The Stacks project

Lemma 114.4.14. Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Let $s$ be an integer. Assume

  1. $A$ has a dualizing complex,

  2. if $\mathfrak p \not\in V(I)$ and $V(\mathfrak p) \cap V(I) \not= \{ \mathfrak m\} $, then $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim (A/\mathfrak p) > s$.

Then there exists an $n > 0$ and an ideal $J \subset A$ with $V(J) \cap V(I) = \{ \mathfrak m\} $ such that $JI^ n$ annihilates $H^ i_\mathfrak m(M)$ for $i \leq s$.

Proof. According to Local Cohomology, Lemma 51.9.4 we have to show this for the finite $A$-module $E^ i = \text{Ext}^{-i}_ A(M, \omega _ A^\bullet )$ for $i \leq s$. The support $Z$ of $E^0 \oplus \ldots \oplus E^ s$ is closed in $\mathop{\mathrm{Spec}}(A)$ and does not contain any prime as in (2). Hence it is contained in $V(JI^ n)$ for some $J$ as in the statement of the lemma. $\square$


Comments (0)


Post a comment

Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DXL. Beware of the difference between the letter 'O' and the digit '0'.