The Stacks project

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. $\square$

Proof. Let $c \geq 1$. It is clear that $f^ c H^ p$ maps to zero in $H^ p(X, \mathcal{F}_ c)$. If $\xi = (\xi _ n) \in H^ p$ is small in the limit topology, then $\xi _ c = 0$, and hence $\xi _ n$ maps to zero in $H^ p(X, \mathcal{F}_ c)$ for $n \geq c$. Consider the inverse system of short exact sequences

and the corresponding inverse system of long exact cohomology sequences

Since the term $H^{p - 1}(X, \mathcal{F}_ c)$ is independent of $n$ we can choose a compatible sequence of elements $\xi '_ n \in H^ p(X, \mathcal{F}_{n - c})$ lifting $\xi _ n$. Setting $\xi ' = (\xi '_ n)$ we see that $\xi = f^ c \xi '$ as desired. $\square$

Proof. By Lemma 20.36.2 we have $M/fM \subset H^0(X, \mathcal{F}_1)$. From (1) and the Noetherian property of $A$ we get that $M/fM$ is a finite $A$-module. Observe that $\bigcap f^ nM = 0$ as $f^ nM$ maps to zero in $H^0(X, \mathcal{F}_ n)$. By Algebra, Lemma 10.96.12 we conclude that $M$ is finite over $A$. Finally, suppose $s = (s_ n) \in M = \mathop{\mathrm{lim}}\nolimits H^0(X, \mathcal{F}_ n)$ satisfies $fs = 0$. Then $s_{n + 1}$ is in the kernel of $\mathcal{F}_{n + 1} \to \mathcal{F}_ n$ by condition (1) of Lemma 20.36.1. Hence $s_ n = 0$. Since $n$ was arbitrary, we see $s = 0$. Thus $f$ is a nonzerodivisor on $M$. $\square$

Proof. Set $I = (f)$. We will use the criterion of Lemma 20.35.1. Observe that $f^ n : \mathcal{F}_1 \to I^ n\mathcal{F}_{n + 1}$ is an isomorphism for all $n \geq 0$. Thus it suffices to show that

is a graded $S = \bigoplus _{n \geq 0} A/(f) \cdot f^ n$-module satisfying the ascending chain condition. If $A$ is not Noetherian, then $H^{p + 1}(X, \mathcal{F}_1)$ has finite length and the result holds. If $A$ is Noetherian, then $S$ is a Noetherian ring and the result holds as the module is finite over $S$ by the assumed finiteness of $H^{p + 1}(X, \mathcal{F}_1)$. Some details omitted. $\square$

Proof. Set $I = (f)$. We will use the criterion of Lemma 20.35.2 involving the modules $N_ n$. For $m \geq n$ we have $I^ n\mathcal{F}_{m + 1} = \mathcal{F}_{m + 1 - n}$. Thus we see that

is independent of $n$ and $\bigoplus N_ n = \bigoplus N_1 \cdot f^{n + 1}$. Thus we conclude exactly as in the proof of Lemma 20.36.4. $\square$