The Stacks project

Lemma 52.8.2. Let $I, J$ be ideals of a Noetherian ring. Let $M$ be a finite $A$-module. Let $s$ and $d$ be integers. With $T$ as in ( assume

  1. $A$ has a dualizing complex,

  2. if $\mathfrak p \in V(I)$, then no condition,

  3. if $\mathfrak p \not\in V(I)$, $\mathfrak p \in T$, then $\dim ((A/\mathfrak p)_\mathfrak q) \leq d$ for some $\mathfrak q \in V(\mathfrak p) \cap V(J) \cap V(I)$,

  4. if $\mathfrak p \not\in V(I)$, $\mathfrak p \not\in T$, then

    \[ \text{depth}_{A_\mathfrak p}(M_\mathfrak p) \geq s \quad \text{or}\quad \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > d + s \]

    for all $\mathfrak q \in V(\mathfrak p) \cap V(J) \cap V(I)$.

Then there exists an ideal $J_0 \subset J$ with $V(J_0) \cap V(I) = V(J) \cap V(I)$ such that for any $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$ the map

\[ R\Gamma _{J'}(M) \longrightarrow R\Gamma _{J_0}(M) \]

induces an isomorphism in cohomology in degrees $\leq s$ and moreover these modules are annihilated by a power of $J_0I$.

Proof. Let us consider the set

\[ B = \{ \mathfrak p \not\in V(I),\ \mathfrak p \in T,\text{ and } \text{depth}(M_\mathfrak p) \leq s\} \]

Choose $J_0 \subset J$ such that $V(J_0)$ is the closure of $B \cup V(J)$.

Claim I: $V(J_0) \cap V(I) = V(J) \cap V(I)$.

Proof of Claim I. The inclusion $\supset $ holds by construction. Let $\mathfrak p$ be a minimal prime of $V(J_0)$. If $\mathfrak p \in B \cup V(J)$, then either $\mathfrak p \in T$ or $\mathfrak p \in V(J)$ and in both cases $V(\mathfrak p) \cap V(I) \subset V(J) \cap V(I)$ as desired. If $\mathfrak p \not\in B \cup V(J)$, then $V(\mathfrak p) \cap B$ is dense, hence infinite, and we conclude that $\text{depth}(M_\mathfrak p) < s$ by Local Cohomology, Lemma 51.9.2. In fact, let $V(\mathfrak p) \cap B = \{ \mathfrak p_\lambda \} _{\lambda \in \Lambda }$. Pick $\mathfrak q_\lambda \in V(\mathfrak p_\lambda ) \cap V(J) \cap V(I)$ as in (3). Let $\delta : \mathop{\mathrm{Spec}}(A) \to \mathbf{Z}$ be the dimension function associated to a dualizing complex $\omega _ A^\bullet $ for $A$. Since $\Lambda $ is infinite and $\delta $ is bounded, there exists an infinite subset $\Lambda ' \subset \Lambda $ on which $\delta (\mathfrak q_\lambda )$ is constant. For $\lambda \in \Lambda '$ we have

\[ \text{depth}(M_{\mathfrak p_\lambda }) + \delta (\mathfrak p_\lambda ) - \delta (\mathfrak q_\lambda ) = \text{depth}(M_{\mathfrak p_\lambda }) + \dim ((A/\mathfrak p_\lambda )_{\mathfrak q_\lambda }) \leq d + s \]

by (3) and the definition of $B$. By the semi-continuity of the function $\text{depth} + \delta $ proved in Duality for Schemes, Lemma 48.2.8 we conclude that

\[ \text{depth}(M_\mathfrak p) + \dim ((A/\mathfrak p)_{\mathfrak q_\lambda }) = \text{depth}(M_\mathfrak p) + \delta (\mathfrak p) - \delta (\mathfrak q_\lambda ) \leq d + s \]

Since also $\mathfrak p \not\in V(I)$ we read off from (4) that $\mathfrak p \in T$, i.e., $V(\mathfrak p) \cap V(I) \subset V(J) \cap V(I)$. This finishes the proof of Claim I.

Claim II: $H^ i_{J_0}(M) \to H^ i_ J(M)$ is an isomorphism for $i \leq s$ and $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$.

Proof of claim II. Choose $\mathfrak p \in V(J')$ not in $V(J_0)$. It suffices to show that $H^ i_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ for $i \leq s$, see Local Cohomology, Lemma 51.2.6. Observe that $\mathfrak p \in T$. Hence since $\mathfrak p$ is not in $B$ we see that $\text{depth}(M_\mathfrak p) > s$ and the groups vanish by Dualizing Complexes, Lemma 47.11.1.

Claim III. The final statement of the lemma is true.

By Claim II for $i \leq s$ we have

\[ H^ i_ T(M) = H^ i_{J_0}(M) = H^ i_{J'}(M) \]

for all ideals $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$. See Local Cohomology, Lemma 51.5.3. Let us check the hypotheses of Local Cohomology, Proposition 51.10.1 for the subsets $T \subset T \cup V(I)$, the module $M$, and the integer $s$. We have to show that given $\mathfrak p \subset \mathfrak q$ with $\mathfrak p \not\in T \cup V(I)$ and $\mathfrak q \in T$ we have

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

If $\text{depth}(M_\mathfrak p) \geq s$, then this is true because the dimension of $(A/\mathfrak p)_\mathfrak q$ is at least $1$. Thus we may assume $\text{depth}(M_\mathfrak p) < s$. If $\mathfrak q \in V(I)$, then $\mathfrak q \in V(J) \cap V(I)$ and the inequality holds by (4). If $\mathfrak q \not\in V(I)$, then we can use (3) to pick $\mathfrak q' \in V(\mathfrak q) \cap V(J) \cap V(I)$ with $\dim ((A/\mathfrak q)_{\mathfrak q'}) \leq d$. Then assumption (4) gives

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

Since $A$ is catenary this implies the inequality we want. Applying Local Cohomology, Proposition 51.10.1 we find $J'' \subset A$ with $V(J'') \subset T \cup V(I)$ such that $J''$ annihilates $H^ i_ T(M)$ for $i \leq s$. Then we can write $V(J'') \cup V(J_0) \cup V(I) = V(J'I)$ for some $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$. Replacing $J_0$ by $J'$ the proof is complete. $\square$

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 0EFI. Beware of the difference between the letter 'O' and the digit '0'.