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.
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 $D(f) \subset U$. $\square$
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