Lemma 51.10.2. Let $I$ be an ideal of a Noetherian ring $A$. Let $M$ be a finite $A$-module, let $\mathfrak p \subset A$ be a prime ideal, and let $s \geq 0$ be an integer. Assume

1. $A$ has a dualizing complex,

2. $\mathfrak p \not\in V(I)$, and

3. for all primes $\mathfrak p' \subset \mathfrak p$ and $\mathfrak q \in V(I)$ with $\mathfrak p' \subset \mathfrak q$ we have

$\text{depth}_{A_{\mathfrak p'}}(M_{\mathfrak p'}) + \dim ((A/\mathfrak p')_\mathfrak q) > s$

Then there exists an $f \in A$, $f \not\in \mathfrak p$ which annihilates $H^ i_{V(I)}(M)$ for $i \leq s$.

Proof. Consider the sets

$T = V(I) \quad \text{and}\quad T' = \bigcup \nolimits _{f \in A, f \not\in \mathfrak p} V(f)$

These are subsets of $\mathop{\mathrm{Spec}}(A)$ stable under specialization. Observe that $T \subset T'$ and $\mathfrak p \not\in T'$. Assumption (3) says that hypothesis (2) of Proposition 51.10.1 holds. Hence we can find $J \subset A$ with $V(J) \subset T'$ such that $J H^ i_{V(I)}(M) = 0$ for $i \leq s$. Choose $f \in A$, $f \not\in \mathfrak p$ with $V(J) \subset V(f)$. A power of $f$ annihilates $H^ i_{V(I)}(M)$ for $i \leq s$. $\square$

## 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Your email address will not be published.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 0EFE. Beware of the difference between the letter 'O' and the digit '0'.