2, 1.4 and Proposition 6,7.3. (ii) => (iii). Take S = RjI in Proposition 8. (iii) => (iv).

We prove this by induction on s.

For s = I, it is true by (iii). Assume (iv) for all N annihilated by [k(l ,,:;;; k,,:;;; s - 1) and let N be annlhilated by I'. The exact sequence 0 -+ IN -+ N -+ NjlN -+ 0 induces an exact sequence Torf (M, IN) -+ Torr (M, N) -+ Torf (M, NjlN)

The terms at the extremes arc zero by induction and hence To~ (M, N) = O. (iv) => (v). Take S = RjI' in Proposition 8. (v) => (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

o ~ N' ® M R

~N

® M is exact. R

The given sequence induces an exact sequence

N'

o ~ N'n[kN ~

N

[kN for all k.

Since M ® R/[k is R/P flat for all k, the sequence R

N'

is exact,

R

N

R

o~ N'nPN Rllk ® M®jk ~ PN Rllk .® M®[k R R

N'

N

t.e, 0 ~ N'n[kN~ M ~ [kN~M is exact for all k: By Artin Rees Lemma, there exists an integer m such that [k-m

(*)

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

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

~~kN <§' M

R

R

can be identified with the quotient of N'

Then

~M

by

the image of (N' nJ1'N)® M in N' ® M. Hence, the sequence (0) R

R

becomes the exact sequence

o~ Since {[k-mP}

N'®M Ik-mp

[k-m(N'n[mN)

and

[k (N'

N®M

~ Ik(N®M) (**) (k;;" m)

= N'n[kN for all k » m,

® M) induce the same topology on N' ® M. R

R

Taking ['-adic completions, we 'have the exact sequence

A

A

O~N'®~~N®M.

Since Ie N®M

is exact.

c

J(S), the natural maps N'

~ N~

are injective.

®M

~ N~ and

Hence 0

~ N' ®M ~ N ® M R

R

Corollary: Let I:R -e- S be a local homomorphism of Noetherian local rings Rand S with maximal ideals m and n respectively, i.e: I satisfies I(m) c n, Let M be a finitely generated S-module and t ERa non zero divisor. Then M is R-flat if and only if M/tM is R/(t)-flat and t is a non zero divisor of M.

189

HOMOLOGY

Proof:

Consider R/[ where [ = (r),

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

Consider the exact sequence

where A, is multiplication by

t,

This

gives an exact sequence

A,'

0= Tor~ (N, R)~ Tor: (M, R/1)~ M@R~ M®R

Hence Tor: (M, R/[) = 0 if and only if Ker ~~ = 0, i.e. t is a nonzero divisor of M. The result now follows by the equivalence of (i) and (ii). 7.3. EXERCISES -,

1. 2. 3. 4.

5.

Show that if M is a finitely generated modulo over a Noetherian ring R and if paRM = n. then Ext~ (M, R) oF O. Let 0 ~ M' ~ M ~ Mil ~ 0 be an exact sequence of Rmodules. Show tbat pdRM <: Max {pdRM', pdRM"}. Show that a direct product 'liN,. ,. = N of R-modules is injective if and only if each N,. is injective. Two R-modules M and Narc said to be projectively equivalent if M EB P N Ee Q for some projective R-modulell P and Q. Show that for two such modules pdRM = pdRN. Illustrate by an Example that the converse'Is not true. Let M be a.finitely generated module over a Noetherian ring R. Show that • pdRM = Sup pdR,Mp where the supremum is taken over

=

p

•

all prime ideals P of R and can be restricted to all maximal ideals m of R. Deduce that gl. dim R = Sup gl. dim R» = Sup ,.

6.

R with Jacobson radical J.

7.

If:ft is a projective

R/J module

and TorR (M, R/J) = 0, show that M is R-projective. Letj": Rl~S be a ring homomorphism such that S is R-llat. If M and N are R-modules, show that Tor (M, N)®S "",Tor~ (M ®S, N® S) (n;;" 0) •

8.

m

• • gl. dim R m• Let M be a finitely generated module over a Noetherian rmg

Let R

be

R

•.

R

a local ring with maximal ideal m, Show that Tor: (k, k) Of. mimi.

190 9.

COMMUTATIVE ALGEBRA

Show that R =

k(~Jl

is a local ring with

st. dim R = co.

Let O-+M'-+M-+M"-+O be an exact sequence of Rmodules with M projective and M" not projective. Show that pdRM" = 1 + pdRM'. 11. Let R be a ring, a ERa non-unit and a non zero divisor and R* =R/(a). Show that if M is a non-zero R*-module with pdRoM < 00, then pd.M = pdR*M + 1. 12. Let 0 -+ M' -+ M -+ M" -+ 0 be an exact sequence of Rmodules. Show that 10.

(i)

idRM > idRM"

:>

idRM'

(ii) idRM = idRM" => idllM'

(iii) idllM < idllM" 13.

:>

:=

idRM

<: idRM + 1

idRM' = idRM"

+ 1.

For R-modules M, Nand K, consider the adjoint isomorphism (Exercise 6, 1. 3) r/> : Hom. (M, Hom. (N,

K» = HomR (M ® N, K) R

By replacing M by a projective resolution of M and passing to homology show that Exth (M, HomR (N. K»

when K is R-injective.

= HomR (Torf (M,

N). K) (i;> 0)

CHAPTER VIII

DIMENSION

Dimension of a variety is a basic numerical invariant. Its algebraic analogue leads to the notion of dimension for rings and modules. There arc three different formulations of dimension. We outline all of them and prove their equivalence for Noetherian modules over a local ring. One .of these is developed by using Hilbert Samuel polynomial which is less intuitive but more useful in proving properties about dimension. Geometric invariants like degree of a variety or arithmetic genus can be made precise by USing Hilbert Samuel polynomial. Dimension of affine algebras have special properties and these are studied in detail. Dimensio~ is a coarse invariant as varieties of a given dimension are highly non-isomorphic. We introduce another numerical invariant called depth which is closely related to dimension and investigate the relations between them. Rings and modules for which the two coincide form a special class called Cohen Macaulay rings and modules. We study properties of rings and modules which are Cohen Macaulay. 8.1. Hilbert Samuel polynomial

Let N denote the set of natural numbers and Q the field of rational numbers.

Definition: A polynomial function / is a map /:N -+ Q such that there exists a polynomial g[X] E Q[X] with/en) = g(n), for all n greater than or equal to some fixed 110. Clearly such a polynomial g(X) is unique. Its degree and leading coefficient are called the degree and leading coefficient of the polynomial function f.

192

COMMUTATIVB ALGEBRA

Proposition 1: Let f:N -+ Q be a mapping. Thenfis apolynomial function of degree r if and only if tsf': N -+ Q ~fined by 6.f(n) =f(n + i)-fen) is a polynomial function of degree (r-I). 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»

0.)

t.e. fen + I)-qo(n+ 1) = f(n)-q,p = IX (n > > 0); If g(X) = qoX + ee, /(n) = g(n) (n :> > 0). Let r !> 1 and g(X) = aoX,-l + a1Xl'- 1+ ... + a'-l E Q[XJ, ao # 0 be such that6.f(n) = g(n) (n > > 0). i.e, fen + I)-f(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

~ r- 2.

Q[X] is

6..p is a polynomial

By induction, .p is a polynomial funcnr (ao*O) is polynomial

tion of degree r-l. Hence I(n) = .p(n) + tlo

r

function. of degree r. Remark: The above result is also true for r = 0 if we assign the degree - 1 to the zero polynomial. We now consider a special type of polynomial function. Let R = EB I R. be a graded ring. such that Ro is Artinian .~a

and R is finitely generated as an Ro-algcbra generated by r elements au a l ..... a, in R.. Let M = EB :E M. be a finitely generated

....

graded R-module.

Prollosition Z: With the assumptions 'al above. each M" is af.g. Ro-module. Proof: Since R. is Noetherian and R is a fig. Ro algebra. R is Noetherian. Let N = Ea :E Mill C M. Since M isf.g. over R so m~.

is N. 'Choose a generating set

Xl. XI' .... X,

of N over Rand ,,!rite

DIMENSION

X, =YI + Z"Y, E M"andz,E E9 I

III>.

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

Q,= b, +

C"

b, ERa'

C, E

E9 :E R,. Then y '>0

Colollary: IR.(M,,) <

00

=i

b,y,.

1

for each n.

Theorem 1: Let R be a graded ring with Ro Artinian and R f.g. algebra over Ro generated by r elements in R1• Let M be afig. graded R-module.

The map

X(M, ): N -+ Z, given by X(M, n) = IR.(M.) il a polynomial function of degree E:;r - 1.

Proof: ,.The proofis by induction on r, ]f r = 0, R = R a•

Choose

a . ~O~~Ol1I. generating set of Mover R = Ra• If d is the IDlWmUDl

of their degrees. then M" = 0, n> d. Hence . X(M. n)= O'(n!» 0). i.e, X (M. )

~ ~i. a polynomia1functio~ of degree -, 1. Let r !> 0 and assume the result for all r ~ r. Choose a senerating set a a a' R fR R If ~ . I, I . . . ., ,In 1 o

o..ver.,

>..or • M

-+ M·ll. scalar multiplication defined M...1• ~t C_ and X. be tho corresponding the Klrnel respectively 10 that there is an exact

~:~(M.) c _. 1IflUtII!IICe.

~

.' If X

.

.~

i

0 -+ K" -+ M" -+ M'*J, ~ C,,+!-+ 0

= ED

":0 X. and C

(*)

= E9 }) C•• then X and C are Noetherian . "~a graded R-modulcs. Hence X (X, n) and X(C, n) are defined. The exactness of the sequence (*) shows (Exercise 1. 3.4).

X(X. n) -X(M, n) + X(M. n + 1) - X(C. n + I) t.e.

6. X(M, n) = X(C, n

+ 1) -

=0

JC(X, n).

Since a; annihitates both X and C, X and C are finitely generated graded R'·modules where R' is the graded subring of R generated by a1' al ,..... ar_ l • Over R o• By induction. JC(K. n)' and l(C. n) are polynomial functlOns,ofdegree <;; r-2. t.e. 6. X(M, n) is a polynomial function .of degre~ ~ r-2. Hence by Proposition I, X(M. n) is a polYnomial function of degree ~ r-l,

194

COMMUTATIVB ALGBBRA

Corollary: Let R be a Noetherian local ring with maximal ideal = Rjm, Then

m generated by r elements and k

X (gr m (R). n) = /k (m"/m"+!)

is a polynomial function of degree <; r-1. Proof: Apply Theorem I to the graded ring grm(R) = E9 1: m'/ml+ l ,~O

as a graded module over.itself. Definition: The polynomial defining the polynomial function X(M. n) is called the Hilbert polynomial of M. We now consider a special case of the Hilbert polynomial. Definition: Let R be a Noetherian local ring with maximal ideal

m. An ideal I of R is called an ideal of definition of R if m" c I c m for some n > 1. Clearly I i~ an ideal of definition of R if and only if R/ I is anArtinian ring. If M is a finitely generated R-module. M/IM is a finitely generated R/I-module and has finite length over R/I. Consider the I-adic filtrations defined on Rand M respectively. If we denote the corresponding associated graded ring and module by grr (R) = E9

I

/"/[0+1. grr(M).= E9

I

11>0

I"M [»tIM'

then

n>O

grr(M) is a finitely generated grr(R)-module. If lis generated by ~..... a, over R, their images ii!, ii...... ii, in 1/11 generate grr(R) over R/I. Hence X(grr (M). n) is well defined, where

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

::.tf

M

).

Since Supp (M/I"M) = {m},/R(M/I"M) < 00. If we denote Pr(M. n) = MM/I"M). the exactness of the sequence

I"M

M

M

O~ I"+lM ~ I"+lM ~ I"M ~ 0

DIMBNSION

shows that, I1Pr(M. n) = Pr(M. n

+ l)-Pr(M.n) =

195

X(grr(M). n).

Proposition 3: Let R be a Noetherian local ring. M a finitely generated R-module, I an ideal of definition of R generated by r elements.. Then .Pr(M. n) defined by Pr(M. n) = [R(M/lnM) is a polynomial function of degree ~ r, Proof: By Theorem 1. X (grr(M), n) is a polynomial function of degree ~ r-1. The result follows from ProposltionI. Proposition 4: The degree of the polynomial function Pr(M. n) doe. not depend on the ideal of definition I. . Proof: It is sufficient to. show that deg Pr(M. n) is equal to tUgPJ..M, n). Sineem' c I for some I ~ 1, m'N C In c m" for all n: Hence PJ..M. In) >Pr(M. n) >P..(M. n). This shows that Pr(M. ) and P..(M. ) have the same degree. Definition: The degree ofPr(M. n). for any ideal of definition I is denoted by d(M). . Theorem2: Let R be a Noetherian local ring. I an ideal of definition and 0 ~M' ~ M ~ Mn ~ 0 an exact sequenceof finitely generated R-modules. Then Pr(M". n)

+ Pr(M'. n) =Pr(M. n) + R(n)

where R(n) is a polynomial function of degree less than d(M) and the leading coefficientof R(n) is non-negative. Proof: The exact sequence o~ M' ~ M ~ M" ~ 0 induces for each n. an exact sequence of R-modules o~

M' M'

n InM

Lee M' n I"M = M·". so that

M

~ InM ~

we have

Pr(M. n)-Pr(M". n) = /R

Thus

II!..M'/M~")

is a polYnomialfunction.

(Z:J

(*)

196 COMMtlTATIVB ALGEBRA 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 >

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.

>

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

always. Hence by Theorem 2. we have deg P,(M'. n)";;; degP,(M, n).

Let R be a Noetherian local ring with maximal ideal m generated by r elements. We saw in the Corollary to Theorem 1 that 'X (grm(R). n) = lRI,.

(mn::l )

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

.p(Xi) =

ii,

=

a, + m

l

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

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

Assume that

.p is

an isomorphism and let A. be the n-th

DIMENSION

197

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

m"lm·+1 as k·spaces so that

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

I,

8.1). Since n -+

(n

C: ~~ I)

+ r - 1) r-I is a polynomial

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

n~

o -+ B. -+ A. -+ m·lmn+l -+ 0., This implies 'X (grm(R), n)

== (n + r

-

r-I

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). 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. Corollary:

.p is an

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

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

8. I.

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

COMMUTATIVE ALGEBRA

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

(n + r- 1).

_a free R-module of rank

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

a, ( : )

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

S.

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.

= d (M).

8.2. Kroll dimension 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 •.

=1= The integer

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.

DIMENSION

199

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

P,

= {X.,.

,

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

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. 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 P~l

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.

•

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) <: 00. OlJ a•• .... 0, ~ m.

200

COMMUTATIVB ALGBBRA

DefiDition: The Chevalley dimension sCM) of M i. the smallest integer r for which there exist a1• aI' ...• a, E m with IR(M/(al, .... a,)M) <: 00. If M = O. sCM) is defined to be -1.' Theorem 1: (Dimension Theorem). Let R be a Noetherian local ring and M a finitely generated R-module. Then dim (M) = d(M) = sCM).

Proof:

We first show that dim(M)';;:; d(M). If d(M) = - I. Pm(M. n) = 0 for n ~ n,. Hence M = mnM for all n ;;:. no. By Nakayama Lemma. M= O. i.e. dim M = - I. Assume now d(M) ;;:. O. Choose a prime ideal P E Ass(M) with dim M = Coht P = dim R/P. Since R/P is isomorphic to a submodule of M. wehave by Corollary to Theorem 2. 8.1. d(R/P) .;;:; d(M). It is therefore sufficientto show that dim R/P <; d(R/P) for every P E Ass(M). Consider any chain of prime ideals P = p. C P 1 C ... C P, of R. We must show that r.;;:; d(R/P).

* * *

.

The proof is by induction on r, Since R/P:;c O. d(R/P) :> 0 and the result is true for r = O. Assume the result for (r - 1). Choose aE Pl. a f:P and a prime ideal P' E Ass(R/Ra + P) with P' CPl' Then there exists a chain of prime ideals P' C PI c ... cPr' By induction. r-I .;;:; d(R/P'). i.e. r';;:; .I Consider the exact sequence

* *

=1= 1 + d(R/Ra+P). + d(R/P').;;:;

A.

O--,,"R/!' --"" R/P--""R/(Ra +P) --"" 0 where 1.. is scalar multiplication by a. By Theorem 2. 8.1. Pm(R/P. n) + Pm(R/(Ra + P), n) =- Pm(R/P. n) + R(n)

where R(n) is a polynomial function of degree less than d(R/ Pl. Hence d(R/Ra + P) < d(R/P). i.e. r';;:; d(R/P) and the proof is complete. We now show that d(M) <: sCM). If sCM) = - I. then M = 0 and d(M) =- 1. Asiume sCM) = s > 0 and let a1• Om..... a. E m be such that (

QI'

M )M has finite length. . Since al , ...• a.

M ~M® R (al' .... a.)M R (all .... a.)

DlMBNSION

R

by Proposition 8.3.2. Supp (al • .. .. a,)

201

n Supp(M) ... {m}.

LetI = Ann(M) and J = 1+ (a 1 • .. . . a.). Then Supp(R/I) = {m} and in particular Ass(R/I) = {m}, i.e.!. is mprimary. Hence J is an ideal of definition of R. Let R = R/I and J = JII. Then R is a local ring with J as an ideal of definition. generated by ii 1 • .. . . ii.. Considering M as an R-module. it follows from Proposition 3. 8.1 that PJ(M. n) has degree atmost s. Since Ii

(J~) = IR (J~ ). PJ (M.n)

and

(M. n)

PJ

have the same degree andd(M) ~ sCM)· . We now show that sCM) ~ dim M by induction on dim M. If dim M = O. IR(M) < 00. so that s,M) = O. Assume dim M> 0 andletP u Pa.....PkEAss(M) forwhichdimM=Coht P, (1 .:;; i':;;k). Since dim M > O. P, ~ m (1 .;;:; i ~k) so that m ¢ U P" Choose . I a Em. a f: U P, and let M' = M/aM. Then I

Supp(M')

C

Supp(M) - {Pl' P 2'

....

PIc}

and hence dim M' < dim M. Choose a1• . .. . a, E m, t least. such that M'/(ab .... a,)M' has finite length. This implies MI(o. a1•

M'

"'J

o,)M c>< ( Qho .• ,a, )M'

also has finite length and hence sCM) .:;; t + 1. By induction. . t ~ dim M' and hence sCM) .;;:; dim M. Corollary 1: Let R be a Noetherian local ring with maximal ideal m and M a finitelygenerated R-module. Then dim M < 00. In particular dim R < 00 and is equal to the minimum number of generators of an ideal of definition. Proof: dim(M) = d(M) < 00 and dim R = s(R) ber of generators of an ideal of definition.

=

minimum num-

CoroDary 2: Let R be a Noetherian local ring with maximal ideal Then dim R <: dim" (m/m l ) .

m and k = Rlm.

Proof: Choose m/ml. Then 0 1,

0 1, .... arE m •••• a, generate

such that ii 1..... ii, is a k-basis of

m so that dim R ~ r

= dim,,(m/m').

202

COMMUTATlVB ALGEBRA

Corollary 3: ideal m and Proof:

Let R be a Noetherian local ring with maximal

i. its m-adic completion.

Then dim R

= dim R.

By Proposition 3, 6.3. R!mn=.Rlmn(n> 1)

so

that

Pm(R. n) = Pm(R. n). Hence they have the same degree.

Corollary 4: The prime ideals of a Noetherian ring satisfy descending chain condition. Proof: By localising, we reduce to the case of a local ring. The result in this case follows as dim R < 00. Corollary 5: Let R be sa Noetherian ring and P a prime ideal of R. The following conditions are equivalent. 0) ht (P) < n. [ii) There exists an ideal I of R generated by a-elements such that P is a minimal prime ideal of I. Proof: (ii):> (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). 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: 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..

DIMBNSION

203

m' = m/(a).primary ideal, (s=dim R'). Then a,.ol' Qa•• ••• a, generate an m-primary ideal in R. Hence dim R < I + s = I + dim R'.

Corollary 8: Let P be a prime ideal of a Noetherian ring Rand a EPa non-zero divisor. Then ht P!(a) = ht(P)-l. P~oof: Apply Corollary 7 to the ring R».

Corollary 9: If R = k[[X.. X., ... , X,Jj, is the power series ring in Xl' X 2I... , X n over a field k, then dim R = n. Proof: By Corollary I, dim R 0:;;; n. as Xl,.... X n generate the maxi-. mal ideal. But dimR ~ n follows by considering the chain of . prime ideals (XI) C (Xl' X.) C ." C (X., X, .... ' Xn) in R.

=F

=F

=F

We now study the behaviour of height and coheight under ring extensions. 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. 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

204

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

DIMENSION

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

COMMUTATIVE ALGEBRA

8.2. 1.

2. 3.

4.

EXERCISES

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

. In inflni . =F =l= exist mte Iy many such prime ideals. 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 5.

9.

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). 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

DIMENSION

207

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

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. 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.

. 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.

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

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

. = P,S'*'(0 <;'*'i ~ n) '*'and Q.+l = Q. + (X). Then Define Q, Q. c Ql c ... Q. C Q.t1 =I'*' '*'

Is a chain of prime ideals of S and dim S;;;' 1 + dim R.

To prove the reverse inequality, consider a chain Qo c Ql C

... c Q. of =l-

'*'

'*'

prime ideals of S and let P, = Q, n R (0 ~ i ,.;; n).

II all the P, are distinct, dim R ;;;, dim S ;) dim S - 1. If not, choose largcstj such that P J = P)+l' By Proposition: I, QJ "'" PJS and by Proposition 3, ht QJ "'" ht P)' Since ht QJ ;;;'j, we have ht P ~j. But PJ = PJ+l C PJ+I C ... C p. is a chain of prime J =+= '*' 9-

208 COMMUTATIVB ALGEBRA ideals of R. so that ht P J + n - j -1 ~ dim R•.i.e. n - 1 <: dim R, n E;;; 1 + dim R. This implies dim S E;;; 1 + dim R and the proof is complete. Corollary 1:

If R is a Noetherian ring. then dim R[X1 • XI' .... X a] = n + dim R.

Corollary 2: If K is a field. dim K[X1• XI..... X a] Corollary 3:

= n.

Let R = K[X1• XI' .... X a]. K field. Then ht (X.. X•• "', X,)

= t,

(1 E;;; i E;;; n).

Proof: Since dim R = n, ht (Xl' XI' ...• Xa) = n. BY.Cor~lIary 8 to T heorero 1• 8 . 2• ht (Xl' •..• X a-]) = n - 1 and by Iteration we have ht (Xu ...• X,) = i (1 E;;; i El; n). " d al P f We saw in Section 8.2. an Example of a prime l e o a Noetherian ring R such that ht P + Coht P ~s less than dim R. H we will show that this cannot happen In affine k-algebras. 'N Iisati owever We first prove a general version of ~other.s . orma ~n Theorem which says that any ideal of height t In the polynomial . K[v rIng A] . . . . . .X]" a K field • is of the type as in Corollary 3. by a change of variables up to an integral extension. Theorem 1" Let R = K[X1..... X.]. K field and 1 an ideal of R with ht l=h. Then there exist Yl' YI,""Ya E R algebraically independent Ya] and over K such that (i) R Integral over K[YlO YI' (ii) 1 n K[YI' ...• y.]

=

(Yl. YI

, Yh)'

The proof is by induction on h. If h = O. 1 = ~ and d~e j n). If h > Ideal 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:

- v (1 :r;;;. y,-Al

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,

DIMENSION

209

there exist Yh =" f, YhH'·... Y. such that K[y'h• ... . y' a] is integral over 'Key", ...• Ya]. Define Y, = yl, i < h. We show that Yl' YI..... Ya satisfy (i) and "Oi) for I. R is clearly integral over K[Yh Y • .. .. y.]. a Since 1 n Kfyto Ya• ...• Ya] contains (Yl' Ya• .... Yh) which is of height at least h. K[y]. YI..... Ya] = (Yl' YI' .... Yh).

In

Proposition 5: Let R be an affine k-algebra over a field k and P a prime ideal of R. Then ht P + Coht P = dim R. Proof: By the Normalisation Theorem (Theorem 1. 4.4) R is integral over a polynomial algebra. IiI view of Proposition 1.8.2. it is sufficient to prove the result for a polynomial algebra over k, Let R = k[X1 ,' XI' .... X.] and P a prime ideal of R of ht h. By Theorem I, there exist Yl' YI' .... Y. algebraically independent over k such that R is integral over k[Yl' YI..... Y.) . and Q = P n k[yu Y I• ...·.yl] = (Yl. YI' .... YI). By Corollary 3. to Proposition 4, ht Q = h. S' . k[yl ..... Y.] k tu o-;« race Q "'" [Yh+I .... Ya]. Co tQ=dlmk[yh...1o .... Y.]=n_h.

Hence Coht P

= Coht Q

= n-h.

Thus ht P + Coht P

= n ...

dim R.

DefiDltion: Let R be an affi!1e k-algebra with quotient field F. _Then Tr degtR is defined to be equal to Tr deghF.

Pro,oiition 6: Let R be an affine k-a1gebra over a field k. Then Tr tkg,.R

= dim R.

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

COMMUTA11VB ALGEBRA

o

0

1

o ~ SI ® S. o~

KI

1

® K, k

~

~

-1

R 1 ® R,

LI

r

® 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

Since L, is free over K, (i

k

such that x is a zero-divisor of R 1 ~ R.. Coht P

~ CoM Q = dim SI ® S.

Hence =' dim SI+ dim S.

•

= 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)

DIMENSION

211

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

--

\

dimSIQo = dim RIPI + dim RIP•.

Sinceht (QIQJ = dim (SIQo)- dim (SIQ). we have n ;;,'hl ("QIQo) ~ dimRIPI + dim RIP. - dim RIP I

i.e.

.

Coht P >Cohl PI + Coht p.- n.

~; .'~

.

8.3. EXERCISES 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

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. Definition: Let M be a finitely generated module over a Noetherian local ring R with' maximal ideal m. A set of clements au 01' .... a. of m (n = dim M) such that MI(o" .... a.)M has finite length is called a system of parameters for M. 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 R = k [[XI' X., .... X.]]. k field,

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 010

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""

a'_l)

M N 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)
s(~»S(M)-I.

Let t=1.

DIMENSION

~~~;fIttn( M) =s(~);> s(M)-1 =dimM-l. ··'·'·"~L.· . aIM '. 1 M . . N

213

M

·Lt> 1. Then dim = dlmwhere N = ., . (ol.· ..a,)M a (a.,...a,)M

,\<.k' "

lN

~~~duction dimN~ dimM-(t-l) and dim(

M )M = dim NN

al, .. .a,

"'

Suppose now that equality holds, so that dime M )M = n- t. a1 .. ··, at Choose a system of parameters al+l • a'+2' ... , a. E m for the module M N M , so that 0/. has finite ("',.... a,.M (a'+1...,a.)N (al, ...a"a,+, .... a.>M length. This implies that a l ... • a. is a system of parameters for M.

N

Conversely assume that a1.... a' is a part of a system of parameters

al ,. ..a. for M. Then a'+1,

a'+I ... a. are such that

finite length, where N = ( M )M' a1 .... a, dimN =n-t.

N has (al+1 , •••• a.)

Hence dim N:s;;;; n-i-t and

Corollary: Let aI' a2•... at Em be a part of system of parameters. for R such that 1= (al, ... a,) has height t, Then ht(l) + Coht(l) = dimR. Proof:"" From the above Theorem, dim R/I=d-t, where d = dim R; 10 that ht (I) + Coht (1) = dim R. The following Example shows that if a1, ... a, is a part of a . system of parameters for R, 1== (al,. ..a,) need not have height equal to t,

"""le: . '.

s

Let S = k[[X, Y. Zll. k field and R = (XY. XZr Then we~Ye. seen in the section 8.2 that dim R = 2. 1181-(1'. Y. Z) and J= (Z. X + Y). then ;nl= (XI: yl, ZI, Y2) c/. Hence J is an ideal of definition of Rand:Z X + Y il a 1JI,IteJn. ·of parameters for R. Thus {Z} is a part or a system of patIIIleteri for R. Since (X) n (Y Z) = 0 (Y Z)· has htO and boncoh'(~ = O. "" .

214

COMMUTATIVB ALGBBRA

Remark: In the above Example. Z was a zero divisor of R. However if a E m is not a zero divisor, ht(a) = I by Krul1's principal ideal Theorem. Then dimR/(a) = dimR-I shows that {a} will be a part of a system of parameters for R. This leads us to the concept of M-sequences. Definition: Let M be an R-module. R is said to be an M -sequence if (i)

(OJ •••• on)M

(ii)

0,

A set of elements all ... on of

=F- M and

is not a zero divisor of (a1o ...M ,_ JM (1 .,;;; a

j .,;;; n).

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.· Examples: (i) If R = K[Xl , .... X n] the polynomial ring. K field then Xl....' X n is an R-sequenco. (ii> 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. is an M-sequence.

(i)

0l,

On

(ii)

a.,

O, is an M-sequence and a,+1.... 0n is an

(

M )M sequence, for all i, (1 ,;:;; i";;; n).

0 1, ...0,

Proof: If 1 and J are ideals of Rand N = M/IM. then we have M

N

(1 + J)M "'" IN'

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-

215

DIMENSION

:;:.,* R

then 1 = (01"", a,), need' not have height equal to :'>;-'R-sequences are better behaved in this respect. 1: Let R be a Noetherian ring. If 1 = (a....., a,), then ht(l) = t.

Ce.

0 10 O2.... ,

I.

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. .

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

dim

~

=

If t

dim M - 1.

= 1. 0 1 is not a

By Theorem

zero

I, {Ot}

If I > I. by induction, is a part of system of parameters for M and hence

isa part of system of parameters for M.

. fOt. :... Ot_,

dim

where n = dim M. N

=

M

(all ...• 0'-1) M

=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

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

COMMUTATIVE ALGEBRA

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. 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. 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. "

(0 a. is not a zero divisor of M and (ii)

° is not a zero divisor of M/aiM. 1

To prove (ii), let alx

= 0, x e..M/o.M.

Then alx E a.M, i.e.

~, I x = a.z.

0lX = o,.y, JI E M. Since o. is not a zero divisor of

JI E aIM.

t.e.' JI = alz, Z E M. Hence 0IX = a.alz and 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.

0..

°

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).

e

DIMENSION

217

'~'.," .: 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.

"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 --»-

M M --»- 0 where 710 •

0

t

il 100lar" multiplication by at" This induces an exact sequence

M)8

71:.

~ Ext;-l(N. M) ~ Ext;-l N, 0tM ~ Ext'R(N, M) --»-

(

Ext'R (N. M)

(.)

where 71:1 is also mUltiplication by 71 0 1 (Exercise 7, 7.2). Since 0IN = O. 71;' = 0 and hence 8 is surjective.

=- 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'

By induction. ,Ext'R-l(N, M)

by induction

1'''')

II-J

'

Ext~-I(N'a7M) =-Hom N'(a.,.~an)M) ""'HOmR(N·(al... ~n)M) R(

and the proof is complete. If we take N = RII. with ai' a., an E I, we have 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

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). 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). 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 r

e..

Let 1\ F be the roth homogeneous component of the exterior R-algebra 1\ F of F. Define R-Jinear maps

basis el' ...•

a,:

r

r-l

1\ F --*- t.. F. by B,(ei,1\. ei.· .. A ei) =

i (- I)J+l ail ei

J~'

A ... ;i/'" A ei, •

1

(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-' --*

~

.,. X, --* X'-l --* ... --* Xl --* Xo--* 0

DIMBNSION

219

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 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 is acyclic. where do: X'o

= M -e-

i:t

is the natural projection. We

denote by ;flk) the augmented Koszul complex with respect to the elements a l ,.... ak. We prove by induction on k that X(k) is acyclic for each k so that X(n) = X will be acyclic. For k = O. X(O) is the ld

.

complex 0 --* M --* M --* 0 and is clearly acyclic. Assume now that the complex X(1<-l) is acyclic. Consider the truncated complex yek) obtained from X(k) . y(k):

0 -e-

.n

k

)

--* ~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 X(k-"):

,

'"L) '

ye

0 --*

xL:"il ) --* xL":.-,})

0 -e-

--* ....,.. ,rok-ll -e- MIlk_1M --* 0

J., ~ d~" ~ Aic_.

~L ~ II~ ~

-e- Ai

k-l

0

where tbe'vertical maps are induced by multiplication by ek and IkM. . dueed bY mu It'ipl'rcat'Ion byak' M- --* the last map - IS 1D Ik-1M I k _ 1M All the vertical maps are isomorphisms as a1 ... • .ak is an M-sequence.

220

COMMUTATIVE ALGEBRA

Since by induction. the complex above is is also exact. exact the complex below Corollary 1:

If I is an ideal of R generated by an R-sequence = n.

alO••• , an, then pdR(R/I)

Proof:

X:

The complex X for M = R is given by

O-+X -+x n

0-1

. d1 do -+ ...-+ X r -+ ...-+ Xl -+ Xo -+ R/I -+ 0 (*)

It is a free resolution of R/I as an free R-module (r ~ 0).

R-m~dule. as each

X.

=

r

AF ISa .

Hen"ce pdR(R/I).;;. n. Using the projective resolution (*) to compute ExtR(R/I, R/I) we see that Ext; (R/I, R/I) C>!. R/I *- O. Hence pdR(R/I) ~ n, i.e. pdR(R/I) = n. Theorem 5: _Let R be a Noetherian local ring and a ... a an Rsequence . If I -- (aI' a., , Or) • t h e natural map l' .,

"': R/I[X1• X.,

, Xrl-+ grr(R)given by

"'(X,) = a, + /·(1 ';;'i';;' r), "" Rl] = ld,

~an be e~tended to a graded algebra homomorphism and is an

isomorphlsm. We first prove the following Lemma.

Lemma: Let R be a Noetherian local ring and I an ideal of R generated by an R-sequence. If for some c E R I: (c) =1 th Id: (c) = [d for alI d ";p. 1. " en Proof: The proof is by induction on d. The result is true for d.= 1 by assumption. Assume the result for d and all ideals gene~ated by R-sequences. Let I = (01' a., ..., Or) where 01> a., ..., a, ~s an .R-sequence. If bE Id+1: (c), then be E Id+! C Id and by ID~uctlOn, b E [d •• Write b as a homogeneous polynomial of degree dID 01' a., ..., Or In the form b = alul + alu. + ... + aquq, (q.;;. r), u, E (01' 02, ... , a,)4-1.

~t is s~cient t,o show that u, E I d (I .;;. i ~ q). induction on q. Write

b = b'

We prove this by

+ aqu. where b' = a1u + '" + aq-luo- l' 1

DIMENSION

Then cb = cb'

+ caqu.E Id+l =

221

aq/d + Kd+1

where K = (ai' a••••• a0-1' aq+1>" .a r). Then cb = 1-.a. + IL. 1-. E Id. l.l E Kd+l C Kd and

aq{cuq- 1-.) = !" - b'c E Kd. By inductive assumption applied to K, we have

cu. - 1-. E Kd C ]d. i.e; cu. E Id. Since Id: (c) = Id. we have uJ,E [d. Now cb' = cb - caquq E [d+1 and since b' = a1u1 +...+ aqu'_l' by induction on q. u, E Id (I <; i .;;. q - 1). Hence u, Eo [d (1 .;;. i .;;. q) and the proof is complete. Proof of Theorem 5: Clearly the mapping", is a surjective graded algebra homomorphism. It is sufficient to show that Ker ~ = O. l.e. if F(X1, •• 0' X,)is a homogeneous polynomial of degree d with F(a) = F(a1o 01 •..•• a,) E Id+1, then all the coefficients of F are in I. We prove this by double induction on d and r, The result is clear for d = O. It is also true for r= 1 as 01 is a non-zero divisor of R. Consider now a homogeneous polynomial F(X1, XI,"" X r ) of degree d with F(a) = F(a1, all ...• or) E [d+1. Write F(a) = G(a) where G .is a homogeneous polynomial of degree d with coefficients in I. Replacing F by F - G we may assume without loss of generality that F(a) = O. Let F(X 1, •••• Xr) = X1G(X1' .... X,) + H(X., .... Xr) where G and H are homogeneous polynomials of degrees d - 1 and d respectively. Then 0 = F(a) = 0lG(a) + H(a). (*) This implies alG(a) E (a., .•. , ar)4 and by the Lemma, G(a) E (a l ••. . ,a,)4 c Id. Since deg G = d - 1, by induction. G has all its"coefficients in 1. Consider now the relation (.) modulo (a1 ) and denote the cosets by putting bars. We have 0= O(a., .... a,). Since at... .•ar is aD RI(a1)-sequence, we have, by induction on r, the coefficients of lio in (a.,...• ar) . Hence the coefficients of H lie in (01) a...... ar) = I. Thus the coefficients of F are in I and the proof is complete.

o

Corollary: Lee R be a Noetherian local ring. and a1... '. a, an R~sequence. If I = (a.. a...... ar), then Illl"is a free R/I-module of rank r, If S(l/I~denotes the symmetric algebra of /1/1 over RjI. ita n-th homogeneous component 8"(1/[1} is isomorphic to In/pr+l (.. ~ 1).

222

COMMUTATIVB ALGEBRA

Proof: The result follows as '" is a graded algebra isomorphism. 8.4. EXERCISES I.

2.

3.

4.

5.

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 P:J/

P runs through the set of prime ideals of R containing I.

6.

7.

8.

Letf:R 4 S be a local homomorphism of Noetherian local rings with maximal ideals m and n, i.e.fis a ring homomorphism withf(m) c n, M a finitely generated S-module. Show that depthR(M) = depths(M) when Sis alg. R-module. Let R be a local ring with maximal ideal m and I c m. =J= a E m - I, and J = (1. a). Show that for an R-module M. depth/eM) ~ 1 + depthz(M). Let 0 4 M' 4 M 4 M" 4 0 be an exact sequence of finitely generated R-modules. I an ideal of R with 1M' i= M', 1M i= M

DIMENSION

~~

223

dIM" i= M n • Show that

depthz(M') = depthI(M) if depthI(M) < depthr(M"). (ii)depth1(M') = 1 + depth/eM) if depth/eM) > dePt~:(M")' (iii). depthr(M'):> depthI( M) If depthI( M) = depthI(M ).

9.

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.. 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

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

field at the maximal ideal (X, Y) is not

/ C.M. Theorem 1: Let R be a Noetherian local ring and M a finitely generated R-module. Thendepth(M)":;; dim RIP, for every P E Als(M). Proof: Let m be the unique maximal ideal of R. that if a Em and P E Ass(M) there exists some Q E Ass(MlaM) with Q:J P

Since .p E Ass(M), HomR(RIP, M)

* O.

We ,first show

+ Ra,

Consider the exact sequence

o -+

).. M M -+ M -+ M' -+ 0,. where M' = M 'a

and x, is .sealar multiplication by a. Then the sequence A·

o -+HomR(RIP, M) ~ HomR(RIP,M) -+ HomR(RIP,M')

*

is exact. This implies HomR (RIP, M') 0, for otherwise, by Nakayama Lemma HomR (RIP, M) = 0, a contradiction as P E Ass(M). Hence HomR

(P:Ra,M') -HomR (RIP, M') * O.

This implies that there exists Q E AS8(M') with Q :J P + Ra = I, for otherwise 1 ¢ U Q, i.e. COflome a E I, {a} is an M'-scquence, Qe .....s(M'j

DIMENSION

225

>.~ •

{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 <: dim RIP. Hence depth (M) :;:;; dim RIP.

+

Corollary 1: If M is a a.M. module, then depth (M) =.dim RIP"'; dim M, for every P E Ass(M).

Proof: .

depth (M) :;:;; dim RIP ~ '

Sup

dim '(RIQ) = dimM.

QeM'~

Since M is C.M., depth (M) = dim M and the result follows. Corollary 2:

If M is a C.M, module every P E Ass(M) is minimal.

Proof: If r:» P' E Ass(M), then dim RIP = dim RIP' = dimM. Hence P = P', Proposition 1: Let M be a C.M. module and a Em. The following conditions are equivalent. (i) {a} is an M-sequence (ii) dim (MlaM) = dim M - 1.

If any of the above two conditions is satisfied, then MlaM is C.M. Proof: Clearly (i) => (ii) by Exercise 5, 8.2. Assume now condition , (ii). By Theorem 1, for every P E Ass(M), dim(M) = dim (RIP). If " P E Supp(MlaM) , then dim MlaM ~ Coht P = dim M contradicting (ii). Hence P rf Supp(MlaM). Now MlaM ~ M ~ RI(a) and ,

R

, Supp (M/aM) = Supp(M) n Supp(R/(a». Since P E Supp(M) we 'I.. have P rf Supp(Rla), i.e. a rf P for all P E Ass(M). Thus a is non~,iCrO divisor of M and {a} is an M-sequence. i, .Assume now that the condition (i) or (ii) issatisfied. By ".~ 4, 804, depth (M/aM) = depth (M) - I = dim M -1 = , (MlaM) so that MlaM is a.M.

2:

If M is a C.M. R-module, then every system of

_

226

COMMUTATIVB ALGIlBRA

parameters of M is an M-sequence. Conversely if every system of parameters of M is an M-sequence, then M is C.M. Proof: The converse is trivially true for if a1 , ••• , a. is a system of parameters of M, (n = dim M) then it IS an M-sequence so that depth (M) ~n= dim M. Hence depth (M) = dim M', i.e:M is C.M. , To prove the first part assume that M is C.M. and let al' ... , a., be a system of parameters for M. We prove by induction on k, that

01' ... ,Ok

is an M-sequence and that (_ ) M is C.M. at,_..M ot ak = 0 as M is C.M. Assume

(0 ,.;; k ,.;; n). This is clearly true for k

now that

01> ... , 010-1 is

an M-sequence and (ai' ... ,

~

k-l )M

= N. is

C.M.

Now.!!-=~ OkN

)'M and by Theorem 1, 8.4,

(01"'" ak

dim.!£. = dim (M) ,- k ak N

and

dim (N)

= dim (M)

- (k - 1).

Hence dim!!- = dim N - 1 and by 'Proposition 1, {Ok} is not a akN

--!!\M' i.e. ai' ..., ak is an (01' ... ,Ok-V !!~ ----!!--M is C.M. OkN- (ai' ••• , ak)

zero divisor of N= and

M-sequence

Corollary 1: If M is a C.M. module of dimension n and if

0 1, .... a,

(~a )M is

is a part of system of parameters for M, then a l ,

.,.,

r

C.M.

of dimension equal to n - r. Corollary 2: If M is a C.M. module, every maximal M-sequence is a system of parameters for M. Proof: By Proposition 3, 8.4 every M-sequence is a part of a system of parameters and since depth (M) = dim M, the result follows. 1beOreDl2: Let M be a C.M. R-module and P a prime ideal of

.R. Then ,Mp is a C.M. Rp-module. Proof: If P::tl Ann(M) then M p = O. We therefore assume that

D1MBNSION

.P:::> Ann(M). Then dimMp =

227

ht~~ Ann(M) and depthMp-d - epthp (M) •

Th.e proof,is by induction on depthp(M).

If depth (M)

0 th

ined p =,en • eontaine some Q E Asl{M) and since M I'S C M Q' P •IS 1m I I f ' ., IS a mm a. e emen~ 0

Ass(M) so that P

'. dtm M p = ht Ann(M) = QE

o.

Let depthp(M)

P, which is a non-zero divisor of M.

= Q E Ass(M). Hence > 0, so that there existsIf N = ~- then aM'

dimNp = dim ::':p = dim M p - 1 and depth Np = depth Mp-I
,i'

CoroUary 1:

If R is a C.M. local ring and P a prime ideal of

R, Rr is a C.M. ring. C41rollary 2: Let R be a a.M. local ring and I a proper ideal o~ R, Then' ht(I) + Coht(I) = dim R.

-:

Prool~ If the result Is true for Ii prime ideal. it is true for any ideal , I, for If ht 1='. choose a prime ideal p-::J I with ht P = r, Thea

ht

1. + Coht I ....ht P + Coht I

;> ht P + Coht P = dini R. ";

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 .
228

COMMUTATIVE ALGEBRA

The following Example shows that if P is a prime ideal of R, the length of a saturated chain contained in P need not be equal to ht P. Example: Consider the Example given in 8.2 where R

k[[X, Y,Zn k d - X 'V -Z) (XY, XZ)' field, m - ( • L • •

We have seen that Q = (Y. Z) is a minimal prime ideal of Rand the chain Q em = (X. Y. Z') is saturated. but ht m = 2. However this will not happen in a C.M. local ring. Corollary 3: Let R be a C.M. local ring and Q c P be prime ideals of R. Then all saturated chains of prime ideals from Q to P arc of the same length and is equal to httP) - ht(Q). Proof:

It is sufficientto show, that if Q c P such that there is no

=1=

prime ideal strictly lying between Q and P, ht P - ht Q = 1. Clearly by assumption, dim (RpIQRp) = 1. Since R p is C.M., we have by p Corollary 2. ht QRp + dim (RpIQRp) = dim R». Since ht QR = ht Q, we have ht P - ht Q = 1. Corollary 4: Let S be a local domain which is the quotient of a C.M. ring and "Qa prime ideal of S. Then htQ + Coht'Q ... dim S. Proof:

Write S = RIP. where R is a C.M. local ring and P a

prime ideal of R. If Ql ~

Q; c ~ = Q.

are prime ideals of S,

then all saturated chains of prime ideals betweenthem have the same length equal to ht Q.:..... ht QI = ht

(Q.) - ht WI)

as R is C.M. Let Q = QIP be any prime ideal of S. Choose maximal chains of prime ideals of S

Qh = Q c Q~ c .., Q; Coht Q. Since the above chain is saturated,

Qo c Ql c.·· where h = ht Q and t = h

C

+ t = ht @:) -

ht (Qo)

= dim

S

as Q; is maximal and Qo a minimal prime ideal of S.

,.

DIMENSION

229

;~a1tioD: A Noetherian ring R is called a C. M. ring if R . ,,'O.M. for every prime ideal P of R. piS ;'"' ,. ~his ~efinition coincides with the one already given for local ;,', n~ m View of Corollary ,1, Theorem 2. Enmples: The rings (i) k[X]. k field (ii) Dedekind domain and

CO u

k[X, (X' :..... Y] YI)'

k field. are C. M. rings as their Iocalisations are

one dimensionalloeal domains. ~e now show.that Cohen-Macaulay rings arc precisely the class of nngs for which the following classical unmixedness property holds. Definition: An ideal I 0 f a Noetheri . IS . said to be height ' • • oe enan nng :Dl1xed.If ht(I) = ht(P,) for all P, E Ass (R/l). A Noetherian ring let unmlxedness property if every ideal of height r generated by r e ements, and the zero ideal, is height unmixed. Theorc~ 3: Let R be a Noetherian ring. The following conditions arc equivalent, (~~ R h.as the unmixedness property. (11) R p IS C.M. for every prime ideal P (iii) s; is C.M. for every maximal idcai m.

Proof:

We first prove the following.

th If P is a prime i~eal of. ht r, then there exist aI' " ' j ar .~ P such r:::\ht~al"~ a,) =1,1 ~ 1 ~ r",We do this by induction on r. For ., , t P -,Iso .th~t the~e eXists a l E P, a l is not a zero divisor 1 A ssume th at and s principal Ideal Theorem ht(a) tlt byaKrull EPh b 1=,'

...., '-1

ave een constsucted with the property. ht(a 1, ... , a,)=I, 1 ~ I

~

r - 1.

Let Q. be the minimal prime ideals of (n.. a ) Th th nl .... .... 1'-1' en theseQb aQ•• .... ( re e 0 y prime ideals of height (r- 1) containing tlt..... a,-J. Now P ¢ Q" as ht(P) = n> ht Qt{I ~ i ~ I).

Y

, Choose a, E P, ar ¢ Q, (1 ~ i ~ s). It is easy to see that ht(tlt,

By (i), the ideal (a l

, ar ) ,

= r.

a,) is unmixed and hence a'+1' Q

230

COMMUTATIVE ALGEBRA

for any Q E Ass (alt.~.Qj»)' P.

Hence a1• .. .. a, is an R-sequence in

This implies depth R p ;> T = dim Rp• i.e, R p is C.M.

(ii) => (iii) is clear. (iii) => (i). Since unmixedness property is a local property it is sufficient to show that R m has unmixedness property for all m. i.e. if R is a C.M. local ring then R has unmixedness properly. By Corollary z, Theorem 1. the zero ideal is unmixed. Let 1=(a1••••• a,) be an ideal of ht T. Then dim Rjl = dim R-T and hence al •...• a, is a part of a system of parameters (Theorem 1. 8.4). Since R is C.M.• alt.... a, is an R-sequence. Hence Rj(al..... a,) is C.M .• i.e. (a1'.... a,) has unmixedness property.

Theorem 4: C.M.

If a Noetherian ring R is C.M.. then R[X1, .... X.I is

Proof: It is sufficient to prove this for the case /I = 1. We first observe that if m is a maximal ideal of R[X]. then m contains a nonzero divisor. for if m consists entirely of zero divisors. X ¢ m and hence (m, X) = R[X], i.e. 1 c: Xf(X) + g(X). g(X) E m. But g(X) = l-XJtX) has constant term 1. This implies that g(X) cannot be a zero divisor which is a contradiction as gEm. Let m be a maximal ideal of R[X] and P = m n R. Since R is C.M .• so is Rp • t.e. ht P = depthp(R) = T say. By Propositions 1 and 3 of 8.3. itiseasy to see that PR[X],p.m and ht m = T+ 1. .If we show that depthmR[X] = Ttl. then RIXlm will be C.M. for everymaximal ideal m of R[X} and R[X] will be C.M. Let a1.....a, be a maximal R-sequence in P. Then is also an R[X}-sequence in PR[X] c m. Let [ = (a1.....a,) and m the image

it

of m in

:~f1.1 """:rXj.

Then in isa maximal ideal in:

[Xl

and

- must contain anon-zero dimsor -a'+l E [R[Xr R[X] Th hence m en a1, .. ·.a,.

+

a'+l is an R[X]-sequence in m. i.e. depthmR[X]:> 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.

,_ '.

I

~.

.~

DIMENSION

8.5.

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.

+

'.,

CHAPTER IX

REGULAR LOCAL RINGS

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.

REGULAR LOCAL RINGS

233

DefiJiitioD: A generating sot of d elements for m is called a regular system of parameters of R. 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

(;!~ ~)'

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

COMMUTATIVE ALGBBllA

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:

Let R be a local ring and

Then R is regular if and only if

R its m-adic completion.

R is regular.

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,

system of parameters. Then m is gene-

RBGULAR LOCAL RINGS

235

rated by al." .a.; 01+1,· ••• a". Hence R is regular and al>" .a, is a part of a regular system of parameters.

..

Corollary 1: let R be a Noetherian local ring with maximal ideal m. Then R is regular <¢?·m is generated by an R-sequence. I Proof: Let R be a regular and al •••• a" a regular system of parameters. For each 1 (0 <: 1 ~ d) R = Rf(aJ•....a,} is regular and

111+1" ; • Ii"

generate in = (~)' i.e. it is a regular system of 0 1 .... 0/

*"

parameters for R. Slnee R is a domain. a,+! 0 is not a zero divisor of R. i.e, at+1 is not a zero divisor of Rf(al, ... o,) (0 ~ 1 <: d). Hence a., a l •••• , a" is an R-sequence .. Conversely let m be generated by an R.sequence al'" so that d ~ depth R ~ dim R ~ d. Hence dim R = d and R is regular.

.a".

Coronary 2: A regular local ring is C. M. Proof: If m is generated by an R-sequence a b a••...• a". then depth R = dim R = d. Corollary 3: Let R be a regular local ring and P a prime ideal of R. The following conditions are equivalent. (i) RfP Is a regular local ring. . (ii) P is generated by a p~rt of a regular system of parameters. . . Proof: (ii) => (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

+ 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.

~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 ( ~ )

REGULAR LOCAL RINGS

237

9.1. EXERCISES 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
00.

Theorem1: Let R be a regular local ring. Then (i) gi.dim R
Since

paRk = d, TOr:+1

(~ • k) = 0

so that the exact sequence M 0-+ k-+M-+ 7C -+ 0

If t

= O.

238

COMMUTATIVE ALGEBRA

induces an exact sequence 0= Toli+l (~. k)

~

Tor; (k. k)

~ To~ (M. k)

Since Tor~ (k, k) ,t 0 (Theorem 5, 7. 3), Tor~ (M, k) ,t O. Hence pdRM = d and the result is proved. Assume therefore that depthRM = t > O. Then there exists some a E m which is not a zero divisor of M. Let M J = MjaM and consider the exact

x,

sequence 0 ~ M ~ M ~ M 1 ~ 0, where AG is scalar multiplicacation by a. Since deplh RM1 = t - 1. we have by induction pdRM 1 depth RMI = d. Consider the exact sequence

+

~

3

R

~ Torr (M. k) ~ Tor, (M, k) ~ Tort (Ml' k) ~ R ~: R Tort_l(M, k) ---+ Torl-l (M, k)

where ~ is multiplication by a. Since a annihilates k, we have an exact sequence 0 ~ Tor,R(M,k)~Torr(Ml.k} -+ Tort_l(M,k) ~ O. Applying Proposition 7. 7.3, we get pdllM1 = 1 + pdRM. Hence pdRM + deplhRM = d. Corollary 1: Let M be a f.g. R-module over a regular local ring R. Then pdRM = dim R <0' m E Ass(M). Proof:

depthRM

= 0 <0' mE Ass(M).

Corollary 2: Let R be a regular local ring and M af.g. R-module with dim M = d == dim R. Then Mis C.M. <0' M is free. Proof: Mis free
Corollary 3:

= 0<0' depthRM= dim R = dim M:

C.M.

Let R be a regular local ring. dim R

= d, and Sa

~ local ring of dimension dwhich is an R-algebra of finite type,i.e. f.g. as an R-module.

Then S is C.M."* S is free as an R-module.

Proof: By Exercise 3. 8.5. Sis C.M. over S "* S is C.M. over R. By Corollary 2. this is equivalent to S being free over R.

RBGULA~ LOCAL RINGS

239

Corollary 4: Let R be a regular local ring, dim R = d. and S a regular local ring of dimension d which is an R-algebra of finite type. Then S is free over R. Proof:

This follows from Corollary 2 to Proposition I. 9. L

Proposition 1: Let M be an R-module. a ERa non-zero divisor of R as well as M. If pdRM <1 eo, then pdRf(G)(MjaM) 0 and consider the exact sequence ~.

O~R ~ R~Rj(a)~O

..\

where All is scalar multiplication bya. This induces an exact sequence

>.:

O~ Torf(M,R/(a» ~ M~' M ~ MjaM ~ O.

Since a is not's zero divisor of M, Torr (M, R/(a» ;= O. Consider an exact sequence 0 ~ K ~ F ~ M ~ 0, with F free. Then the sequence

0= Torr (M, R/(a» ~ K/aK ~ F/aF ~ M/aM ~ 0 is ex~ct. By Exercise 10, 7.3, pdllK = pdRM - 1. so that induction applies to K because a is a non-zero divisor on KeF. Since pdRM <~, we havepdRK<~. This implies pdRrl.I(K/aK)<. co and hence pdRIC.)(M/aM) < co.

Proposition 2: Let R be a Noetherian local ring with maximal ideal m. (i) For any a Em":'" ml , the exact sequence of R/(a) modules,

o ~ mamaRa -Ra ~ -m ~ -m ~ 0 splits. (ii) If m =F m' and every a E m - ma is a zero divisor any finitely generated R-module M with pdRM < co is free. • Proof: (i) Let d = dim. m/m'. Since a ¢ ma, there exist a1• a,• . .. a"-1 Em such that (a. a1• . .. ad-,) is a minimal generating set for In. If [=(a 1•••• ad-l) then lriRa=Inma. forifb=caEI,

240

COMMUTATIVE ALGBBRA

ERe cannot be a unit for otherwise a E I. which is impossible a'a a is, a minimal generating set. Hence bE I n ma, , H'" d-l . m I + Ra I _ ~~. Hence the natuNow RQ=~""InRa- Iflma- ma

C

as

---!...-

!!!.. -'>-!!!.. maps ~ isomorphicaUy onto mjRa, whose r al map ma Ra ma , inverse gives the required spliting miRa -'>- mlm~. . . . (i) Assume now that each aE m - m IS a zero divisor.Then I m _ ml cUP where the union is taken over all the l. Sincem.pml " . :dal fR'e mcUPum associated prime 1 e s o , I. • p we have m cUP and hence m = P. This implies that there exists

. cti10n ofPk -_ Rim m' R If M a finitely generated R-module . • d an .InJe withpdRM = n,the exact sequence O-'>-k-'>- R -'>-Rik -'>-0. 1D uces an exact sequence 0-'>- TO~+I(M. Rlk) -'>-To~ (M. k)-'>- Tor{ (M.R)

Since

To~ (M, k) .p 0, we have T~~ (M. R) .p O.

This.is possible only if n = O. Hence M is projective free.

and hence

Corollary: Let R be a Noetherian local ring with ~l. dim R <: 00 and a E m - ml , a non-zero divisor of R. Then gl. dim R/(a) <: 00. Proof:

Consider the exact sequence of R/(a)-modules 0-'>- miRa -'>-RI Ra -'>-Rim -l- O.

Since gl. dim R/(a) = pdR/Ca) Rim. it is suffidcient to pdR/(a) miRa <:'00. Since. gl. dim R <: 00, P R m <: 00 have by Proposition 1. pdR/(a) mjma
sh:: t~: d

By Proposition 2. miRa is an R/(a).direct summand of mlma so thatpdR/ca) miRa

<

00.

Theorem 2: Let R be a Noetherian 10l?81 ring with gl. dim R <: 00. Then R is regular.

REGULAR LOCAL RINGS

241

Procd': The proof is by induction on t = dim. mimi. If t = 0. nl = mi. so that by Nakayama Lemma. m = 0 and Ris regular. Assume t > O. If every a E m - ml is a zero divisor. then gl. dim R = pdRk < 00 would imply (Proposition 2) toot k is R~free, i.e• . m = 0. a contradiction. Hence there exists some a E'm - ml which is not a zero divisor. By Corollary to Proposition 2, we have l gl. dim R/(a) < 00. = miRa, is a k-vector space of dlmension t - I. so that by induction RI(a) is regular, i.e, is generated by an R/(a)-sequence al • . .. , ii,_I' Then a. aI' .... a, is an R-sequence which generates m and R is regular.

mrm

Ifrn

m

Corollary: Let R'be a regular local ring and P any prime ideal of R. Then R p is a regular local ring. Proof: Let gl. dim R = n so that pdRRIP ,;;; n. Hence it.has a free resolution of length t ,;;; n, say O-,>-F,--+ F'_I -'>-... -l-Fo-,>-RIP-l-O.

Since R p is R~ftat, the sequence O-'>-Rp

® F, -'>-R p ® F'~I -'>- ... -'>-Rp ® Fo-'>-Rp ® R

R

is exact and each R p

®' Flo

R

R

R/P-l-O

(0 ,;;; i ,;;; t) is Rp-free.

R

Hence pdR , ( n,

~

RIP) ,;;; n. i.e. PdR,(p't ) ,;;; n. Hence

gl. dim R p ,;;; n. i.e. Rp'is a regular local ring. ~tion: A ring R is called regular if Rp is a regular local ring for every prime ideal P of R. In view of the above Corollary to Theorem 2, a regular local riag will also be regular in the sense of the above definition.

Theorem 3: If R is a regular ring, the polynomial ring RIXI• XI, ... ,X.) is also regular. Proof: It is sufficient to prove the result for n = I. Let Q be a p e ideal of S = R[Xj lying above P = R n Q. Since Sa is a localisation of Rp[XJ and Rp is a regular local ring, we may assume without loss of generality that R is a regular local ring with maximal ideal m. We have SimS"" R[X] ® k "'" k[XJ. k = Rim. Since Q ::>

»s, we have Q =

R

'

mS or Q ... mS + (f), f( X) E S[Xi. a monic

·242

COMMUTATIVE ALGEBRA

polynomial. Since R is regular local. m is generated by d = dim R elements al• am' .... ad. Hence Q is generated over S by d or d + 1 elements. according as Q = mS or Q = mS + (I). Since ht(Q) ;;.;. d. We have ht(Q) = d if Q = mS or ht(Q) = d + I if Q = mS + (I). Hence Sa is a regular local ring. Corollary 1: If k is a field. the polynomial ring k[X1• regular.

•• ••

X.l is

Corollary 2: If R is Noetherian regular. then gl. dim R = dim R. Proof: gl. dim R = sup. st. dim RIO (Exercise 5, 7.3) 10

= =

.

sup. dim RIO (Theorem I. 9.2) dim R (Exercise 6. 8.3)

Corollary 3: gl. dim k[Xl • X 2,

....

X.] = n, k field.

Proof: dimk[Xl • XI..... X o] = PI, by Corollary 2 to Proposition 4,8.3. We have already proved that a regular local ring is a domain (Corollary 1 to Theorem 1, 9.1). We now show that a regular local ring is a DFD. PropositioD3: Let R be a Noetherian domain. Then R is a DFD if and only if every ht 1 prime ideal of R is principal. Ploof: Assume R is a DFD and let P be a prime ideal of ht 1. Let a E P, a 0, and pan ineducible factor of a. Then Rp c P and since ht P = 1 and Rp is prime, Rp = P, i.e, P is principal. Conversely assume that every ht I prime ideal is principal. Since R is Noetherian. each a E R which is a non-unit can be expressedas a product of irreducible elements. It is sufficient to show that every irreducible element is prime. Let a E R be irreducible and P a minimal prime ideal of (a). Then ht P = 1 and by assumption P is principal. i.e, P = Rp, PER, P prime. Clearly p is a prime dividing the irreducible element a so that p = ua, u unit. Hence a is a prime element and the proof is complete.

*

PropositioD 4: Let R be a ring and M a projective R-module. (i) If M has a finite free resolution

RBGULAR LOCAL RINGS

243

0-+ Fo-+ FII-1 -+ '" -+Fo -+ M -+ 0

where each Pi free of finite rank, (0 ~ i ~ n), there exists a finitely generated free R-module F with M E9 F free of finite rank. (ii) . If M Ef) R' ce R"+l, then M"",- R. Proof: (i) The proof is by induction on n. .If n = O. M ""'- Fo is finitely generated free. Let n > 0 and let K = Ker (F. -+ M). Since M is projective. Fo =. M E9 K and K has a free resolution of length n - 1. By induction there exists a finitely generated free module F' with K E9 F' = F, finitely generated free. Then . M E9F= ME9KE9F' =.FoE9F' is finitely generated free. (ii) Let Mbesuchthat MEf) R°=.R"+l. ThenMpE9R~o
(R/\ M)po
/\

R R

i

M ""'-/\ (Rp®M) R~

R

0< /\

R~

As this is true for every primeideal P, 0+1

"+1

Now R = /\ Ro+1"",- /\ (M e RO) 0< o

=. Ii M

AM =

EB:E

0, for i > I.

I

}

A M ® I>. Ro

l+j-II+1 1 ,.

11+1

® /\

M p = O(i> 1).

R°Ef) /\ M® 1\ Ro

=.M®R=.M. . R

Corollary: Let I be a non-zero projective ideal of R having a finite free resolution. Then I is principal. Proof: By (i), Proposition 4, there exists a finitely generated free module F such that I E9 F = F', where F' is free of finite rank. Since ·lp Ef) Fp =.~, I is of constant rank equal to rank F' -rank F. Since I.i~ a non-zero projective ideal, rank 1 = 1. Hence by (ii), Proposition 4, 10
244

COMMUTATIVB AL01lJJRA

ideal of ht 1. Since d> 1, m;6m l and we can choose a Em - m 2• Since {a} is a part of a regular system of parameters (Exercise 4, 9.1) the ideal Ra is a prime ideal i.e. a is a prime element. If a E P, then Ra c P and since P is of ht 1. P = Ra is principal. Assume therefore that a ¢ P and consider the multiplicatively closed set S = {an! n:> O}. If R' = R s, then P' = PR' is a prime ideal of R' of ht 1. We show that P' is principal. Since gl. dim R <: co. P admits a finite free R-resolution and hence P' = PR' admits a finite free R'-resolution. We show that P' is a projective ideal in. R'. Let Q' = QR' be any prime Ideal of R' where Q is a prime ideal of R. Then Q :f. m as a ¢ Q. Then R~, "'" R Q is a regular local ring of dimension less than dim R. Hence by induction, R~, is a UFD. Since P'R~, is a prime ideal of ht I. it is principal Hence pdR,P' = S;P pdR'Q'P'R~, = O. Hence P' is R'-projective. Now by Corollary to Proposition 4. P' is a principal ideal in R'. Write P' = R'p, pEP. By removing the highest power of a dividing p, we may assume without loss of generality that a docs not divide p. We claim that P =Rp. Cleatly Rp

c

P.

Let x E P

= P' n R.

Since a is a prime element and i.e. x E Rp showing P = Rp,

Write x =

!!.. p, a'"

i.e., d"x = bp.

a does not divide p. we have am Ib,

9.2. EXERCISES I.

2.

Let R be a regular local ring and I an ideal of R. Then Rjl is Cohen Macaulay if and only if the ideal I satisfies the condition ht(!) = pdR(RII). Let M be a finitely generated module over a Noetherian ring Rand N a submodule of M. Show that- if N is expressed as a finite intersection of irreducible submodules N = n N,• with

n Nle ¢

Ie,,}

3.

I

Nj, then the number of factors in the decomposition

depends only on N (called the-index of irreducibility (110.) of NiD. M). . Let R be a C.M. local ring with dim R = d. If a1 .... ad and bi •• ••bdare two systems of parameters for R, show that the ideals (a". "', ad) and (bI • bl..... bd) have the same I.I.D. (called indu of kreducibtllty of R). .

RBGULAll LOCAL lUNGS

245

4.' Let R be a regular local ring of dimension d and aI'

,ad, a ad) is system of parameters. Show that the ideal (a1 irreducible. 5. Let R be a local ring such thJ!,t any ideal generated by a system of parameters is irreducible. Show that R is C.M. 6. Let R be a C.M. ring whose index of irreducibility is one. Then R is called a Gorenstein ring. Show that a local ring R with dim R = n is a Gorenstein ring ~ idRR = n. (H. Bass: On the ubiquity of Gorenstein Rings: Math. Zeitschrifr, Vol. 82 (1963».

9.3.

Normality conditions

Let R be a Noetherian ring. Consider the following conditions. (k;;;. 0) (S,,): depth R p ;;;'Min (k.dim R p) '(Ric): dim R p .::;; k implies R p is regular. Clearly (So) is always true. The following Proposition shows that (SJ is equivalent to the condition that R has no embedded primes. Proposition 1: R satisfies (SI) if and only if R has no embedded primes. Proof: Assume that R satisfies (81) and suppose P is a prime ideal of R which is not minimal. Then dim R p :> 1 and hence by (SJ. depth Rp~ 1. This implies P¢Ass(R) for if PEAss(R). then PR p E As.r(Rp ) , a contradiction, as depth R p ~ 1. Hence every P E Ass(R) is minimal. Conversely assume that every P E Ass(R) is minimal. If P:= Ass(~), then depth R p ,>O=Min (I, dim Rp). If P¢Ass(R), Min (I. dim Rp) = 1 and depth Rp > 1, for if depth R» = 0, then PRp E Ass(Rp), i.e. P E Ass(R), a contradiction. Hence (St) is satisfied. Proposition 2: R is reduced,' i.e. the nil radical N(R) only if R satisfies (81) and (Ro).

= 0 if and

Proof: Clearly (RJ is equivalent to the condition that R p is a field for all minimal prime ideals P of R. . 'Assume that R is reduced. If P is a minimal prime ideal of R, ~p is an Artinian local ring and N(R p) = N(R)p = O. so that RplS a field, t.e. R satisfies (Ro).

246

COMMUTATIVE ALGBBRA

Suppose R does not satisfy (SI)' By Proposition I, there exists PE AII(R) which is not minimal. Let PI' p., ... , P, be the minimal prime ideals of R. Then P ¢ U P, and choose a E P,

,

a rf: Ph (1 ~ i ~ F). If ba = 0, bE R, then (bajI) = 0 in Rp,. Since (ajl) is a unit in Rp" (b/I) = O. i.e. s,b = 0, I, rf: P, (I ~ i ~ F). This implies bEn P, = 0, a contradiction as a is a zero divisor.

,

P 1>

Conversely assume that R satisfies (SI) and (R o)' Let p., ..., P, be minimal prime ideals ofR. If n P, =1= 0, there exists

some a E

n, P" a "* O.

i.e. sa = 0, s, rf: P, (1 hence Ann(a) ¢

I

Since Rp, is a field (1 ~ i ~ F), P,Rp, = 0,

~ i ~ F). Thus Ann(a)

¢ P, (1 ~ i ~ F) and

Ur; By (SI)' R has no embedded primes so that

i-I

UP,

=

U

P. This implies Ann(a) ¢

PEA.,(R)

tion.

Hence

n, P, =

U

P, a contradic-

PEA.,(R)

O.

DefiDitioD: A ring R is called normal if R p is an integrally closed domain for every prime ideal P of R. Theorem 1: Let R be a ring and Q(R) its total quotient ring. The following conditions are equivalent. (i) R is normal. (ii) R is reduced and integrally closed in Q(R). (iii) R is a finite product ofintegrally closed domains. Proof: (i) ~ (ii). Let R be normal. If a E R is nilpotent, ti' = 0 in R p , for every prime ideal P and since R p is a domain, a = 0 In Rp for every P, t.e. a = O. Hence R is reduced. Let ot = alb E Q(R) be integral over R. For any maximal ideal m of R, ot E Q(Rm ) is integral over R", and since R", is integrally closed, ot E R m • This implies ca E Rb, for some c rf: m. Let I = Vol AE R, >'a E Rb}. If I were a proper ideal, t c: mo' for some maximal ideal mo. Now, for some Co rf: mo, E Rb, a contradiction. Hence I E I and ot = alb E R. (ii) ~ (iii). Since R is reduced, by Proposition 2, R satisfies SI' Let PI' p.,..., P, be the minimal prime ideals of R. Then n P, = 0 and by (SI) these are all the prime ideals of Ass(R). If T is the set of non-zero divisors of R, R T = Q(R). Now P,RT(I ~ i ~ r)

coa

REGULAR LOCAL RINGS

are mutually coprime and Theorem Q(R)

n P,R r =

= RT =

O.

By the Chinese Remainder

Rr

r

'~1 P,RT "'" ~K,

where K, is the quotient field of RIP, (1 ~ i Consider the composite of injections R -+

247

~1 RIP, -+ ~, K, "'"

~

F).

Q(R).

By (ii), R is integrally closed in Q(R). If e, = (0,0,..,1,0,... 0), with 1 in the i-th place, then e~ = ei and eie] = 0, j. Since e, is integral over R, e, E R and hence

i"*

R =- ~ Rei. Each Re, is integrally closed as R is integrally closed 1

in Q(R). (iii)

Hence R is a finite product of integrally closed domains.

* (i).

Let R =

~R, be a finite product of integrally closed 1

domains R/(I ~ i ~ F). It is easy to see that any prime ideal P of

R is of the type P = ~P" where P,

,

....,

f

= R, for

i"* j for

some j and p)

=- (R))p)

is an integrally closed

Theorem 2: Let R be a Noetherian ring. and (S.) if and only if R is normal.

Then R satisfies (R 1 )

is a prime ideal of R). Then R p domain. Hence R is normal.

Proof: Assume R is normal and let P be a prime ideal of R Then Rp is an integrally closed domain and if dimRp ~ 1, then Rp is either a field or a DVR and hence R p is regular, i.e. (R 1) is satisfied. To prove (S.) we have to prove, in addition, depth R p ~ 2 whenever dim R p ~ 2. Let a E PR p , a O. Since Rp is a domain, a is non-zero divisor, i.e. {a} is an Rp-sequence. Since R p is integraIly closed, the ideal (a) has no embedded components (Exercise I, 8.5). Thus, if Q E Ass(RplaR p ) , by KruII's principal ideal Theorem, Q PR p as ht PR p;;:' 2. Hence there exists b E PR p with brf: U Q, where Q runs over the set Ass(RplaRp). Thus {a, b} is an Rp-sequence and

"*

"*

depth Rp>2.

Conversely assume that R satisfies (R1) and (S.). This implies that R satisfies (Ro) and (8 1 ) , i.e, R is reduced by Proposition 2. Let PI' p., ... , P, be the minimal prime ideals of R.

. 248

C014MUTATIVE ALGJiBllA

,

Then by Theorem I. the total quotient ring of R is Q(R) !Y.1tK, where K, is the quotient field of RjP, (I ,,;; i ,,;; F). Consider the natural injection R -+ Q(R) given by a -+ (a + P,) and identify R as a subring of Q(R). We will show that R is integrally closed in Q(R). Let IX E Q(R) be expressed as IX = alb, a E R, bET. Suppose IX satisfies an integral relation of the type rx.n

+ C11Xn-'l + ...+ Cn = 0,

c, E R (1 ,,;; i ,,;; n).

an + I c,on-'b' = O. I

Then

Let P be a prime ideal of R of ht 1. By (R 1) . Rp is regular and hence integrally closed. If a = (ajI) and b = (b/I) in R p • the relation an + I c,an-lb' = 0, shows that ajb is integral over Rp and

,

hence belongs to R p , i.e. a E '6Rp , for every prime ideal P of ht 1. Since R satisfies (8 2) and b E R is a non-zero divisor, RlbR satisfies (SI) and by Proposition I, RjbR has no embedded primes. Thus (bR) is unmixed of ht I and since a E bRp for every prime P of ht1, aE bR, i.e. oc = albE R. . Since R is reduced and integrally closed in Q(R),R is normal. by Theorem 1.

9.3.

1.

EXERCISES

Show that if for every prime ideal P of R, depth R»

:> k

dimRp .;;;; k implies R p is C.M., then R satisfies (Sk).

and State

2.

and prove the converse. Show that R satisfies (Sk) if and only if for every t E Rand R, regular sequence ai' a l •.•. , a, (F < k), rhe R,-module

3.

R, )R has no embedded primes. al " al" .. 'J Or t A normal domain R is called a Krull ring if (i) for each a E R. the number of ht I prime ideals containing (a) is finite and (ii) R = n R p• Show that-a Noetherian normal ring is a Krull

4.

ring but not conversely. Show that if R is a Noetherian domain, the integral closure R in its quotient field is a Krull ring.

(

htP-l

of

lUlGULAIl LOCAL IlINOS

249

9.4. Complete local rings Let R be a local ring with maximal ideal m and residue field Rim = k, Definition: R is said to be of equicharactertstic type if ch R = ch k. Otherwise it is said to be of inequicharactertsttc type. . If R is of inequicharactertsttc type. then ch R = 0, el, k = p > 0 with p E m. Definition: Let R be of tnequicharacteristic type. R is said to be unramified if p Em-mi. Otherwise R is said to be ramified. We propose to obtain the structure Theorem for complete regular local rings. This Theorem will be a consequence of the following Theorem on the existence of a coefficient ring. Theorem 1: (Cohen's Theorem). Let R be a complete local riag with maximal ideal m, k = Rlm and ch k = p ~ O. Then Rcontainl a subring S satisfying (i) m n S = (p)

.

(ii) S is a complete local ring and m

S

n S ~ k.

Proof: We first consider the case when p = O. Let (>:R -+ Rim be the natural projection. Si~ec" k = 0, we have ch R = 0 and R::l Z, Z n m = O. Then R ::l Q. the rational number field and by Zorn's Lemma, R contains a maximaIsubfield L. We claim that L is the required field, t.e. (> maps L isomorphically onto k = Rjm. Clearly (>(L) is a subfleld of k and if k is not algebraic over (L). there exists some IX E R-L such that (1%) is transcendental over I/>(L). This implies. m n L[IX] = 0, i.e. n :: L(oc), a contradiction to the maximality of L. Hence k is algebraic and separable over (>(L). If ~ E k, then (3 is separable over KL), i.e. there exists some monic polynomial.ttX) E R[XJ. such that (3 is a simple root ofthe irreducible poIYllQmiall(X) E (>(L)[X). . By Hansel's Lemma,/(X) has a simple: root ~ E R Iltld! that (>(~) = (3. The' isomOrphism (> : L -+ I/>(L) can be extended to an isomorphism 1> : L(~)-+(>(L)(~). By maximality. ; ELand then ~ E r/>(L). i.e. k = ~L). Before we consider the case when ch k = p > O. w~ prove some preliminary Lemmas, Definition: Let ch k = p

> O.

A p-basis for k is a subset B

c

k

250

COMMUTATlVB ALGEBRA

such tbat (i) kP(B) =k (jj) All monomlals of the type b;' b; ...b;'. biEB. 0 ~ e, ~ p-I are linearly independent over k». .

Any field k with chk=p > 0 has a p.basis B satisfying the conditions

Lemma 1:

(i) kP"(B)

= k for all n :> 1.

(ii) [k .... (b1• bl ••••• b,): k....] = P'''. b, E B (I ~ I ~ r).

Proof: By Zorn's Lemma. there exists a maximalp·independent set. t.e. a set satisfying the condition (ii)· of a p-basis. It will also satisfy condition (i) in view of tbe maximality, Hence k bas a p-basis B. Let.p: k ~ kp be the isomorpbism given by ,pea) = ar, Assume by induction that k = kprr-l(B). Applying.p. we have k p = kpn(BP). Hence k = b(B) = kp"(B). This proves (i). The relation (ii) is also proved by induction on n. It is clear for" = I. Now

r: :> [kP"(bu . • •• b,): kP"] = [kp·(b1.....b,): kPn(bf..... b:}J [kP" (b~.....if,): kp·]

:> p' .p
as [kP"(bf..... bn: kp·]=[kpll-l(b1.....b,): kprr-l)= p 1n-l )' by induction. Hence [k....(bu .... b,): k n) = pM. CoroUary: If B is a p-basis of k, any a E k can be expressed uniquely as a linear combination over kpn of monomials of the type b~lb2" . . .

Proof:

W, 0 ~ e, ~ pn-l. b, E

R. (I ~ I ~ r),

Follows from the relations (i) and (ii).

Let m be an ideal of Rand p E m, p prime. If a - b E m. a. b E R then as" - bP" E mo+1 for all n :> O.

Lemma 2:

Proof: Tbe proof is by induction on n. Wc prove first for 11.= I. Let a = b + c. c E m. Thcn ap-bP=

,;J~)bP-lc'EmS. asp I

(n.

I!>O.

REGULAR LOCAL RINGS

a.... - b.... E m.o+1 •

Assume now that Then

251

as" - bP" = d. dE m-+1•

Now aPn+1 - b""+l = I

(P) (bP")p-I d' E m + n

'>0 i

l•

as p

I

(l!). I

Corollary: If m is a maximal ideal of Rand m' = O. the mapping a ~ a. n (n + I ~ t) induces a welI defined injective map of Rim into R.

an:

Rlm n

k=mlm n

~ RIm-·

We denote the image a.(k) by A.. We note that A. is closed under multiplication but not. addition. Let .

Tj:Rlm n

~ :ft':::.t::!.k

be the natural projection. Then Tj(A~) = kplll and by Corollarary to Lemma 2. "I I A.: A. ~ k plll is bijective. Let B be the set of residue classes of elements of B modulo m· and T. the set of polynomials in ·B with coefficients in An such that the degree in each variable of B is atmost p'n. The bijection "I: A. ~ kplll can be extended to a bijection "I: Tn = A.[B] ~ k.Ut(B*) = k, Let S. =T.+ pT. +...+pn-lT•. We'claim that Sn is a subring of RIm-. Clearly the product of two elcments of Tn is an clement of T. and thus S. is closed under multiplication. To show that S. is closed under addition.

:'1

252

COMMUrATlVE ALGEBRA

it is sufficient to show that the S\lIIl of two elements of Tn is in 8 n- Let G DC the additive subgroup generated by Tn in Rjm", We claim that G c T; + pG. Each element ~ of Tn is of the type }; aMM, aM E An and M a monomial of the type b~l ... b:'t 0.:::; ]1". Write aM = ~n, cME. RIm". By taking the image of eM under the natural map Rjm" -+ RIm = k and taking its inverse image under TJ we have eLM = dt·, dM E T•. We may assume without loss of generality that eM E Tn. Let ~t E. T. where

e, .: :;

}; dK,1JI ~l b:',liM E. r; f: + ~t = }; (eM + dM),on b~'...b:' + p( ~' =

Then

...

where the expression in the bracket is a sum of products of elements 'of Tn, i.e. an element of G. By iteration, we have Gc

r, +pTn+ p'G C r, +pTn +... + P-1T.=8n

as p".G = O. Thus the sum of two elements of Tn is in 8 n and S; is a subring of RIm·. Clearly p8n C mn8. where in = mlm". If ~ E. mn8n, then f: = f:o + p~t, f:o n T.. Since TJ is all. injection on r; we have m n T. = (0). Hence ~o = 0 and pSn = m n 8 n• Since the mapping 'I) restricted to T« is a surjective map of Tn onto k, 8 nlm n 8 n C>!. k. Thus p8n is a maximal ideal of 8 n and 8 nlp8n C>!. k, Let 1tn: Rlmn -+ Rlmll-1 be the natural map. Then 1tn(Tn) C 8 n- 1 so that 1tn define a homomorphism 1tn: S« -+ 811-1' By definition of Tn. it is easy to see that each element of T 1I-1:is in the image 1t.(T.) so that 1tn: 8. -+ 8.-1 is surjective. Thus (8•• 1t.) is an inverse system of rings with 1t. surjective for each n. Let

Em

lim

8

= ~- Sn.

We claim that 8 is the required ring.

n

Since each 8 n is a subring of RIm· and R is m-adicailY complete. by passing to the inverse limit. 8 can be identified as a subring R. Since (mlm n ) 8. = p8n, by passing to the inverse limit we have p8=mnS. Consider the exact sequence

n

o -+ p8n -+ 8 n -+ 8.lp8n -+0: By passing to the inverse limit we have (Proposition 3. 6.2) exact sequence o -+ p8 -+ 8 -+ k -+ O.

lUI

253

REGULAR LOCAL RINGS

.showing that p8 = m n 8 is a maximal ideal of 8 and S[m n 8 C>!. k, We claim that 8 is a local ring which is complete under the (p)-adic topology. In fact the (p)-adic the completion of 8 lim

8

= ~ pk8

lim lim C>!.

*;;-

r:

8.

P"8.

lim lim 8 ' n ""'-(--+---n k pk8 n

Now 8 n is (p)-adlcally complete as p8. is nilpotent. lim

Hence the right hand side is isomorphic to +-- 8 n C>!. 8 show. /I

.-

ing that' 8 is complete. Moreover. since 8 is (p)-adically complete. p8 is contained in the Iacobson radical of 8. Hence p8 is the unique maximal ideal of 8. Thus 8 is a complete local ring with pS ~ m nS and 81m n 8 C>!. k,

Dellnition: The ring 8 is called a coefficient ring of R. Corollary 1:

The coefficient ring S is Noetherian.

Proof: Since nrS c n'!In = (6), we have by Exercise 5. 5.1. every non-zero ideal of S is -of the type (r). n ~ 1. Hence S is Noetherian. Corollary 2: domain.

The coefficiellt ring S is a field or a DVR when it is a

Proof: If R is of -equieharacteristtc type. m nS = (0) and 8 "'" k, In the other case 8 is a Noetherian local domain With maximal ideal principal. Hence 8 is a DVR (Theorem 2. 5.2). Corollary 3: Let R be a complete local domain. Then R is isomorphic to a quotient of the power series ring S[[X1• X..... Xdll. where 8 is either a field or a DVR. Proof: Let S. be a coefficient ring of R. Choose a generating set Ql' a2..... ad for m and consider the ring homomorphism given by

254

COMMUTATIVE ALGI!BRA

r/>: S[Xu X•• ..., Xd]-+R

r/>(X,) = a, (1 ~ i ~ d)

and r/> I S = Id. Since R is complete. r/> can be extended to a ring homomorphism r/>: S[[XI , Xl ••·•• X,,]]-+ R. Now gr(r/»:S[XI , .•• , Xd]-+gr..(R) is surjective. Since S[[Xu ...• X d ]] is complete and n m' = (0), r/> is surjective (See Proof of Theorem 1. 6.4). Hence R is a quotient of S[[X], Xl •...• Xd]] where S is a field or a DVR. Proposition 1: Let R be a complete local ring of equicharacteristic type, k = RIm and au a" .•.• ad a system of parameters for R. Then R is a module of finite type over R' = k[[a], . . . • ad]]' Proof: R' being the quotient of a power series ring. is a complete local ring with maximal ideat m' = (a] ••.. ,a,,). If I = (a] • ...•a,,) R. then I is an m-primary ideal. To rahow that R is of finite type over R' it is sufficient to show. (Theorem 1. 6.4) that grI(R) is a moduie of finite type over gr.., (R'). Since grI(R) is naturally isomorphic to a quotient of the polynomial ring RII [Xu ... ,X,,], it is sufficient to show that RII is a module of finite type over R'[m' = k, Since I contains a power of m and RIm. mImi. ml/m s•... are finite dimensional k-spaces, RII is a finite dimensional k-space. Corollary:

Let R be an equicharacteristic complete local ring and

. k = RIm. If a]. 0 1•• • • • ad is a system of parameters for R, then a], a" •..• ad are analytically independent over k.

Proof: By Proposition 1. R is a finite type module over R' = kUal • a., "', a,,]]. Hence by Theorem 1,4.1, .R is integral over R'. This implies dim R' = dim R = d. Consider the nng homomorphism r/>: k[[X]. X•• . . . • X,,]] -+R' given by . .rf>(X/)

=

a/ (1

~ i ~ d). '" I k = Id.

'" is surjective and if .I. - I.J. 0 di k[[X].....X,,]] < d.• k er'l'-r- • 1m I

i.e. dim R' < d a contradiction, Hence '" is an isomorphism showing that al' adaro analytically independent ov~r k,

q.. ....

REGULAR LOCAL RINGS

255

Theorem 2: Let R be an equicharacteristic complete local ring and k = RIm. Let K be an extension field of k. Then there exists a complete local ring R' with residue field R'[m' = K such that (i) R' is faithfully fiat over R. (ii) m' = mk', (iii) dim R' = dim R. Proof: Express R as the quotient of a power series ring kIT X]• .... X.ll with Kernel I so that we have an exact sequence

0-+ 1-+ k[[Xl • X.> .. '. X.ll-+R -+ 0 If I' be the ideal generated by I in K[[X], XI' ... , X.J] we have a commutative diagram 0-+1-+ k[[XI • X., ... X.ll-+R-+O

1 11

11

0-+1' -+K[[X].X., ... X.ll-+ R' -+ 0

Where R'

=

K[[X" ;'" ... X.n and

I is induced by the inclusion

k c K. ClearlyR' is a local ring with maximal ideal m' = mR' which proves (ii). Now dim R' ,.;; dim R as any m-ptimaty ideal of R generates an m'-primary ideal of R'. We now prove (i). Since 1 is a local homomorphism. it is sufficient to show that R' is R·fiat. If A = k[[XI • . • • and B = K[[XI .... ,X.J]. it is sufficient to show that B is A-fiat for then it will follow that B ® All"", R'

X.n

will be A ® AIIO! R-fiat.

'" that

'"

To show that B is A-fiat it is sufficient

to show Tori "'(B. k) = O. (Theorem 4, 7.3). Consider the Koszul resolution of k as an A-module. By tensoring this resolution with B over A. we get the Koszul resolution of k as a Bmodule. In particular the tensored complex is acyclic showing that Tor] "(B, k) = 0.' This proves (i). Since R is faithfully fiat over R, by Corollary to Proposition 2, 8.2. dIm R' ~ dim R showing (iii). The following Theorem gives the structure of complete regular local rings. Theorem 3: (Cohen's Structure Theorem). Let R be a complete regular local ring of dimension d. Then R is exactly one of the following types.

256

COMMUTATIVE-ALGE8RA

(i) R"", k[[Xl • . .. . X,,]J. k field (R equicharacteristic) (ii) R"", S[[X••...• Xdll. S complete unramified DVR (R inequicharacteristic unramified)

(iii)R"",

~(~)' R o is of type (ii) and

[(X) .is an Eisenstein

polynomial over R o i.e, [(X) = X, + C1X ,- 1 +... + c, with c, E mo. c, ¢ m~ (R inequi. characteristic ramified). Before proving the Theorem we first prove the following Lemma. Lemma:

I

Let R be a complete local ring. with maximal ideal m. DO

ManR-modulesuch that M/mM isfg. overk=R/m and nm'M=O.

'-I

. Then M is/g. over R.

Proof: Choose a basis of MlmM over k = RIm and lift the basis elements to X:t. x......x, E M. If N is the submodule of M generated by {Xl. Xl ••••• x,} then we show that M = N. Now

M=N+mM=N+mN+m"M= ...

= N + mN + ...+ ml-1N + m'M. I(

xEM. write X=YO+Y1+ ... +Yr-1+Z with YJEmJNand t

zEm·!.i. We can exptessYJ= I i1l1x"aJ,EmJ. Since Ris complete,

'-I

DO

for each fixed i, I

J-O

all

t

converges to b, E R (say).

Let w = I b,x,.

Then W-X E m'M for all s and hence w = x. Thus M We now return to the Proof of Theorem 3.

'_1

= N.

Case (i): Let R be of equicharacteristic type. By Corollary.Z, Theorem 1. Seek, Choose a regular system of parameters Xl' X••••••X" for R and consider the natural k-algebra homomorphism ~:k[X1"'" Xd]~R given by r/>(X,l = xi; 1 ::;:; i ::;:; d. r/> I k = Id. Since R is complete. r/> can be extended to a k·algebra homomorphism ~

: k [[li..... X,,]] ~ R

Now gr(~: k[X1.... , Xli] .,.,)0 grm(R) is surjective. 6.4. if' is surjective as gr('f) is surjective. Since di""R

= dim k [[.1"1""'.1",,]] =

d,

if' is an

By Proposition 2, isomorphism.

REGULAR LOCAL RINGS

Case (i1): Let R be tnequicharactertsttc unramtfied type.

2S7 Since

p E m -m", we can choose a regular system of parameters Xv x.....,X"

with Xl = p. Let S be the coefficients ring. By Corollary 2 to Theorem 1, S is a complete unramified DVR. Consider the S-algebra 2,;;; i';;; d, homomorphism r/>: S[X., .... X d] ~ R given by r/>(X,) = r/> I S = Id and extend it to an S-algebra homomorphism 'f): S [[X., .. :.Xd]] ~ R. Then ~ is an isomorphism as in case (i).

x,.

Oase (iii): ' R is inequicharacteristle unramified. We claim that there exist X•• X•• . .. , Xd EO m which is a part of a regular system of parameters for R such that (p, x• ... Xd) is an 'm-primary ideal. This is proved by induction on d. If d= I, there is nothing to prove as (p) ; is m-primary. Assume d > 1. Since R is C.M. (Corollary 2, Proposition 1. 9.1). every P E Ass(p) is minimal of height one and hence m ¢ Ass(p). This implies m ¢ m" U P where the union is taken p

over' all P E Ass(p) for ifm c m" pU P, ,then m = ma or m = P for some P E Ass(p). This implies m = 0 or d = I, a contradiction. Choose XI E m - m·, XI ¢ P, for all P E Ass(p). IfR = R/(xl ) , there exist, by induction, x.' x" .... X" which is a part of a ~gular system of parameters for R such that ... , Xd} has the required properties. Let S be a coefficient ring of Rand R o = S [[x., ,,,,xdll. Since R is complete, Ro is a complete local sabring of R with maximal ideal mo = (p, XI' ... , Xd)' By the Lemma, R is a finite Ro-module as m"R is m-primary and R/mo:R is a finite Rolmo"'" k-module. In particular, R is integral over Ro, dim R o = d and R o is isomorphic to the power series ring in d variables .1"1>..1"2' ... , Xd. i.e. to S [[X2' ... , Xd]]' Thus R o is a complete unramified local ring of dimension d. We now show that R is an Eisenstein extension of Ro• Since Xi"'" Xd is a part of a regular system of parameters for R, choose a regular system of parameters Xl' XI> ... , X" for R. We claim that R = R O[.x1] . Clearly d

R = Ro+rn=Ro+ I Rx,

'-1 R o + }; (R o + m) X, d

=

. tal

258

COMMUTATIVE ALGEBRA d

C R o + 1_1 I Rox, + m" d

d

C R O + I RoX/ + I

'0 }-, RoX,x} + m

,

C R o + Rox, +... Roxtl

3

+ moR

where t is chosen to be the smallest positive integer with xf E moR which exists in view of the fact that moR = (p, XI' ...• Xd) is m-primary. By Nakayama Lemma, R = R o + ROXI

+.0. +

ROX~-l

i.e, R = Ro[x,). Since RIR o is integral, x, satisfies a monic polynomialf(X) over R o of least degree r, say f(X) = X, + cJ X'-l +... 't Crt c, E R o. We claim thatf(X) is an Eisenstein polynomial over Ro• Clearly c, E R o n m = m. as Xl E m, Suppose -some c/ If; m a• Choose a least j with this property. Then

-+:, C1X,-1 +... + C;X' (modulo mo> = X'(X,-J + cl X,- '- l +... +c,). C, O.

l(X) = X,

*

Since 1<X) is a product·· or two coprime monic polynomials over This is a contradiction. It remains to show now that c, ¢ m~. Suppose c,Em:. Therelationx",+cJx1-l+ ... +c,=O gives, in view of the fact that c, E m3, the relation

R./mo.!(X) is reducible over R o by Hensel's Lemma.

xi =

-(CIX~-l +...+ c,) E mum C Rpx ,

+ (x .....xd)R.

Going modulo (xl,,,,xd)R, wehavex{E Rp X,. Since x~ E R, write x'l = bo + b,xl +...+ b'_IX~-I. bl E RD. By minimality of the choice of r, t » rand henceX:E RpXl' Since Xl isa non-zero divisor of R, we have Xf-l E Rp, i.e. xi-I E moR, a contradiction to the choice of t. This completes the proof that f(X) is an Eisenstein polynomial over RD. Consider the ring homomorphism Ro[X] -+ Rgiven by

+:

;(X) = Xl'

+1 Ro=Id•

.It is surjective and induces a surjective ring homomorphism

REGULAR LOCAL lUNGS

RolX)

9i: (f(X» -+

259

Ro[X)

R.

Now

(f{X»

is an integral extension of R. and hence has dimension equal to dim Ro == dim R.· Thus ~ is an isomorphism.

9.4. 1.

2.

EXERCISES

Let (R, m) c (R', m') be complete local rings such that R' is integral over R and R'lm' is a separable extension of Rim. Show that for any coefficient ring S of R there exists a coefficient ring S' of R' containing S and that it is uniquely determint:d by S. Let R be a complete regular local ring with quotient field K and L a finite field extenslon of K. Show that the integral closure R' of R in'L is a finite R-module. Deduce that if R is a local domain whose completion has no non-zero nilpotent elements (analytically unramijied), the integral closure K of R in its quotient field is a finite R-module.

CHAPTER X

SOME CONJECTURES

The homology theory developed in Chapter VII was applied in the subsequent chapters for the study of dimension and regular local rings. We gave a homological characterisation of regular local rings and as a consequence obtained its UFD property. We also showed that R p is a regular local ring when R is regular local and P a prime ideal of R. In this chapter we give some more applications of homology in the solution of some outstanding conjectures in Commutative Algebra. 10.1. Big Cohen-Macaulay modules conjecture M. Hochster announced (Bull. Am. Math. Soc. Vol. 80, 1974) the following conjecture. Big Cohen-Macaulay modules conjecture: If R is a local ring and al • a l •.•.• a. a system of parameters of R. there exists an R-module M (not necessarily finitely generated) such that av al....,a. is an M-sequence. . The solution was given by M. Hochster in the case of equicharacteristic local rings. (Deep Local Rings: Preprint, Arhus University Preprint Series. 1973 or M. Hochster : Topics in the homological theory of modules over Commutative Rings, Regional Conference series in Maths. AMS, Vol. 24 (1975». We outline the proof in the case ch R = P > O. The idea of the proof is to start with the given ring R and modify it successively to kill all bad relations between all tit,... ,a.. By passing to the direct limit we get the required module. We start with some preliminaries. By a local ring we mean a Noetherian local ring and m always denotes its maximal ideal;

SOME CONJBCrURBS .

261

L t R be a local ring and ai' 0 1" .. ,a. a system of parameters (s.o.P.) f:r R, M an R-module, (not necessarily of finite type) and x E M. Defiuition:

A type k-relation for a1, • • •, a. on M is a tuple 1c+1

x

p= ( Xl,X2,'''' HI

) E MIe+1 such that 1; a,x, = 0 (0";; k ,-I

< n).

Given such a relation on M, we define a modification of (M, x) as follows. Let y = XIe+1 EB (al • ai' .... ale) EM EB Ric and M

,

MEBRIe

= --p:y- .

If x' is the image of x under the natural projection M --+ M', (M', x') is called a modification of (M, x), for ~. a2,.. ·,a~ of type .k (0 .:;;; k < n) and is denoted by the symbol (M, x) --+ (M , x'). Suppose now we have a sequence '. (M, x)

= (Mo. xo) --+

(M 1,

Xl)

--+...-7- (M"

X r)

is a modification of (M,. x,) with respect to a (M'+1. X) h were 1+1 ) • II d · P , on M , of type k,+1 (0 "" ~ i ~ r). Then (M" x, IS ca e reIanon an roth modification of (M, x) of type (k 1 • k l , · .. • kr) . Lemma 1: Let R be a local ring and a1, · ... a. an s.o.p. for R. If for every modification (M. x) of (R. 1), x ¢ (~, ... ,a?)M, R has ~ome e an R-module M for which a1, .. ·.a. IS an a-regu Iar module M • i ., M-seluence. Proof:

Choose a sequence of modifications of (R. 1)

(R. 1) = (M o; xo) --+ (MI' Xl) --+...--+ (M" X,) --+... satisfying the following conditions. (i) For each i, there is a finite generating set for type k tioll on M, (0 E;; k' < n) (R is Noetherian). . ..(li) "'For every i and k, (0 .:;;; k < n) and each generator p ~or .'.~ k-relation on M J • there exists j > i such that the relation , ;)til the passage (Mjo x}) --+ (M}+u x}+J is the image of the . .. ·p.under the natural map

.

8m

Let M =-- l o M, and I

.

y the image of 1 in M under the natural

262

COMMUTATIVE ALGEBRA

lim

mapMo=R -+ M= --+ M,. ;

. It is clear that fo~ each t, tbere exists j ;;> i such that the image of (a!•...•a,)M,: Ra'+I1D M J is contained in (010 r;•...• a,)Mj (O~ t< 1J). Hence (a1, · ...a,)M: Ral+1 = (a.,... ,o,)M. (0 ~ t -s 1J). To show that a.....,a. is an M-sequence. it is DOW sufficient to 0 )M' show that y ¢ (a1... ·,a.)M. Since-by assumption x rl. (a ~ .. • t IF HO··I" I or each I. we have by passing to the limit y ¢ (ah ... ,0.)M. Definition: An s.o .p .• a1• • ... a. for R is called amiable if there exists so~e. c E¥ R. non-nilpotent. such· that for all i (0 ~ i < n) and positive Integers t.

c«a~. a~, •.• a!} : Ra/H

) C

(of, a:.....aDR.

Lemma 2: Let Ro be an integrally closed. C.M. local domain and ·~n. s.?p. for Ro' If there exists a local domain R :::J Ra such that (I) R IS e f.g, Ro-module and (ii) the quotient .field K of R is a separa~le extension of the quotient field K o of RD. then a1....a. is an amiable s.a.p. for R. 01...

Proof: . Since KIKo is separable. K = Ko(oc), where oc can be chosen to be ID R. If f(X) is the minimum polynomial of oc over K f( X) has coefficients in R o as R o is integrally closed (Corollary 2 t~ Proposition 3. 4.3) and Ro[oc] =

t:ii~}

Since R o is C.M .• Ro[X] is C.M. ~Theorem 4. 8.5). and. si?ce fis a non-zero divisor of RCX]. Ro[oc] IS also C.M. By (I) R IS Integral over R o (Theorem I. 4.1) ~nd hence over RoCoc]. Thus RoCoc] is a local ring. Clearly a1• O2 , ....0,; IS an s.o.p, for RoCoc] as well as for R. Since R is a fg· Ra[oc] -module and. has the same quotient field as. Ratoc]. there exists some c E R - {O}.such that cR C Ro[a.]. We claim that 0 10 O2.....0. is an amiable S.D.p. for R with respect to c. Let

bE (ot. 0;.....01) : Ra/H (0.<;;; i

,

,

< n) so that

CbO'H = ~ (cbJ)aJ and cb, cbl> .... cb, E Ro[oc]. J-l

Since RJoc] is C.M. and at>

a...... a. is an s.a.p. for Ro[oc]. we have

cb E (o~. t4.....aDRo[oc] C (af,.... aDR and the proof is complete.

SOME CONJECTURES

263

Before proving the next Lemma. we recall some preliminaries on the Frobenius functor. LetR be a local ring of ch p > O. P prime and e a positive integer. The ring_homomorphism R -+ R given by a -+ a~' is called the Frobenius map. R considered as a R-algebra through this map is denoted by rR, For any R-module M. the module M' = M ®'R is also anR-module for the operation R

.

o(x ® b) = x ® ab, x E M. a. b E R. The ~orr~spondence M -: M' is a functor called Frobenius functor. It IS right exact, carnes R to R and changes the entries of the matrix of a map offree modules by raising them to the p·-th power. Such a functor is considered by C. Peskine and L. Szpiro (Dimension projective flnie et cohomologie locale. I.RE.S. Pub. Math. Vol. 42, (1973».

Lemma 3: Let R be a local ring of chp>O.p prime and all a 2 .....a. an amiable s.o.p, for R. Then there exists an a-regular module

M.

.\

Proof: Suppose there does not exist any a-regular module. By Lemma I, there exists a modification (M" x,) of (R, I) such that x, E (a.. .... a.)M,. We obtain a contradiction by applying the Frobenius functor to the sequence of modifications from (R. 1) to (M" x,). . . Since 0 1, a...... a. is an amiable s.o.p. for ~ •. th~re extsts some c E R, non-nilpotent, such that for all i and positive integers t • c«o~. a~

Since

n (of'.

ar

ol> : Rd,+l) C (a:. a~ .....a:)R. a:') R c n m'- = (0) and c' O. there exists

*

.. • • "ep~ I' some sufficiently large e such that c' ¢ (a1 • 02 , .... 0. ).

Consider the sequence of modifications (R. 1) = (Mo. x o) .-; (M.. xJ -+ ...-+ (M" x,) -+ ...-+(M" x,) and apply the Frobenius functor F with respect to the e chosen ahove. .. F is right exact and the natural map M -+ M' = FM given by . l® is quasi-linear with respect to the Frobenlus map 1i.-+tJI'. If(~. x;) is the image of (M" x,),,where x; is the image of x,UDder tho natural map M, -+ FM, = M;, we have a sequence ofm(ldificiltions (R. 1) = (R', 1') = (~. x~) -+ ...-+(M;. x;) -+ ... X;) where (M;+1. X;+l) is a modification of (M:. x;) with respect to the at', Since x, E (a...... 0n)M, we have

:-. 'x..+:

<M:. •

x

a.D.p.

ar.....e:

264 ,

COMMUTATIVE ALGI!BRA ~

e'i'

E (<11 .....a: )M,. We shall show that this leads to a contradiction by -proving the exlSlence of the following,commutative diagram

X,

" ) (Ma.xo -+(M1>x1)-+ ... -+(M"x,) I

1

+0

with.po = ld.

I

,

1

+1 c

+, c

1

(0)

c

-+ (R. I) -+

(R. I)

,

-+ (R. I)

Clearly. if such a diagram exists. we have a contradiction as

X; E implies c'

(af· ..... ar)

M;

= .p, (x~) E (af·.....a:") R

contradicting the choice of c. . The.existen~e of
PI c

c

I

'"

-+ ... -+ (R, I). We have to define PI+l: (M:+l , x;+!) -+ (R. I) such that (R. I)

(M;.

x;>

-+ (M;+!, X;tl)

< P I l l