Lemma 20.36.1. 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. Consider the conditions

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 $\mathcal{F}_ n = \mathcal{G}[f^ n]$, and

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

Then (4) $\Rightarrow $ (3) $\Leftrightarrow $ (2) $\Leftrightarrow $ (1).

**Proof.**
We omit the proof of the equivalence of (1) and (2). We omit the proof that (3) implies (1). Given $\mathcal{F}_ n$ as in (1) to prove (3) 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). The map $f : \mathcal{G} \to \mathcal{G}$ is surjective as the image of $\mathcal{F}_{n + 1} \subset \mathcal{G}$ is $\mathcal{F}_ n \subset \mathcal{G}$ by the short exact sequence (1). Thus $\mathcal{G}$ is an $f$-divisible $\mathcal{O}_ X$-module with $\mathcal{F}_ n = \mathcal{G}[f^ n]$.

Assume given $\mathcal{F}$ as in (4). The map $\mathcal{F}/f^{n + 1}\mathcal{F} \to \mathcal{F}/f^ n\mathcal{F}$ is always surjective with kernel the image of the map $\mathcal{F}/f\mathcal{F} \to \mathcal{F}/f^{n + 1}\mathcal{F}$ induced by multiplication with $f^ n$. To verify (2) it suffices to see that the kernel of $f^ n : \mathcal{F} \to \mathcal{F}/f^{n + 1}\mathcal{F}$ is $f\mathcal{F}$. To see this it suffices to show that given sections $s, t$ of $\mathcal{F}$ over an open $U \subset X$ with $f^ ns = f^{n + 1}t$ we have $s = ft$. This is clear because $f : \mathcal{F} \to \mathcal{F}$ is injective as $\mathcal{F}$ is $f$-torsion free.
$\square$

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