A COURSE IN HOMOLOGICAL ALGEBRA CHAPTER **: GORENSTEIN RINGS & MODULES E. L. Lady (March 13, 1998)

References: Kaplansky, Commutative Rings (First Edition), Chapter 4, Sections 4 through 6. H. Bass, The Ubiquity of Gorenstein Rings, Math Zeitschrift 82 (1963) 8–28. Sharp, Math Zeitschrift 112(1969), 340–56, Math Zeitschrift 115(1970), 117-39. Important: Kaplansky, Commutative Rings (Second Edition), Chapter 4, Sections 5 and 6, pp. 156–168.

The second edition of Kaplansky contains proofs for some of these results simpler than those given here. Unfortunately, Kaplansky uses the word “grade” to mean depth, conflicting with Bass’s more standard use of that word (as indicated below).

Assume throughout that R is a commutative noetherian ring. E(M) denotes the injective envelope of an R-module M.

Lemma. Let R be a noetherian local ring with maximal ideal 𝔪. Let E be the injective envelope of R/𝔪. Then the functor Hom_{R}( , E) restricted to the category of finite-length R-modules preserves length. (Kaplansky, Second Edition, Theorem 220.) Conversely, if M is an R-module such that Hom_{R}(M, E) has finite length, then M has finite length. proof: Consider first a module S with length 1 (i. e. a simple module). Then S ≈ R/𝔪 (so E is isomorphic to the injective envelope of S). Now if φ ∈ Hom_{R}(S, E), then φ(S) is simple (unless trivial) and φ(S) ∩ (R/𝔪) ≠ 0 since E is an essential extension of R/𝔪. Thus φ(S) = R/𝔪, since both these modules are simple. From this we see that Hom_{R}(S, E) ≈ Hom_{R}(R/𝔪, R/𝔪) ≈ R/𝔪, so that length Hom_{R}(S, E) = 1. Since the functor Hom_{R}( , E) preserves (or, more precisely, reverses) short exact sequences, it now follows by induction that for all R-modules M with finite length, length Hom_{R}(M, E) = length M.

Now suppose that M is an R-module such that Hom_{R}(M, E) has finite length. Consider first the case where M is finitely generated. M contains submodules N such that M/N has finite length (for instance, N = M). By the preceding, for such an N, Hom_{R}(M/N, E) has the same length as M/N. But the surjection M → M/N induces a monomorphism Hom_{R}(M/N, E) → Hom_{R}(M, E), so length M/N ≤ length Hom_{R}(M, E). Thus there is a bound on the length of M/N for those N such that M/N has finite length. But if M/N has finite length and N ≠ 0, then since N is noetherian, it contains a maximal proper submodule N′ and length M/N′ = 1 + length M/N. Therefore eventually we must have N = 0, so that M has finite length.

If M is not finitely generated and Hom_{R}(M, E) has finite length, consider a finitely generated submodule M′ of M. The inclusion map from M′ to M induces a surjection Hom_{R}(M, E) → Hom_{R}(M′, E) (because E is injective). Therefore Hom_{R}(M′, E) has finite length and so M′ has finite length by the preceding paragraph, and in fact the lengths of the finitely generated submodules of M are all bounded by length Hom_{R}(M, E). Furthermore, if M′′ is also finitely generated and M′ ⊊ M′′ then length M′′ > length M′. It follows that there exists a maximally finitely generated submodule of M. But clearly this is only possible if this submodule is M itself. Therefore M is finitely generated and hence, by the preceding paragraph, has finite length. ✓

Theorem. Let R be a commutative noetherian local ring and let 𝔪 be its unique maximal prime ideal. The following conditions are equivalent:

(1) R is artinian.

(2) R has finite length (as an R-module).

(3) 𝔪 is the only prime ideal in R.

(4) Ass R = {𝔪}.

(5) 𝔪^{k} = 0 for some positive integer k.

(6) The injective envelope E of R/𝔪 is finite generated.

(7) There exists a finitely generated injective R-module. proof: (2) ⇒ (1): Clear.

(1) ⇒ (4): If R is artinian and 𝔭 ∈ Ass R, then R contains a copy of R/𝔭. Therefore R/𝔭 is artinian. But R/𝔭 is an integral domain, and an artinian integral domain is a field. Therefore 𝔭 is a maximal ideal, so 𝔭 = 𝔪.

(4) ⇒ (5): If Ass M = {𝔪} then for some k, 𝔪^{k}1 = 0 ∈ R so 𝔪^{k} = 0.

(5) ⇒ (2): For each i from 0 to k, 𝔪^{i}/𝔪^{i+1} is finitely generated and a vector space over the field R/𝔪, hence is a finite-length R-module. Thus by induction, R/𝔪^{i} has finite length. If 𝔪^{k} = 0, this shows that R has finite length.

(5) ⇒ (3): If 𝔪^{k} = 0 and 𝔭 is a prime ideal, then 𝔪^{k} ⊆ 𝔭 so 𝔪 ⊆ 𝔭. Since 𝔪 is maximal, 𝔭 = 𝔪.

(3) ⇒ (4): Clear since Ass R ≠ {∅}.

(2) ⇔ (6): If E = E(R/𝔪) is finitely generated, then it has finite length, since all its associated primes are maximal. And conversely, if it has finite length then it is certainly finitely generated. But E ≈ Hom_{R}(R, E). Therefore the preceding lemma implies that E has finite length if and only if R has finite length.

(6) ⇒ (7): Clear.

(7) ⇒ (6): Let M be a non-trivial finitely generated injective module and let 𝔭 ∈ Ass M. Then M contains a copy of R/𝔭. Since R/𝔭 is an essential submodule of R_{𝔭}/𝔭R_{𝔭}, M therefore contains a submodule isomorphic to R_{𝔭}/𝔭R_{𝔭}. Therefore R_{𝔭}/𝔭R_{𝔭} is finitely generated. Now let x ∉ 𝔭. Then multiplication by x is an isomorphism on R_{𝔭}/𝔭R_{𝔭}. If x ∈ 𝔪, this contradicts Nakayama’s Lemma. Therefore x ∉ 𝔪 and we conclude that 𝔭 = 𝔪. This shows that 𝔪 ∈ Ass M, and so M contains a submodule isomorphic to R/𝔪. Since M is injective, it thus contains a copy of E = E(R/𝔪). Therefore E is finitely generated.

(6) ⇒ (4) [Alternate proof]: Since Ass E = {𝔪}, it follows that for each e ∈ E, 𝔪^{k}e = 0 for some k. Thus if E is finitely generated, then 𝔪^{k}E = 0 for some k. It now suffices to show that E is faithful in order to conclude that 𝔪^{k} = 0. Suppose in fact, by way of contradiction, that rE = 0 for some r ≠ 0 ∈ R. The principal ideal (r) is isomorphic to R/𝔞, where 𝔞 = ann r. Since 𝔞 ⊆ 𝔪, there is a surjection (r) → R/𝔪 (so φ(r) ≠ 0). Since E = E(R/𝔪), this extends to a map φ: R → E. Then φ(r) = rφ(1) = 0 since rE = 0. This is a contradiction, showing that no such r exists and therefore E is faithful. Since 𝔪^{k}E = 0, it follows that 𝔪^{k} = 0. ✓

Definition. An submodule N of an R-module M is called irreducible if it is not the intersection of two submodules which properly contain it.

Recall the following characterization of indecomposable injective modules.

Lemma. If M is an injective module over a commutative noetherian ring R, the following conditions are equivalent:

(1) M is indecomposable.

(2) M is the injective envelope of every non-trivial submodule.

(3) The zero submodule is irreducible in M.

(4) For some prime ideal 𝔭, Ass M = {𝔭} and M is isomorphic to the injective envelope of 𝔭,

Proposition. If E is an injective module, then E has a direct summand isomorphic to R/𝔭 if and only if 𝔭 ∈ Ass E. proof: Ass E includes 𝔭 if and only if E contains a submodule isomorphic to R/𝔭. If this is the case, the it will also contain a submodule isomorphic to the injective envelope of E, which will necessarily be a direct summand since it is injective. ✓

Corollary. A local ring R is injective as an R-module if and only if R is artinian and the zero ideal is irreducible in R. Furthermore, in this case, R ≈ E(R/𝔪). proof: Since R is a finitely generated R-module, if it is injective then R is artinian by the Theorem above and Ass R = {𝔪}. Thus R contains a submodule (ideal) isomorphic to R/𝔪, and so if R is injective, it contains a submodule isomorphic to E = E(R/𝔪). But E ≈ Hom_{R}(R, E), so by a lemma above, E and R have the same length. Thus if R is injective, it is isomorphic to E(R/𝔪). Then by the lemma above, the zero ideal is irreducible in R.

Conversely, if R is artinian then it contains a minimal non-zero ideal I, which must be isomorphic to R/𝔪. If moreover the zero ideal in R is irreducible, then R is an essential extension of I, so R is isomorphic to a submodule of E(I) ≈ E(R/𝔪). But, as in the previous paragraph, we see that E(R/𝔪) and R have the same length. Thus R = E(I), hence R is injective. ✓

Definition. If 𝔭 is a prime ideal in R, we write k(𝔭) to denote R_{𝔭}/𝔭R_{𝔭}. (This is the quotient field of R/𝔭.)

Definition. A minimal injective resolution of a module M is a complex

0 ⟶^{d_{−1}} M ⟶ E_{0} ⟶^{d_{0}} E_{1} ⟶^{d_{1}} …

such that each E_{i} is the injective envelope of d_{i−1}(E_{i−1}).

Minimal injective resolutions are unique up to isomorphism. Since R is noetherian, each injective module E_{i} is a direct sum (often infinite) of indecomposable injective modules. By the preceding lemma, each such indecomposable summand is isomorphic to E(R/𝔭), for some prime ideal 𝔭.

Notation. If in a given decomposition of E_{i} as a direct sum of indecomposable modules, k of these indecomposable modules are isomorphis to E(R/𝔭), we write μ_{i}(𝔭, M) = k. We also write μ_{i}(𝔭) to denote μ_{i}(𝔭, R).

It is easy to see that μ_{i}(𝔭, M) is the largest integer k such that E_{i} contains a submodule isomorphic to (R/𝔭)^{k}. In particular, we have the following.

Note. μ_{0}(𝔭, M) = dim_{k(𝔭)} Hom_{R_{𝔭}}(k(𝔭), M_{𝔭}).

Proposition. The injective dimension of an R-module M is the smallest positive integer n such that Ext_{R}^{n+1}(R/𝔞, M) = 0 for every ideal 𝔞 of R.

Lemma. For any multiplicative set S, S^{−1}E(M) is the injective envelope of S^{−1}M as an S^{−1}R-module.

Corollary. Minimal injective resolutions localize. Therefore inj. dim_{S^{−1}R} S^{−1}M ≤ inj. dim_{R} M and inj. dim M is the supremum of inj. dim_{R_{𝔭}} M_{𝔭} over all maximal ideals 𝔭 of R.

Corollary. If 𝔭 does not blow up in in S^{−1}R then μ_{i}(S^{−1}𝔭, S^{−1}M) = μ_{i}(𝔭, M).

Lemma. μ_{i}(𝔭, M) = dim_{k(𝔭)} Ext_{R_{𝔭}}^{i}(k(𝔭), M_{𝔭}) = dim_{k(𝔭)} Ext_{R}^{i}(R/𝔭, M)_{𝔭}. proof: Localizing, we may suppose that 𝔭 is maximal. Then k(𝔭) = R/𝔭 is a simple module, and Hom_{R}(R/𝔭, E(R/𝔮)) = 0 for 𝔮 ≠ 𝔭, since 𝔭 ∉ Ass R/𝔮 = {𝔮}. Furthermore Hom_{R}(R/𝔭, E(R/𝔭) ≈ Hom_{R}(R/𝔭, R/𝔭) ≈ R/𝔭, since E(R/𝔭) is an essential extension of the simple module R/𝔭. It then follows that μ_{i}(𝔭, M) = dim_{k(𝔭)} Hom_{R}(R/𝔭, E_{i}) = dim_{k(𝔭)} Hom_{R}(k(𝔭), E_{i}).

Now if φ ≠ 0 ∈ Hom_{R}(k(𝔭), E_{i}), then φ(k(𝔭)) is a simple submodule of E_{i} and since E_{i} is an essential extension of d_{i−1}(E_{i−1}), it follows that φ(k(𝔭)) ⊆ d_{i−1}(E_{i−1}). Therefore d_{i}φ(k(𝔭)) = 0. This shows that the complex

… ⟶^{d_{i−1}^{∗}} Hom_{R}(k(𝔭), E_{i}) ⟶^{d_{i}^{∗}} Hom_{R}(k(𝔭), E_{i+1}) ⟶^{d_{i+1}^{∗}} …

has trivial differentiation. Therefore Ext_{R}^{i}(k(𝔭), M_{𝔭}) ≈ Hom_{R}(k(𝔭), E_{i}), and

μ_{i}(𝔭, M) = dim_{k(𝔭)} Hom_{R}(k(𝔭), E_{i}) = dim_{k(𝔭)} Ext_{R}^{i}(k(𝔭), M_{𝔭})_{𝔭}. ✓

Corollary. If M is finitely generated, then μ_{i}(𝔭, M) < ∞.

Hauptlemma Let 𝔮 ⊊ 𝔭 be adjacent primes and let M be finitely generated. If μ_{k}(𝔮, M) ≠ 0, then μ_{k+1}(𝔭, M) ≠ 0. (In other words, if 𝔮 ∈ Ass E_{k} then 𝔭 ∈ Ass E_{k+1} .) proof: We may suppose that R is local with maximal ideal 𝔭. Let k = R/𝔭.

(1) We claim that there exists a finite length R-module C such that Ext_{R}^{k+1}(C, M) ≠ 0. In fact, let B = R/𝔮 and C = B/xB = R/(𝔮, x), where x is an element of 𝔭 not belonging to 𝔮. Then x is regular on B, so we have an exact sequence

0 ⟶ B ⟶^{x} B ⟶ C ⟶ 0,

which induces a long exact sequence, part of which is

… ⟶ Ext_{R}^{k}(B, M) ⟶^{x} Ext_{R}^{k}(B, M) ⟶ Ext_{R}^{k+1}(C, M) ⟶ …

Now Ext_{R}^{k}(B, M) ≠ 0 since μ_{k}(𝔮, M) ≠ 0, and by Nakayama’s Lemma, multiplication by x is not surjective, so Ext_{R}^{k+1}(C, M) ≠ 0.

Now among such C, choose one with smallest possible length. We easily see that in fact, C = k. Thus Ext_{R}^{k+1}(k, M) ≠ 0, so that μ_{k+1}(𝔭, M) ≠ 0. ✓

Corollary. Let M be a finitely generated R-module with finite injective dimension r. Let 𝔭 ∈ Supp M. Then

(1) height_{M} 𝔭 ≤ inj. dim M_{𝔭}.

(2) If 𝔭 is a prime ideal such that μ_{r}(𝔭, M) ≠ 0, then 𝔭 is maximal. proof: (1) If 𝔮 ⊆ 𝔭 and 𝔮 is minimal in Supp M, then 𝔮 ∈ Ass M_{𝔭} ⊆ Ass E(M_{𝔭}), so μ_{0}(𝔮, M_{𝔭}) ≠ 0. Now use induction and the Hauptlemma to see that if height_{M} 𝔭 = n then μ_{n}(𝔭, M_{𝔭}) ≠ 0, so that n ≤ inj. dim M_{𝔭}.

(2) Clear from the Hauptlemma. ✓ Remark. In particular, inj. dim R is at least as large as the Krull dimension of R.

Corollary. If R is local with maximal ideal 𝔪 and M is an R-module with finite injective dimension, then inj. dim M is the largest integer r such that μ_{r}(𝔪, M) ≠ 0. proof: Clearly μ_{i}(𝔪, M) = 0 for i > inj. dim M. And if r = inj. dim M, then E_{r} ≠ 0 so μ_{r}(𝔭, M) ≠ 0 for some prime 𝔭. Since R is local, by the preceding corollary 𝔭 = 𝔪. ✓

At this point, we recall a theorem from the chapter on projective dimension.

Theorem. Let (R, 𝔪) be local and x_{1}, …, x_{r} a regular M-sequence. Then

proj. dim M/(x_{1}, …, x_{r})M = r + proj. dim M.

In particular, if I is an ideal generated by a regular R-sequence of length s, then proj. dim R/I = s.

Corollary. Let R be local with maximal ideal 𝔪. Let M be a finitely generated R-module. Then either inj. dim M = depth_{𝔪} R or inj. dim M = ∞. proof: Assume that M has finite injective dimension. Let I be an ideal generated by a maximal regular R-sequence in 𝔪. By the theorem above, depth_{𝔪} R = depth_{I} R = proj. dim R/I. Therefore it suffices to prove that inj. dim M = proj. dim R/I. Now since M is finitely generated, it has a maximal proper submodule M′ and M/M′ ≈ k, so there is a surjection M → k. Now if s = proj. dim R/I, then we have seen previously that Ext_{R}^{s}(R/I, k) ≠ 0. Furthermore, the functor Ext_{R}^{s}(R/I, ) is right exact. Therefore Ext_{R}^{s}(R/I, M) ≠ 0 and depth_{𝔪} R = proj. dim R/I can be characterized as the largest integer s such that Ext_{R}^{s}(R/I, M) ≠ 0.

Now let r = inj. dim M. Then Ext_{R}^{r}( , M) is right exact. Now μ_{r}(𝔭, M) ≠ 0 for some prime ideal 𝔭, and from the Hauptlemma we see that 𝔭 must be maximal, i. e. 𝔭 = 𝔪. This shows that that Ext_{R}^{r}(k, M) ≠ 0 (see a lemma above). Now since I is generated by a maximal regular R-sequence in 𝔪, 𝔪 consists of zero divisors on R/I, so 𝔪 ∈ Ass R/I, and therefore R/I contains a copy of k = R/𝔪, and it follows from the right exactness of Ext_{R}^{r}( , M) that Ext_{R}^{r}(R/I, M) ≠ 0. From this it follows that inj. dim M can be characterized the largest integer r such that Ext_{R}^{r}(R/I, M) ≠ 0. But comparing this with the characterization of proj. dim R/I in the paragraph above, we that inj. dim M = proj. dim R/I = depth_{𝔪} R. ✓

We now recall two more theorems from the chapter on projective dimension.

Theorem. Let R be a noetherian ring and M a finitely generated R-module and 𝔞 an ideal such that 𝔞M ≠ M. If r is the smallest integer such that Ext_{R}^{r}(R/𝔞, M) ≠ 0, then every regular M-sequence in 𝔞 has length r. Thus depth_{𝔞} M = r.

Theorem. If x_{1}, …, x_{r} is a regular M-sequence in 𝔞 and r = depth_{𝔞} M, then

Ext_{R}^{r}(R/𝔞, M) ≈ Hom_{R}(R/𝔞, M/(x_{1}, …, x_{r})M).

Corollary. If 𝔭 ∈ Supp M, then depth_{𝔭} M ≤ depth_{𝔭R_{𝔭}} M_{𝔭} = depth_{𝔭} M_{𝔭}. If depth_{𝔮} M > depth_{𝔭} M for every prime 𝔮 strictly containing 𝔭 (for instance if 𝔭 = 𝔪), then depth_{𝔭} M = depth_{𝔭} M_{𝔭}. proof: Since Ext_{R_{𝔭}}^{r}(R_{𝔭}/𝔭R_{𝔭}, M_{𝔭}) ≈ Ext_{R}^{r}(R/𝔭, M)_{𝔭}, it follows from the first of the two theorems above that depth_{𝔭} M ≤ depth_{𝔭R_{𝔭}} M_{𝔭}.

Now if s = depth_{𝔭} M and I is an ideal generated by a maximal regular M-sequence in 𝔭, then by the second theorem above, Ext_{R}^{s}(R/𝔭, M)_{𝔭} = Hom_{R}(R/𝔭, M/IM)_{𝔭}. By the maximality of I, 𝔭 consists of zero divisors on M/IM, so that 𝔭 is contained in some prime 𝔮 in Ass M/IM. If depth_{𝔮} M > depth_{𝔭} M for every prime ideal 𝔮 strictly containing 𝔭, then 𝔮 ∉ Ass M/IM and it follows that 𝔭 ∈ Ass M/IM. Then there is a monomorphism φ: R/𝔭 → M/IM. Then φ_{𝔭} ≠ 0 ∈ Hom(R/𝔭, M/IM)_{𝔭} (see the second theorem above), so that Ext_{R}^{s}(R_{𝔭}, M)_{𝔭} ≠ 0 and so s = depth_{𝔭} M ≥ depth_{𝔭R_{𝔭}} M_{𝔭}. Thus depth_{𝔭} M = depth_{𝔭} M_{𝔭}. ✓

Corollary. For 𝔭 ∈ Supp M, depth_{𝔭} M_{𝔭} is the smallest integer r such that μ_{r}(𝔭, M) ≠ 0. proof: It was shown above that μ_{i}(𝔭, M) = dim_{k(𝔭)} Ext_{R}^{i}(R/𝔭, M)_{𝔭}. and also that depth_{𝔭} M_{𝔭} is the smallest i such that Ext_{R}^{i}(R/𝔭, M)_{𝔭} ≠ 0. Thus depth_{𝔭} M_{𝔭} is the smallest i such that

μ_{i}(𝔭, M) ≠ 0. ✓

In an earlier corollary it was shown that if M_{𝔭} has finite injective dimension, then inj. dim M_{𝔭} is the largest r such that μ_{r}(𝔭, M) ≠ 0. Another corollary stated that if M has finite injective dimension, then inj. dim M_{𝔭} = depth_{𝔭} M_{𝔭}. Combining these observations with the corollary above and the standard fact that depth_{𝔭} M_{𝔭} ≤ height_{M} 𝔭 and the fact that height_{M} 𝔭 ≤ inj. dim M_{𝔭} (because of the Hauptlemma), we get the following result.

Corollary. If 𝔭 is a prime in Supp M then depth_{𝔭} M_{𝔭} ≤ height_{M} 𝔭 ≤ inj. dim M_{𝔭}, with equality if M_{𝔭} has finite injective dimension.

Definition. An R-module M is a Gorenstein module if depth_{𝔭} M = inj. dim M_{𝔭} for all primes 𝔭 in Supp M. Equivalently (see the preceding corollary), M is Gorenstein if and only if M has finite injective dimension. If R is a Gorenstein module, then we call R a Gorenstein ring.

Lemma. For M to be a Gorenstein module, it suffices that depth_{𝔭} M = inj. dim M_{𝔭} for all maximal ideals 𝔭 in Supp M.

Theorem. If R is a ring with finite Krull dimension (for instance if R is local [and noetherian]), then R is a Gorenstein ring if and only if R has finite injective dimension. Furthermore, in this case the injective dimension of R is the same as its Krull dimension. proof: If R is a Gorenstein ring, then inj. dim R_{𝔪} = depth_{𝔪} R = height 𝔪 < ∞ for all maximal ideals 𝔪. Since the injective dimension of R is the supremum of inj. dim R_{𝔪} for all maximal ideals 𝔪, it follows that if R has finite Krull dimension then it also has finite injective dimension.

Conversely, if R has finite injective dimension, then so does R_{𝔭} for all prime ideals 𝔭. By a result above, inj. dim R_{𝔭} = depth_{𝔭} R = height 𝔭. Thus R is a Gorenstein ring and since inj. dim R is the maximum of inj. dim R_{𝔭} over all prime ideals 𝔭, thus inj. dim R = Krull dim R. ✓

Corollary. If 𝔭 and 𝔮 are prime ideals with 𝔮 ⊆ 𝔭 and if R_{𝔭} is a Gorenstein ring, then so is R_{𝔮}. proof: Since injective resolutions localize, if R_{𝔭} has finite injective dimension than so does R_{𝔮}. ✓

Note. A not necessarily commutative noetherian ring Λ has finite injective dimension if and only all Λ-modules with finite projective dimension also have finite injective dimension. proof: If Λ-modules with finite projective dimension also have finite injective dimension, then in particular Λ, which is projective, has finite injective dimension. On the other hand, suppose that Λ has finite injective dimension. Then all projective Λ-modules have finite injective dimension, and it follows by induction that all Λ-modules with finite projective dimension have finite injective dimension. ✓

Theorem. Let 𝔭 be a prime in Supp M. The following conditions are equivalent:

(1) depth_{𝔭R_{𝔭}} M_{𝔭} = inj. dim M_{𝔭}.

(2) μ_{i}(𝔭, M) ≠ 0 only when i = height_{M} 𝔭.

(3) M_{𝔭} is a Gorenstein module. proof: As seen previously, depth_{𝔭} M is the smallest integer r such that μ_{r}(𝔭, M) ≠ 0 and inj. dim M is the largest r such that μ_{r}(𝔭, M) ≠ 0 and depth_{𝔭} M_{𝔭} ≤ height_{M} 𝔭 ≤ inj. dim M_{𝔭}. This shows that (1) and (2) are equivalent. Furthermore, the lemma above asserts that M_{𝔭} is a Gorenstein module if and only if depth_{𝔭} M_{𝔭} = inj. dim M_{𝔭}, i. e. (1) and (3) are equivalent. ✓

Definition. An R-module M is Cohen-Macauley if depth_{𝔭} M = height_{M} 𝔭 for all primes 𝔭 in Supp M.

Proposition. Gorenstein modules are Cohen-Macauley. proof: See the previous theorem. ✓

Proposition. If M is Cohen-Macauley then Ass M consists of the minimal primes in Supp M. proof: A prime ideal 𝔭 ∈ Supp M belongs to Ass M if and only if it consists of zero divisors on M, i. e. if and only if depth_{𝔭} M = 0. If M is Cohen-Macauley, then these are precisely the prime ideals with height 0, i. e. the minimal primes in Supp M. ✓

Proposition. M is Cohen-Macauley if and only if for all 𝔭 ∈ Supp M, μ_{i}(𝔭, M) = 0 for i < height_{M} 𝔭. proof: As seen previously, depth_{𝔭} M is the smallest i such that μ_{i}(𝔭, M) ≠ 0. ✓ Corollary. For M to be Cohen-Macauley, it suffices that depth_{𝔪} M = height_{M} 𝔪 for all maximal ideals 𝔪 ∈ Supp M.

Corollary. If M is Cohen-Macauley then Supp M is catenary.

Lemma. If R_{𝔭} does not have finite injective dimension then μ_{i}(𝔭, M) ≠ 0 for all i ≥ height 𝔭. proof: If there exists a prime 𝔮 properly contained in 𝔭 such that inj. dim R_{𝔮} = ∞, then this follows by induction from the Hauptlemma. On the other hand, if inj. dim R_{𝔮} < ∞ for all 𝔮 < 𝔭, then inj. dim R_{𝔮} = height 𝔮. So if i ≥ height 𝔭 and E_{i} is the ith term in the injective resolution of R, then (E_{i})_{𝔮} = 0 for all 𝔮 ⊊ 𝔭, i. e. μ_{i}(𝔮) = 0. But (E_{i})_{𝔭} ≠ 0, so we conclude that μ_{i}(𝔭) ≠ 0. ✓

Lemma. Suppose that x_{1} …, x_{n} is both a regular R-sequence and a regular M-sequence. Let I be the idea generated by x_{1}, …, x_{n}. Let E be a minimal injective resolution of M and let E′_{i} = Hom_{R}(R/I, E_{i+n}). Then E′ is a minimal injective resolution for the R/I-module M/IM. proof: E′_{i} is injective since if X is an R/I-module, then there is a natural isomorphism

Hom_{R/I}(X, E′_{i}) = Hom_{R/I}(X, Hom_{R}(R/I, E_{i+n})) ≈ Hom_{R}(X, E_{i+n}),

showing that Hom_{R/I}( , E′_{i}) is an exact functor.

By induction, it suffices to consider the case where n = 1 and x_{1} is regular on both R and M. Then proj. dim R/(x) = 1, so Ext_{R}^{i}(R/(x), M) = 0 for i > 1. Also Ext_{R}^{1}(R/(x), M) = M/xM. Thus

0 → M/xM → E′_{1} → …

is exact. ✓

Corollary. If I is the ideal generated by a regular R-sequence contained in 𝔭 which is also a regular M-sequence, then μ_{i}(𝔭/I, M/IM) = μ_{i+s}(𝔭, M). In particular, if M is a Gorenstein [Cohen-Macauley] R-module, then M/IM is a Gorenstein [Cohen-Macauley] R/I-module. Conversely, if M/IM is a Gorenstein module then so is M_{𝔭}.

Theorem. For any prime ideal 𝔭 of R, the following statements are equivalent:

(1) R_{𝔭} is a Gorenstein ring.

(2) height 𝔭 = depth_{𝔭R_{𝔭}} R_{𝔭} = inj. dim R_{𝔭}.

(3) R_{𝔭} has finite injective dimension.

(4) μ_{i}(𝔭) = 0 for all i > height 𝔭.

(5) μ_{i}(𝔭) = 0 for at least one i > height 𝔭.

(6) μ_{i}(𝔭) = 0 for all i < height 𝔭 and μ_{h}(𝔭) = 1 for h = height 𝔭.

(7) For all primes 𝔮 ⊆ 𝔭, μ_{i}(𝔮) = 0 for i ≠ height 𝔮 and μ_{i}(𝔮) = 1 for i = height 𝔮. proof: We have seen the equivalence of (1), (2), and (3) above. Furthermore, a lemma above shows that if R_{𝔭} is not Gorenstein (i. e. inj. dim R_{𝔭} = ∞), then μ_{i}(𝔭) ≠ 0 for all i > height 𝔭. On the other hand, if R_{𝔭} is Gorenstein then inj. dim R_{𝔭} = height 𝔭, so μ_{i}(𝔭) = 0 for all i > height 𝔭. Thus (1), (4) and (5) are equivalent.

(1) ⇒ (6): Suppose that R is Gorenstein, and without loss of generality suppose that 𝔭 is maximal. Let h = depth_{𝔭} R = height 𝔭. Then μ_{i}(𝔭) = 0 for i < h. Let I be an ideal generated by a maximal regular R-sequence (necessarily of length h) in I. Then by the corollary above, R/I is a Gorenstein ring of dimension 0 and μ_{h}(𝔭) = μ_{0}(𝔭/I, R/I). But this says that R/I is an injective R/I-module. By a corollary near the beginning of the chapter, this implies that R/I = E((R/I)/(𝔭/I)) = E(R/𝔭), so that μ_{h}(𝔭) = μ_{0}(𝔭/I, R/I) = 1.

(6) ⇒ (1): By an earlier proposition, the statement that μ_{i}(𝔭) = 0 for i < height 𝔭 is equivalent to the fact that R_{𝔭} is Cohen-Macauley. Assuming this to be true, the exists a regular R-sequence in R_{𝔭} with length equal to height 𝔭. Let I be the ideal in R_{𝔭} generated by this R-sequence. Then by the previous Corollary, μ_{0}(𝔭R_{𝔭}/I) = μ_{h}(𝔭), where h = height 𝔭. If μ_{h}(𝔭) = 1, this says that I is an irreducible ideal in R_{𝔭}, and so by a result at the beginning of the Chapter, R_{𝔭}/I is self-injective. Thus in particular R_{𝔭}/I is a Gorenstein ring. It then follows from the previous corollary that R_{𝔭} is a Gorenstein ring.

(6) ⇔ (7): This is now clear from the equivalent of (1), (4), and (6) plus the fact that a localization of a Gorenstein ring is a Gorenstein ring. ✓

Definition. A system of parameters for a local ring R is a sequence x_{1}, …, x_{n} contained in the maximal ideal 𝔪 such that n = height 𝔪 and 𝔪 is the only prime ideal containing x_{1}, …, x_{n} (or, equivalently, Ass R/(x_{1}, …, x_{n}) = {𝔪}).

Proposition. Every noetherian local ring has a system of parameters.

Proposition. A noetherian local ring R is Cohen-Macauley if and only if every system of parameters is a regular R-sequence.

Lemma. Let R be a local ring of dimension n. Then μ_{n}(𝔪) = 1 if and only if some [every] system of parameters generates an irreducible ideal in R. proof: Let I be generated by a system of parameters. Then Ass R/I = {𝔪} and so E(R/I) ≈ E(R/𝔪)^{t}, where t = μ_{0}(R/I). Now μ_{n}(I) = 1 if and only if μ_{0}(R/I) = 1, i. e. if and only if E(R/I) is indecomposable. As shown by a proposition in the beginning of the chapter, this is true if and only if I is irreducible. ✓

Theorem. Let R be a Cohen-Macauley ring with maximal ideal 𝔪. The following are equivalent. (1) R is Gorenstein.

(2) R is Cohen-Macauley and some system of parameters generates an irreducible ideal.

(3) Every system of parameters in R generates an irreducible ideal.

(4) μ_{i}(𝔪) = 0 for i < n and μ_{1}(𝔪) = 1.

(5) For every prime ideal 𝔮 in R, μ_{i}(𝔮) = 0 for i < height 𝔮 and μ_{h}(𝔮) = 1 for h = height 𝔮.

(6) If E_{i} is the ith term of a minimal injective resolution for R, then E_{i} is the direct sum of the injective envelopes of R/𝔮 for all primes 𝔮 with height i.

(7) R is Cohen-Macauley and for all finite length modules M, there is a natural isomorphism

Ext_{R}^{n}(M, R) ≈ Hom_{R}(M, E(k)).

proof: As shown earlier, R is Cohen-Macauley if and only if μ_{i}(𝔪) = 0 for i < height 𝔪. Thus the equivalence of (1), (2), (3), and (4) follows from the preceding Lemma and the Theorem before it.

Lemma. Let M be a Cohen-Macauley module and I an ideal such that IM ≠ M. Then depth_{I} M is the minimum of height_{M} 𝔮 for all primes 𝔮 in Supp M which do not contain I.

Theorem. Let M be a finitely generated module over a local ring R. Then M is Gorenstein if and only if it has finite injective dimension and every regular R-sequence is also a regular M-sequence. In this case, M is also Cohen-Macauley.

Definition. If M is an R-module then Cousin Complex C^{∗}(M) is defined inductively as follows: C^{n} = 0 for n < 0. We write M^{0} = M and C^{0} = ∐{M_{𝔭}^{0} | height_{M} 𝔭 = 0}. Let β^{0} : M^{0} → C^{0} be the canonical map.

Now if M^{n−1} , C^{n−1} and β^{n−1} have been defined, let M^{n} = Coker β^{n−1} , C^{n} = ∐{M_{𝔭}^{n} | height_{M} 𝔭 = n} and let β^{n}: M^{n} → C^{n} be the canonical map. Let d^{n} be the composite of β^{n} with the canonical map from C^{n+1} to M^{n+1} = Coker β^{n+1} :

C^{n} ⟶ M^{n+1} ⟶^{β^{n+1}} C^{n+1}.

Theorem. (1) {C^{∗}(M), d^{∗}} is a cochain complex.

(2) If M_{𝔭}^{n} ≠ 0 then height_{M} 𝔭 ≥ n.

(3) If height_{M} 𝔭 = n − 1 then every element of (M^{n−1})_{𝔭} is annihilated by some power of 𝔭.

(4) H^{n}(C) ≈ Ker β^{n+1} and height_{M} 𝔭 ≥ n + 2 for all 𝔭 ∈ Supp H^{n}(C).

(5) If x is an element of R which annihilates M or annihilates C^{n}(M), then x annihilates C^{t}(M) for all t ≥ n.

(6) If x is an element of R such that xM = M or xC^{n}(M) = C^{n}(M), then xC^{t}(M) = C^{t}(M) for all t ≥ n.

(7) If M is non-zero and finitely generated and if x ∈ R is regular on M, then xC^{n} = C^{n} for all n.

(8) If n > dim M then C^{n} = 0. proof: (1) is clear.

(2) By induction. The map C^{n−1} → M^{n} is surjective so if M_{𝔭}^{n} ≠ 0 then C_{𝔭}^{n−1} ≠ 0. By the constuction of C^{n−1} , it follows that there exist primes 𝔮 with height 𝔮 = n − 1 such that (M_{𝔮}^{n−1})_{𝔭} ≠ 0. Then height 𝔭 ≥ n − 1 by induction. Now if height 𝔭 = n − 1 then ∅ ≠ Ass(M_{𝔭}^{n−1})_{𝔮} ⊆ {𝔮} ∪ {𝔭} so 𝔮 = 𝔭, since the primes in Ass M^{n−1} have height at least n − 1. Thus if height 𝔭 = n − 1 then the only non-zero term M_{𝔮}^{n−1} in the coproduct defining C^{n−1} is M_{𝔭}^{n−1} . This means that β^{n−1} : M_{𝔭}^{n−1} → C_{𝔭}^{n−1} is the identity map and so M_{𝔭}^{n} = 0.

(3) If height_{M} 𝔭 = n and 𝔭 ∈ Supp M_{𝔭}^{n−1} , then 𝔭 must be minimal in Supp M_{𝔭}^{n−1} , so that M_{𝔭}^{n−1} is 𝔭-primary.

(4) H^{n}(C) = (Ker d^{n})/d^{n−1}(C^{n−1}) = (Ker d^{n})/β^{n}(M^{n}) ≈ Ker β^{n+1} ⊆ M^{n+1} . Now as shown in the proof of (2), if height 𝔭 = n + 1 then localizing the map β^{n+1} : M^{n+1} → C^{n+1} at 𝔭 yields the identity map, so that H^{n}(C)_{𝔭} = 0 for all 𝔭 with height 𝔭 ≤ n + 1.

(5) Easy.

(6) Clear.

(7) If x is regular on M and height_{M} 𝔭 = 0 then 𝔭 ∈ Ass M, so x ∉ 𝔭 and thus xM_{𝔭} = M_{𝔭}. Thus xC^{0} = C^{0} . Now apply (6).

(8) Trivial. ✓

Theorem. For any multiplicative set S, the identity map on S^{−1}M induces an isomorphism of complexes from S^{−1}(C(M)) to C(S^{−1}M). proof: Construct the isomorphism recursively. ✓

Remark. If 𝔭 has height n in Supp M then C_{𝔭}^{i}(M) = 0 for i > n and C_{𝔭}^{n}(M) ≈ M_{𝔭}.

Lemma. Let B be a finitely generated modules such that height_{M} 𝔭 ≠ n for all 𝔭 ∈ Supp B. Then Ext_{R}^{i}(B, C^{n}(M)) = 0 for all i. proof: Since C^{n}(M) = ∐{M_{𝔭}^{n} | height_{M} 𝔭 = n} and B is finitely generated, it follows that

Ext_{R}^{i}(B, C^{n}(M)) = ∐{Ext_{R}^{i}(B, M_{𝔭}^{n}) | height_{m} 𝔭 = n} = ∐{Ext^{i}(B_{𝔭}, M_{𝔭}^{n}) | height_{M} 𝔭 = n} = 0. ✓

Theorem. Suppose that β^{0} induces an isomorphism M ≈ H^{0}(C(M)), and that H^{i}(C(M)) = 0 for all i ≤ n − 2. Let B be a finitely generated R-module such that height_{m} 𝔭 ≥ n for all 𝔭 ∈ Supp B. Then Ext_{R}^{i}(B, M) = 0 for i ≤ n − 1, Extr^{n}(B, M) ≈ Hom_{R}(B, M^{n}), Ext_{R}^{j}(B, M) ≈ Ext^{j−n}(B, M^{n}) for j ≥ n. proof: By item (4) of the previous theorem, the sequence 0 → M^{i} → C^{i} → M^{i+1} → 0 is exact for i ≤ n − 1. Furthermore Ext_{R}^{i}(B, C^{r}) = 0 for all i whenever r ≤ n − 1. ✓ Theorem. The following statements are equivalent:

(1) C(M) is acyclic and β^{0} induces an isomorphism from M to H^{0}(C).

(2) Ext_{R}^{i}(R/𝔭, M) = 0 for all 𝔭 ∈ Supp M with height_{M} 𝔭 > i.

(3) M is Cohen-Macauley.

(4) If B is a finitely generated module such that height_{M} 𝔭 ≥ n for all p ∈ Supp B, then Ext_{R}^{i}(B, M) = 0 for all i < n. proof: It is easy to see that (1) implies (4), and (4) implies (2), and (2) is equivalent to (3).

(2) ⇒ (1): We will show by complete induction that the maps β^{i} are all monic. Suppose now that β^{1}, …, β^{n−1} are monic and suppose x ≠ 0 ∈ Ker β^{n} = H^{n−1}(C). Choose 𝔭 ∈ Ass R_{x}. Now if H^{n+1}(C)_{𝔭} ≠ 0 then height_{m} 𝔭 ≥ n + 1, so that Ext_{R}^{n}(R/𝔭, M) ≈ Hom_{R}(R/𝔭, M^{n}) ≠ 0, a contradition. Thus Ker β^{n} = 0. ✓

Lemma. For all n, C^{n} is an essential extension of β^{n}(M^{n}).

Theorem. Let M be a finitely generated R-module. The following assertions are equivalent: (1) C^{∗}(M) is an injective resolution for M.

(2) For all finitely generated R-modules B and 0 ≤ i < j, Ext_{R}^{i}(Ext_{R}^{j}(B, M), M) = 0.

(3) M is a Gorenstein module.

(4) For all primes 𝔭 ∈ Supp M, C(M)_{𝔭} is an injective resolution for the R_{𝔭}-module M_{𝔭}.

(5) M is Cohen-Macauley and if n = dim M then C^{n}(M) is injective.