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 $\mathcal{G}$. $\square$
Comments (0)
There are also: