Lemma 47.20.1. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring with normalized dualizing complex $\omega _ A^\bullet $. Then $\text{depth}(A)$ is equal to the smallest integer $\delta \geq 0$ such that $H^{-\delta }(\omega _ A^\bullet ) \not= 0$.

## 47.20 Cohen-Macaulay rings

Cohen-Macaulay modules and rings were studied in Algebra, Sections 10.102 and 10.103.

**Proof.**
This follows immediately from Lemma 47.16.5. Here are two other ways to see that it is true.

First alternative. By Nakayama's lemma we see that $\delta $ is the smallest integer such that $\mathop{\mathrm{Hom}}\nolimits _ A(H^{-\delta }(\omega _ A^\bullet ), \kappa ) \not= 0$. In other words, it is the smallest integer such that $\mathop{\mathrm{Ext}}\nolimits _ A^{-\delta }(\omega _ A^\bullet , \kappa )$ is nonzero. Using Lemma 47.15.2 and the fact that $\omega _ A^\bullet $ is normalized this is equal to the smallest integer such that $\mathop{\mathrm{Ext}}\nolimits _ A^\delta (\kappa , A)$ is nonzero. This is equal to the depth of $A$ by Algebra, Lemma 10.71.5.

Second alternative. By the local duality theorem (in the form of Lemma 47.18.4) $\delta $ is the smallest integer such that $H^\delta _\mathfrak m(A)$ is nonzero. This is equal to the depth of $A$ by Lemma 47.11.1. $\square$

Lemma 47.20.2. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring with normalized dualizing complex $\omega _ A^\bullet $ and dualizing module $\omega _ A = H^{-\dim (A)}(\omega _ A^\bullet )$. The following are equivalent

$A$ is Cohen-Macaulay,

$\omega _ A^\bullet $ is concentrated in a single degree, and

$\omega _ A^\bullet = \omega _ A[\dim (A)]$.

In this case $\omega _ A$ is a maximal Cohen-Macaulay module.

**Proof.**
Follows immediately from Lemma 47.16.7.
$\square$

Lemma 47.20.3. Let $A$ be a Noetherian ring. If there exists a finite $A$-module $\omega _ A$ such that $\omega _ A[0]$ is a dualizing complex, then $A$ is Cohen-Macaulay.

**Proof.**
We may replace $A$ by the localization at a prime (Lemma 47.15.6 and Algebra, Definition 10.103.6). In this case the result follows immediately from Lemma 47.20.2.
$\square$

Lemma 47.20.4. Let $A$ be a Noetherian ring with dualizing complex $\omega _ A^\bullet $. Let $M$ be a finite $A$-module. Then

is an open subset of $\mathop{\mathrm{Spec}}(A)$ whose intersection with $\text{Supp}(M)$ is dense.

**Proof.**
If $\mathfrak p$ is a generic point of $\text{Supp}(M)$, then $\text{depth}(M_\mathfrak p) = \dim (M_\mathfrak p) = 0$ and hence $\mathfrak p \in U$. This proves denseness. If $\mathfrak p \in U$, then we see that

has a unique nonzero cohomology module, say in degree $i_0$, by Lemma 47.16.7. Since $R\mathop{\mathrm{Hom}}\nolimits _ A(M, \omega _ A^\bullet )$ has only a finite number of nonzero cohomology modules $H^ i$ and since each of these is a finite $A$-module, we can find an $f \in A$, $f \not\in \mathfrak p$ such that $(H^ i)_ f = 0$ for $i \not= i_0$. Then $R\mathop{\mathrm{Hom}}\nolimits _ A(M, \omega _ A^\bullet )_ f$ has a unique nonzero cohomology module and reversing the arguments just given we find that $D(f) \subset U$. $\square$

Lemma 47.20.5. Let $A$ be a Noetherian ring. If $A$ has a dualizing complex $\omega _ A^\bullet $, then $\{ \mathfrak p \in \mathop{\mathrm{Spec}}(A) \mid A_\mathfrak p\text{ is Cohen-Macaulay}\} $ is a dense open subset of $\mathop{\mathrm{Spec}}(A)$.

**Proof.**
Immediate consequence of Lemma 47.20.4 and the definitions.
$\square$

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