The Stacks project

Lemma 20.35.1. Let $I$ be an ideal of a ring $A$. Let $X$ be a topological space. Let

\[ \ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1 \]

be an inverse system of sheaves of $A$-modules on $X$ such that $\mathcal{F}_ n = \mathcal{F}_{n + 1}/I^ n\mathcal{F}_{n + 1}$. Let $p \geq 0$. Assume

\[ \bigoplus \nolimits _{n \geq 0} H^{p + 1}(X, I^ n\mathcal{F}_{n + 1}) \]

satisfies the ascending chain condition as a graded $\bigoplus _{n \geq 0} I^ n/I^{n + 1}$-module. Then the inverse system $M_ n = H^ p(X, \mathcal{F}_ n)$ satisfies the Mittag-Leffler condition1.

Proof. Set $N_ n = H^{p + 1}(X, I^ n\mathcal{F}_{n + 1})$ and let $\delta _ n : M_ n \to N_ n$ be the boundary map on cohomology coming from the short exact sequence $0 \to I^ n\mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1} \to \mathcal{F}_ n \to 0$. Then $\bigoplus \mathop{\mathrm{Im}}(\delta _ n) \subset \bigoplus N_ n$ is a graded submodule. Namely, if $s \in M_ n$ and $f \in I^ m$, then we have a commutative diagram

\[ \xymatrix{ 0 \ar[r] & I^ n\mathcal{F}_{n + 1} \ar[d]_ f \ar[r] & \mathcal{F}_{n + 1} \ar[d]_ f \ar[r] & \mathcal{F}_ n \ar[d]_ f \ar[r] & 0 \\ 0 \ar[r] & I^{n + m}\mathcal{F}_{n + m + 1} \ar[r] & \mathcal{F}_{n + m + 1} \ar[r] & \mathcal{F}_{n + m} \ar[r] & 0 } \]

The middle vertical map is given by lifting a local section of $\mathcal{F}_{n + 1}$ to a section of $\mathcal{F}_{n + m + 1}$ and then multiplying by $f$; similarly for the other vertical arrows. We conclude that $\delta _{n + m}(fs) = f \delta _ n(s)$. By assumption we can find $s_ j \in M_{n_ j}$, $j = 1, \ldots , N$ such that $\delta _{n_ j}(s_ j)$ generate $\bigoplus \mathop{\mathrm{Im}}(\delta _ n)$ as a graded module. Let $n > c = \max (n_ j)$. Let $s \in M_ n$. Then we can find $f_ j \in I^{n - n_ j}$ such that $\delta _ n(s) = \sum f_ j \delta _{n_ j}(s_ j)$. We conclude that $\delta (s - \sum f_ j s_ j) = 0$, i.e., we can find $s' \in M_{n + 1}$ mapping to $s - \sum f_ js_ j$ in $M_ n$. It follows that

\[ \mathop{\mathrm{Im}}(M_{n + 1} \to M_{n - c}) = \mathop{\mathrm{Im}}(M_ n \to M_{n - c}) \]

Namely, the elements $f_ js_ j$ map to zero in $M_{n - c}$. This proves the lemma. $\square$

[1] In fact, there exists a $c \geq 0$ such that $\mathop{\mathrm{Im}}(M_ n \to M_{n - c})$ is the stable image for all $n \geq c$.

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Your email address will not be published.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