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. $\square$
Comments (2)
Comment #7468 by Qilin,Yang on
Comment #7618 by Stacks Project on