Lemma 61.28.6. Let $X$ be a scheme. Let $\Lambda $ be a ring and let $I \subset \Lambda $ be a finitely generated ideal. Let $\mathcal{F}$ be a sheaf of $\Lambda $-modules on $X_{pro\text{-}\acute{e}tale}$. If $\mathcal{F}$ is derived complete and $\mathcal{F}/I\mathcal{F} = 0$, then $\mathcal{F} = 0$.
Proof. Assume that $\mathcal{F}/I\mathcal{F}$ is zero. Let $I = (f_1, \ldots , f_ r)$. Let $i < r$ be the largest integer such that $\mathcal{G} = \mathcal{F}/(f_1, \ldots , f_ i)\mathcal{F}$ is nonzero. If $i$ does not exist, then $\mathcal{F} = 0$ which is what we want to show. Then $\mathcal{G}$ is derived complete as a cokernel of a map between derived complete modules, see Proposition 61.21.1. By our choice of $i$ we have that $f_{i + 1} : \mathcal{G} \to \mathcal{G}$ is surjective. Hence
\[ \mathop{\mathrm{lim}}\nolimits (\ldots \to \mathcal{G} \xrightarrow {f_{i + 1}} \mathcal{G} \xrightarrow {f_{i + 1}} \mathcal{G}) \]
is nonzero, contradicting the derived completeness of $\mathcal{G}$. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: