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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( (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" .• 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 (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 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 '" :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 (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)
(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) ~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 ( 1. Then Q' n R = P contains 1Jm since a E Q/ and R is integrally closed in K. But P' n R = P does not contain bm since P' is prime and o(a) ¢ P' for all a E G. This is a contradiction. To complete the proof, we have to consider the case when G is infinite. Consider the set :E of all pairs (La, a,,) where L" is a normal extension of K and a" E G(LJK) such that La eLand a..(P' nL,,)=Q' nL". Introduce an order-s; in:E by defining (L,.,a..)
x;
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 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( 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
= (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 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' -- 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'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 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= 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) + fJo 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(-
--+
'"
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. , 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;;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 =
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.
HOMOLOGY
183
Proposition 5: Fo~ R-modul~s M and N, there exists an isomorphism Tor: (M, N) c: (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
:=
idRM
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 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) 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 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 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(M)-1 =dimM-l. ··'·'·"~L.· . aIM '. 1 M . . N
213
M
·Lt> 1. Then dim = dlmwhere N = ., . (ol.· ..a,)M a (a.,...a,)M
,\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" .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 .~ •
{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 ... , 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 . 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 (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
