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

\[ U = \{ \mathfrak p \in \mathop{\mathrm{Spec}}(A) \mid M_\mathfrak p\text{ is Cohen-Macaulay}\} \]

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

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(M, \omega _ A^\bullet )_\mathfrak p = R\mathop{\mathrm{Hom}}\nolimits _{A_\mathfrak p}(M_\mathfrak p, (\omega _ A^\bullet )_\mathfrak p) \]

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

Your email address will not be published.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)