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 10.103 and 10.104.

**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 47.15.3 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 10.72.5.

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 10.104.6). 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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)