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 $H^{-\delta }(\omega _ A^\bullet ) \not= 0$.

**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. $\square$

## 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Your email address will not be published.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)