INJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA


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1 INJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA These notes are intended to give the reader an idea what injective modules are, where they show up, and, to a small extent, what one can do with them. Let R be a commutative Noetherian ring with an identity element. An R module E is injective if Hom R (, E) is an exact functor. The main messages of these notes are Every Rmodule M has an injective hull or injective envelope, denoted by E R (M), which is an injective module containing M, and has the property that any injective module containing M contains an isomorphic copy of E R (M). A nonzero injective module is indecomposable if it is not the direct sum of nonzero injective modules. Every injective Rmodule is a direct sum of indecomposable injective Rmodules. Indecomposable injective Rmodules are in bijective correspondence with the prime ideals of R; in fact every indecomposable injective Rmodule is isomorphic to an injective hull E R (R/p), for some prime ideal p of R. The number of isomorphic copies of E R (R/p) occurring in any direct sum decomposition of a given injective module into indecomposable injectives is independent of the decomposition. Let (R, m) be a complete local ring and E = E R (R/m) be the injective hull of the residue field of R. The functor ( ) = Hom R (, E) has the following properties, known as Matlis duality: (1) If M is an Rmodule which is Noetherian or Artinian, then M = M. (2) If M is Noetherian, then M is Artinian. (3) If M is Artinian, then M is Noetherian. Any unexplained terminology or notation can be found in [1] or [3]. Matlis theory of injective modules was developed in the paper [4], and may also be found in [3, 18] and [2, 3]. These notes owe a great deal of intellectual debt to Mel Hochster, whose lecture notes are very popular with the organizers of this workshop. However, the organizers claim intellectual property of all errors here. 1. Injective Modules Throughout, R is a commutative ring with an identity element 1 R. All Rmodules M are assumed to be unitary, i.e., 1 m = m for all m M. 1
2 2 INJECTIVE MODULES Definition 1.1. An Rmodule E is injective if for all Rmodule homomorphisms ϕ : M N and ψ : M E where ϕ is injective, there exists an Rlinear homomorphism θ : N E such that θ ϕ = ψ. Exercise 1.2. Show that E is an injective Rmodule E if and only if Hom R (, E) is an exact functor, i.e., applying Hom R (, E) takes short exact sequences to short exact sequences. Theorem 1.3 (Baer s Criterion). An Rmodule E is injective if and only if every Rmodule homomorphism a E, where a is an ideal, extends to a homomorphism R E. Proof. One direction is obvious. For the other, if M N are Rmodules and ϕ : M E, we need to show that ϕ extends to a homomorphism N E. By Zorn s lemma, there is a module N with M N N, which is maximal with respect to the property that ϕ extends to a homomorphism ϕ : N E. If N N, take an element n N \N and consider the ideal a = N : R n. By hypothesis, the composite homomorphism a n N ϕ E extends to a homomorphism ψ : R E. Define ϕ : N + Rn E by ϕ (n + rn) = ϕ (n ) + ψ(r). This contradicts the maximality of ϕ, so we must have N = N. Exercise 1.4. Let R be an integral domain. An Rmodule M is divisible if rm = M for every nonzero element r R. (1) Prove that an injective Rmodule is divisible. (2) If R is a principal ideal domain, prove that an Rmodule is divisible if and only if it is injective. (3) Conclude that Q/Z is an injective Zmodule. (4) Prove that any nonzero Abelian group has a nonzero homomorphism to Q/Z. (5) If ( ) = Hom Z (, Q/Z) and M is any Zmodule, prove that the natural map M M is injective. Exercise 1.5. Let R be an Aalgebra. (1) Use the adjointness of and Hom to prove that if E is an injective Amodule, and F is a flat Rmodule, then Hom A (F, E) is an injective Rmodule. (2) Prove that every Rmodule can be embedded in an injective R module. Hint: If M is the Rmodule, take a free Rmodule F with a surjection F Hom Z (M, Q/Z). Apply ( ) = Hom Z (, Q/Z). Proposition 1.6. Let M 0 and N be Rmodules, and let θ : M N be a monomorphism. Then the following are equivalent: (1) Every nonzero submodule of N has a nonzero intersection with θ(m). (2) Every nonzero element of N has a nonzero multiple in θ(m). (3) If ϕ θ is injective for a homomorphism ϕ : N Q, then ϕ is injective.
3 INJECTIVE MODULES 3 Proof. (1) = (2) If n is a nonzero element of N, then the cyclic module Rn has a nonzero intersection with θ(m). (2) = (3) If not, then ker ϕ has a nonzero intersection with θ(m), contradicting the assumption that ϕ θ is injective. (3) = (1) Let N be a nonzero submodule of N, and consider the canonical surjection ϕ : N N/N. Then ϕ is not injective, hence the composition ϕ θ : M N/N is not injective, i.e., N contains a nonzero element of θ(m). Definition 1.7. If θ : M N satisfies the equivalent conditions of the previous proposition, we say that N is an essential extension of M. Example 1.8. If R is a domain and Frac(R) is its field of fractions, then R Frac(R) is an essential extension. More generally, if S R is the set of nonzerodivisors in R, then S 1 R is an essential extension of R. Example 1.9. Let (R, m) be a local ring and N be an Rmodule such that every element of N is killed by a power of m. The socle of N is the submodule soc(n) = 0 : N m. Then soc(n) N is an essential extension: if n N is a nonzero element, let t be the smallest integer such that m t n = 0. Then m t 1 n soc(n), and m t 1 n contains a nonzero multiple of n. Exercise Let I be an index set. Then M i N i is essential for all i I if and only if i I M i i I N i is essential. Example Let R = C[[x]] which is a local ring with maximal ideal (x), and take N = R x /R. Every element of N is killed by a power of the maximal ideal, and soc(n) is the 1dimensional Cvector space generated by [1/x], i.e., the image of 1/x in N. By Example 1.9, soc(n) N is an essential extension. However N soc(n) N N is not an essential extension since the element ( [1/x], [1/x 2 ], [1/x 3 ],... ) N does not have a nonzero multiple in N soc(n). (Prove!) Proposition Let L, M, N be nonzero Rmodules. (1) M M is an essential extension. (2) Suppose L M N. Then L N is an essential extension if and only if both L M and M N are essential extensions. (3) Suppose M N and M N i N with N = i N i. Then M N is an essential extension if and only if M N i is an essential extension for every i. (4) Suppose M N. Then there exists a module N with M N N, which is maximal with respect to the property that M N is an essential extension. Proof. The assertions (1), (2), and (3) elementary. For (4), let F = {N M N N and N is an essential extension of M}. N
4 4 INJECTIVE MODULES Then M F so F is nonempty. If N 1 N 2 N 3... is a chain in F, then i N i F is an upper bound. By Zorn s Lemma, the set F has maximal elements. Definition The module N in Proposition 1.12 (4) is a maximal essential extension of M in N. If M N is essential and N has no proper essential extensions, we say that N is a maximal essential extension of M. Proposition Let M be an Rmodule. The following conditions are equivalent: (1) M is injective; (2) M is a direct summand of every module containing it; (3) M has no proper essential extensions. Proof. (1) = (2) = (3) is left as an exercise, and we prove the implication (3) = (2). Consider an embedding M E where E is injective. By Zorn s lemma, there exists a submodule N E which is maximal with respect to the property that N M = 0. This implies that M E/N is an essential extension, and hence that it is an isomorphism. But then E = M + N so E = M N. Since M is a direct summand of an injective module, it must be injective. Proposition Let M and E be Rmodules. (1) If E is injective and M E, then any maximal essential extension of M in E is an injective module, hence is a direct summand of E. (2) Any two maximal essential extensions of M are isomorphic. Proof. (1) Let E be a maximal essential extension of M in E and let E Q be an essential extension. Since E is injective, the identity map E E lifts to a homomorphism ϕ : Q E. Since Q is an essential extension of E, it follows that ϕ must be injective. This gives us M E Q E, and the maximality of E implies that Q = E. Hence E has no proper essential extensions, and so it is an injective module by Proposition (2) Let M E and M E be maximal essential extensions of M. Then E is injective, so M E extends to a homomorphism ϕ : E E. The inclusion M E is an essential extension, so ϕ is injective. But then ϕ(e) is an injective module, and hence a direct summand of E. Since M ϕ(e) E is an essential extension, we must have ϕ(e) = E. Definition The injective hull or injective envelope of an Rmodule M is a maximal essential extension of M, and is denoted by E R (M). Definition Let M be an Rmodule. A minimal injective resolution of M is a complex 0 E 0 E 1 E 2... such that E 0 = E R (M), E 1 = E R (E 0 /M), and E i+1 = E R (E i / image(e i 1 )) for all i 2.
5 INJECTIVE MODULES 5 Note that the modules E i are injective, and that image(e i ) E i+1 is an essential extension for all i Injectives over a Noetherian Ring Proposition 2.1 (Bass). A ring R is Noetherian if and only if every direct sum of injective Rmodules is injective. Proof. We first show that if M is a finitely generated Rmodule, then Hom R (M, i N i ) = i Hom R (M, N i ). Independent of the finite generation of M, there is a natural injective homomorphism ϕ : i Hom R (M, N i ) Hom R (M, i N i ). If M is finitely generated, the image of a homomorphism from M to i N i is contained in the direct sum of finitely many N i. Since Hom commutes with forming finite direct sums, ϕ is surjective as well. Let R be a Noetherian ring, and E i be injective Rmodules. Then for an ideal a of R, the natural map Hom R (R, E i ) Hom R (a, E i ) is surjective. Since a is finitely generated, the above isomorphism implies that Hom R (R, E i ) Hom R (a, E i ) is surjective as well. Baer s criterion now implies that E i is injective. If R is not Noetherian, it contains a strictly ascending chain of ideals a 1 a 2 a Let a = i a i. The natural maps a R R/a i E R (R/a i ) give us a homomorphism a i E R(R/a i ). The image lies in the submodule i E R (R/a i ), (check!) so we have a homomorphism ϕ : a i E R (R/a i ). Lastly, check that ϕ does not extend to homomorphism R i E R (R/a i ). Theorem 2.2. Let E be an injective module over a Noetherian ring R. Then E = i E R (R/p i ), where p i are prime ideals of R. Moreover, any such direct sum is an injective Rmodule. Proof. The last statement follows from Proposition 2.1. Let E be an injective Rmodule. By Zorn s Lemma, there exists a maximal family {E i } of injective submodules of E such that E i = ER (R/p i ), and their sum in E is a direct sum. Let E = E i, which is an injective module, and hence is a direct summand of E. There exists an Rmodule E such that E = E E. If E 0, pick a nonzero element x E. Let p be an associated prime of Rx. Then R/p Rx E, so there is a copy of E R (R/p) contained in E and E = E R (R/p) E, contradicting the maximality of family {E i }. Definition 2.3. Let a be an ideal of a ring R, and M be an Rmodule. We say M is atorsion if every element of M is killed by some power of a.
6 6 INJECTIVE MODULES Theorem 2.4. Let p be a prime ideal of a Noetherian ring R, and let E = E R (R/p) and κ = R p /pr p, which is the fraction field of R/p. Then (1) if x R \ p, then E x E is an isomorphism, and so E = E p ; (2) 0 : E p = κ; (3) κ E is an essential extension of R p modules and E = E Rp (κ); (4) E is ptorsion and Ass(E) = {p}; (5) Hom Rp (κ, E) = κ and Hom Rp (κ, E R (R/q) p ) = 0 for primes q p. Proof. (1) κ is an essential extension of R/p by Example 1.8, so E contains a copy of κ and we may assume R/p κ E. Multiplication by x R\p is injective on κ, and hence also on its essential extension E. The submodule xe is injective, so it is a direct summand of E. But κ xe E are essential extensions, so xe = E. (2) 0 : E p = 0 : E pr p is a vector space over the field κ, and hence the inclusion κ 0 : E p splits. But κ 0 : E p E is an essential extension, so 0 : E p = κ. (3) The containment κ E is an essential extension of Rmodules, hence also of R p modules. Suppose E M is an essential extension of R p modules, pick m M. Then m has a nonzero multiple (r/s)m E, where s R \ p. But then rm is a nonzero multiple of m in E, so E M is an essential extension of Rmodules, and therefore M = E. (4) Let q Ass(E). Then there exists x E such that Rx E and 0 : R x = q. Since R/p E is essential, x has a nonzero multiple y in R/p. But then the annihilator of y is p, so q = p and Ass(E) = {p}. If a is the annihilator of a nonzero element of E, then p is the only associated prime of R/a E, so E is ptorsion. (5) For the first assertion, Hom Rp (κ, E) = Hom Rp (R p /pr p, E) = 0 : prp E = κ. Since elements of R\q act invertibly on E R (R/q), we see that E R (R/q) p = 0 if q p. In the case q p, we have Hom Rp (κ, E R (R/q) p ) = 0 : prp E R (R/q) p = 0 : prp E R (R/q). If this is nonzero, then there is a nonzero element of E R (R/q) killed by p, which forces q = p since Ass E R (R/q) = {q}. We are now able to strengthen Theorem 2.2 to obtain the following structure theorem for injective modules over Noetherian rings. Theorem 2.5. Let E be an injective over a Noetherian ring R. Then E = E R (R/p) αp, p Spec R and the numbers α p are independent of the direct sum decomposition.
7 INJECTIVE MODULES 7 Proof. Theorem 2.2 implies that a direct sum decomposition exists. Theorem 2.4 (5), α p is the dimension of the R p /pr p vector space Hom Rp (R p /pr p, E p ), which does not depend on the decomposition. The following proposition can be proved along the lines of Theorem 2.4, and we leave the proof as an exercise. Proposition 2.6. Let S R be a multiplicative set. (1) If E is an injective Rmodule, then S 1 E is an injective module over the ring S 1 R. (2) If M N is an essential extension (or a maximal essential extension) of Rmodules, then the same is true for S 1 M S 1 N over S 1 R. (3) The indecomposable injectives over S 1 R are the modules E R (R/p) for p Spec R with p S =. Definition 2.7. Let M be an Rmodule, and let E be a minimal injective resolution of R where E i = E R (R/p) µ i(p,m). p Spec R Then µ i (p, M) is the ith Bass number of M with respect to p. The following theorem shows that these numbers are welldefined. Theorem 2.8. Let κ(p) = R p /pr p. Then µ i (p, M) = dim κ(p) Ext i R p (κ(p), M p ). Proof. Let E be a minimal injective resolution of M where the i th module is E i = E R (R/p) µ i(p,m). Localizing at p, Proposition 2.6 implies that E p is a minimal injective resolution of M p over the ring R p. Moreover, the number of copies of E R (R/p) occurring in E i is the same as the number of copies of E R (R/p) in E i p. By definition, Ext i R p (κ(p), M p ) is the ith cohomology module of the complex 0 Hom Rp (κ(p), E 0 p) Hom Rp (κ(p), E 1 p) Hom Rp (κ(p), E 2 p)... and we claim all maps in this complex are zero. If ϕ Hom Rp (κ(p), E i p), we need to show that the composition κ(p) ϕ Ep i δ Ep i+1 is the zero map. If ϕ(x) 0 for x κ(p), then ϕ(x) has a nonzero multiple in image(ep i 1 Ep). i Since κ(p) is a field, it follows that ϕ(κ(p)) image(e i 1 p and hence that δ ϕ = 0. By Theorem 2.4 (5) E i p), Hom Rp (κ(p), E i p) = κ(p) µ i(p,m), By
8 8 INJECTIVE MODULES so Ext i R p (κ(p), M p ) is the ith cohomology module of the complex 0 κ(p) µ 0(p,M) κ(p) µ 1(p,M) κ(p) µ 2(p,M).... where all maps are zero, and the required result follows. Remark 2.9. We next want to consider the special case in which (R, m, K) is a Noetherian local ring. Recall that we have natural surjections... R/m 3 R/m 2 R/m 0, and that the madic completion R of R is the inverse limit of this system, i.e., { lim(r/m k ) = (r 0, r 1, r 2,... ) } R/m k r k r k 1 m k 1. k k Morally, elements of the aadic completion of R are power series of elements of R where higher terms are those contained in higher powers of the ideal a. There is no reason to restrict to local rings or maximal ideals for topological purposes, completions at other ideals can be very interesting; see, for example, [5]. Note that R/m k R = R/m k. Consequently if M is mtorsion, then the Rmodule structure on M makes it an Rmodule. In particular, E R (R/m) is an Rmodule. Theorem Let (R, m, K) be a local ring. Then E R (K) = E R(K). Proof. The containment K E R (K) is an essential extension of Rmodules, hence also of Rmodules. If ER (K) M is an essential extension of Rmodules, then M is mtorsion. (Prove!) If m M is a nonzero element, then Rm E R (K) 0. But Rm = Rm, so E R (K) M is an essential extension of Rmodules, which implies M = E R (K). It follows that E R (K) is a maximal essential extension of K as an Rmodule. Theorem Let ϕ : (R, m, K) (S, n, L) be a homomorphism of local rings such that ϕ(m) n, the ideal ϕ(m)s is nprimary, and S is modulefinite over R. Then Hom R (S, E R (K)) = E S (L). Proof. By Exercise 1.5, Hom R (S, E R (K)) is an injective Smodule. Every element of Hom R (S, E R (K)) is killed by a power of m and hence by a power of n. It follows that Hom R (S, E R (K)) is a direct sum of copies of E S (L), say Hom R (S, E R (K)) = E S (L) µ. To determine µ, consider Hom S (L, Hom R (S, E R (K))) = Hom R (L S S, E R (K)) = Hom R (L, E R (K)). The image of any element of Hom R (L, E R (K)) is killed by n, hence Hom R (L, E R (K)) = Hom R (L, K) = Hom K (L, K)
9 INJECTIVE MODULES 9 and L µ = Hom K (L, K). Considering vector space dimensions over K, this implies µ dim K L = dim K L, so µ = 1. Corollary Let (R, m, K) be a local ring and let S = R/a where a is an ideal of R. Then the injective hull of the residue field of S is Hom R (R/a, E R (K)) = 0 : ER (K) a. Since every element of E R (K) is killed by a power of m, we have E R (K) = t N (0 : ER (K) m t ) = t N E R/m t(k). This motivates the study of E R (K) for Artinian local rings. 3. The Artinian case Recall that the length of a module M is the length of a composition series for M, and is denoted l(m). The length is additive over short exact sequences. If (R, m, K) is an Artinian local ring, then every finitely generated Rmodule has a composition series with factors isomorphic to R/m. Lemma 3.1. Let (R, m, K) be a local ring. Then ( ) = Hom R (, E R (K)) is a faithful functor, and l(m ) = l(m) for every Rmodule M of finite length. Proof. Note that (R/m) = Hom R (R/m, E R (K)) = K. If M is a nonzero Rmodule, we need to show that M is nonzero. Taking a nonzero cyclic submodule R/a M, we have M (R/a), so it suffices to show that (R/a) is nonzero. The surjection R/a R/m yields (R/m) (R/a), and hence ( ) is faithful. For M of finite length, we use induction on l(m) to prove l(m ) = l(m). The result is true for modules of length 1 since (R/m) = K. For a module M of finite length, consider m soc(m) and the exact sequence 0 Rm M M/Rm 0. Applying ( ), we obtain an exact sequence 0 (M/Rm) M (Rm) 0. Since Rm = K and l(m/rm) = l(m) 1, we are done by induction. Corollary 3.2. Let (R, m, K) be an Artinian local ring. Then E R (K) is a finite length module and l(e R (K)) = l(r). Theorem 3.3. Let (R, m, K) be a Artinian local ring and E = E R (K). Then the map R Hom R (E, E), which takes a ring element r to the homomorphism multiplication by r, is an isomorphism. Proof. By the previous results, l(r) = l(e) = l(e ), so R and Hom R (E, E) have the same length, and it suffices to show the map is injective. If re = 0, then E = E R/Rr (K) so l(r) = l(r/rr), forcing r = 0.
10 10 INJECTIVE MODULES Theorem 3.4. Let (R, m, K) be a local ring. Then R is an injective R module if and only if the following two conditions are satisfied: (1) R is Artinian, and (2) soc(r) is 1dimensional vector space over K. Proof. If R = M N then K = (M R K) (N R K), so one of the two summands must be zero, say M R K = 0. But then Nakayama s lemma implies that M = 0. It follows that a local ring in indecomposable as a module over itself. Hence if R is injective, then R = E R (R/p) for some p Spec R. This implies R that is ptorsion and it follows that p is the only prime ideal of R and hence that R is Artinian. Furthermore, soc(r) is isomorphic to soc(e R (K)), which is 1dimensional. Conversely, if R is Artinian with soc(r) = K, then R is an essential extension of its socle. The essential extension K R can be enlarged to a maximal essential extension K E R (K). Since l(e R (K)) = l(r), we must have E R (K) = R. 4. Matlis duality Theorem 4.1. Let (R, m, K) be a local ring and let E = E R (K). Then E is also an Rmodule, and the map R Hom R (E, E), which takes an element r R to the homomorphism multiplication by r, is an isomorphism. Proof. Since E = E R(K), there is no loss of generality in assuming that R is complete. For integers t 1, consider the rings R t = R/m t. Then E t = 0 : E m t is the injective hull of the residue field of R t. If ϕ Hom R (E, E), then ϕ(e t ) E t, so ϕ restricts to an element of Hom Rt (E t, E t ), which equals R t by Theorem 3.3. The homomorphism ϕ, when restricted to E t, is multiplication by an element r t R t. Moreover E = t E t and the elements r t are compatible under restriction, i.e., r t+1 r t m t. Thus ϕ is precisely multiplication by the element (r 1 r 2 ) + (r 2 r 3 ) + R. Corollary 4.2. For a local ring (R, m, K), the module E R (K) satisfies the descending chain condition (DCC). Proof. Consider a descending chain of submodules E R (K) = E E 1 E Applying the functor ( ) = Hom R (, E) gives us surjections R = E E 1 E Since R is Noetherian, the ideals ker( R Et ) stabilize for large t, and hence Et Et+1 is an isomorphism for t 0. Since ( ) is faithful, it follows that E t = E t+1 for t 0. Theorem 4.3. Let (R, m, K) be a Noetherian local ring. The following conditions are equivalent for an Rmodule M. (1) M is mtorsion and soc(m) is a finitedimensional Kvector space;
11 INJECTIVE MODULES 11 (2) M is an essential extension of a finitedimensional Kvector space; (3) M can be embedded in a direct sum of finitely many copies of E R (K); (4) M satisfies the descending chain condition. Proof. The implications (1) = (2) = (3) = (4) follow from earlier results, so we focus on (4) = (1). Let x M. The descending chain Rx mx m 2 x... stabilizes, so m t+1 x = m t x for some t. But then Nakayama s lemma implies m t x = 0, and it follows that M is mtorsion. Since soc(m) is a vector space with DCC, it must be finitedimensional. Example 4.4. Let (R, m, K) be a discrete valuation ring with maximal ideal m = Rx. (For example, R may be a power series ring K[[x]] or the ring of padic integers Ẑp, in which case x = p.) We claim that E R (K) = R x /R. To see this, note that soc(r x /R) is a 1dimensional Kvector space generated by the image of 1/x R x, and that every element of R x /R is killed by a power of x. The next result explains the notion of duality in the current context. Theorem 4.5. Let (R, m, K) be a complete Noetherian local ring, and use ( ) to denote the functor Hom R (, E R (K)). (1) If M has ACC then M has DCC, and if M has DCC then M has ACC. Hence the category of Rmodules with DCC is antiequivalent to the category of Rmodules with ACC. (2) If M has ACC or DCC, then M = M. Proof. Let E = E R (K). If M with ACC, consider a presentation R m R n M 0. Applying ( ), we get an exact sequence 0 M (R n ) (R m ). Since (R n ) = E n has DCC, so does its submodule M. Applying ( ) again, we get the commutative diagram with exact rows (R m ) (R n ) M 0 R m R n M 0. Since R R is an isomorphism, it follows that M M is an isomorphism as well. If M has DCC, we embed it in E m and obtain an exact sequence 0 M E m E n. Applying ( ) gives an exact sequence (E n ) (E m ) M 0. The surjection R n = (E m ) M shows that M has ACC, while a similar commutative diagram gives the isomorphism M = M.
12 12 INJECTIVE MODULES Remark 4.6. Let M be a finitely generated module over a complete local ring (R, m, K). Then Hom R (K, M ) = Hom R (K R M, E R (K)) = Hom R (M/mM, E R (K)) = Hom K (M/mM, K), so the number of generators of M as an Rmodule equals the vector space dimension of soc(m ). References [1] M. F. Atiyah and I. G. Macdonald, Introduction to commutative algebra, Addison Wesley Publishing Co., Reading, Mass.LondonDon Mills, Ont., [2] W. Bruns and J. Herzog, CohenMacaulay rings, Cambridge Studies in Advanced Mathematics 39, Cambridge University Press, Cambridge, [3] H. Matsumura, Commutative algebra, W. A. Benjamin, Inc., New York, [4] E. Matlis, Injective modules over Noetherian rings, Pacific J. Math. 8 (1958) [5] R. Hartshorne, On the de Rham cohomology of algebraic varieties, Inst. Hautes Études Sci. Publ. Math. 45 (1975), 5 99.
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