The Stacks project

Lemma 52.12.3. Let $I \subset \mathfrak a$ be ideals of a Noetherian ring $A$. Let $\mathcal{F}$ be a coherent module on $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Assume

  1. $A$ is $I$-adically complete and has a dualizing complex,

  2. if $x \in \text{Ass}(\mathcal{F})$, $x \not\in V(I)$, $z \in V(\mathfrak a) \cap \overline{\{ x\} }$, then $\dim (\mathcal{O}_{\overline{\{ x\} }, z}) > \text{cd}(A, I) + 1$,

  3. for $x \in U$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$ we have $\text{depth}(\mathcal{F}_ x) \geq 2$,

Then we obtain an isomorphism

\[ H^0(U, \mathcal{F}) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F}) \]

Proof. Let $\hat s \in \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$. By Proposition 52.12.2 we find that $\hat s$ is the image of an element $s \in \mathcal{F}(V)$ for some $V \subset U$ open containing $U \cap V(I)$. However, condition (3) shows that $\text{depth}(\mathcal{F}_ x) \geq 2$ for all $x \in U \setminus V$ and hence we find that $\mathcal{F}(V) = \mathcal{F}(U)$ by Divisors, Lemma 31.5.11 and 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Your email address will not be published.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 0EIG. Beware of the difference between the letter 'O' and the digit '0'.