Lemma 52.3.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $f \in \Gamma (X, \mathcal{O}_ X)$. Let

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

be inverse system of $\mathcal{O}_ X$-modules. The following are equivalent

for all $n \geq 1$ the map $f : \mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1}$ factors through $\mathcal{F}_{n + 1} \to \mathcal{F}_ n$ to give a short exact sequence $0 \to \mathcal{F}_ n \to \mathcal{F}_{n + 1} \to \mathcal{F}_1 \to 0$,

for all $n \geq 1$ the map $f^ n : \mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1}$ factors through $\mathcal{F}_{n + 1} \to \mathcal{F}_1$ to give a short exact sequence $0 \to \mathcal{F}_1 \to \mathcal{F}_{n + 1} \to \mathcal{F}_ n \to 0$

there exists an $\mathcal{O}_ X$-module $\mathcal{G}$ which is $f$-divisible such that $\mathcal{F}_ n = \mathcal{G}[f^ n]$.

If $X$ is a scheme and $\mathcal{F}_ n$ is quasi-coherent, then these are also equivalent to

there exists an $\mathcal{O}_ X$-module $\mathcal{F}$ which is $f$-torsion free such that $\mathcal{F}_ n = \mathcal{F}/f^ n\mathcal{F}$.

**Proof.**
We omit the proof of the equivalence of (1) and (2). The condition that $\mathcal{G}$ is $f$-divisible means that $f : \mathcal{G} \to \mathcal{G}$ is surjective. Thus given $\mathcal{F}_ n$ as in (1) we set $\mathcal{G} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ n$ where the maps $\mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to \ldots $ are as in (1). This produces an $f$-divisible $\mathcal{O}_ X$-module with $\mathcal{F}_ n = \mathcal{G}[f^ n]$ as can be seen by checking on stalks. The condition that $\mathcal{F}$ is $f$-torsion free means that $f : \mathcal{F} \to \mathcal{F}$ is injective. If $X$ is a scheme and $\mathcal{F}_ n$ is quasi-coherent, then we set $\mathcal{F} = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. Namely, for an affine open $U \subset X$ the transition maps $\mathcal{F}_{n + 1}(U) \to \mathcal{F}_ n(U)$ are surjective by vanishing of higher cohomology. This produces an $f$-torsion free $\mathcal{O}_ X$-module with $\mathcal{F}_ n = \mathcal{F}/f^ n\mathcal{F}$ (Lemma 52.2.4).
$\square$

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