Lemma 47.16.11. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring with normalized dualizing complex $\omega _ A^\bullet$. Let $\mathfrak p$ be a minimal prime of $A$ with $\dim (A/\mathfrak p) = e$. Then $H^ i(\omega _ A^\bullet )_\mathfrak p$ is nonzero if and only if $i = -e$.

Proof. Since $A_\mathfrak p$ has dimension zero, there exists an integer $n > 0$ such that $\mathfrak p^ nA_\mathfrak p$ is zero. Set $B = A/\mathfrak p^ n$ and $\omega _ B^\bullet = R\mathop{\mathrm{Hom}}\nolimits _ A(B, \omega _ A^\bullet )$. Since $B_\mathfrak p = A_\mathfrak p$ we see that

$(\omega _ B^\bullet )_\mathfrak p = R\mathop{\mathrm{Hom}}\nolimits _ A(B, \omega _ A^\bullet ) \otimes _ A^\mathbf {L} A_\mathfrak p = R\mathop{\mathrm{Hom}}\nolimits _{A_\mathfrak p}(B_\mathfrak p, (\omega _ A^\bullet )_\mathfrak p) = (\omega _ A^\bullet )_\mathfrak p$

The second equality holds by More on Algebra, Lemma 15.99.2. By Lemma 47.16.2 we may replace $A$ by $B$. After doing so, we see that $\dim (A) = e$. Then we see that $H^ i(\omega _ A^\bullet )_\mathfrak p$ can only be nonzero if $i = -e$ by Lemma 47.16.5 parts (1) and (2). On the other hand, since $(\omega _ A^\bullet )_\mathfrak p$ is a dualizing complex for the nonzero ring $A_\mathfrak p$ (Lemma 47.15.6) we see that the remaining module has to be nonzero. $\square$

Comment #7468 by on

The proof of Lemma 47.16.11 is not true. We can use Lemma 47.16.10 and 47.16.09 to give a right proof.

Comment #7618 by on

OK, I carefully checked the proof and I cannot fault it. @everyone: if you find a mistake in a proof, please in your comment point explicitly to the first step in the proof that does not work. Thanks very much!

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