The Stacks project

Lemma 52.13.2. Let $A$ be a Noetherian ring. Let $f \in \mathfrak a \subset A$ be an element of an ideal of $A$. Let $M$ be a finite $A$-module. Let $s \geq 0$. Assume

  1. $A$ is $f$-adically complete,

  2. $H^ i_\mathfrak a(M)$ is annihilated by a power of $f$ for $i \leq s + 1$.

Then with $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$ the map

\[ H^ i(U, \widetilde{M}) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^ i(U, \widetilde{M/f^ nM}) \]

is an isomorphism for $i < s$.

Proof. By induction on $s$. If $s = 0$, the assertion is empty. If $s = 1$, then the result is Lemma 52.12.7. Assume $s > 1$. By induction it suffices to prove the result for $i = s - 1 \geq 1$. We may apply Lemma 52.3.2 to $U$ and $\mathcal{F} = \widetilde{M}|_ U$ because $\mathcal{F}$ is a Noetherian object in the category of coherent $\mathcal{O}_ U$-modules. Observe that $H^ j(U, \mathcal{F}) = H^{j + 1}_\mathfrak a(M)$ for all $j$ by Local Cohomology, Lemma 51.8.2. Thus for $j = s = (s - 1) + 1$ this is annihilated by a power of $f$ by assumption. Thus it follows from Lemma 52.3.2 that $\mathop{\mathrm{lim}}\nolimits H^{s - 1}(U, \mathcal{F}/f^ n\mathcal{F})$ is the usual $f$-adic completion of $H^{s - 1}(U, \mathcal{F})$. Then again using that this module is killed by a power of $f$ we see that the completion is simply equal to $H^{s - 1}(U, \mathcal{F})$ as desired. $\square$

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Your email address will not be published.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