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. $\square$
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