The Stacks project

Proof. Recall that $\textit{Coh}(U, I\mathcal{O}_ U)$ is an abelian category, see Cohomology of Schemes, Lemma 30.23.2. Over affine opens of $U$ the object $(\mathcal{F}_ n)$ corresponds to a finite module over a Noetherian ring (Cohomology of Schemes, Lemma 30.23.1). Thus the kernels of the maps $f^ N : (\mathcal{F}_ n) \to (\mathcal{F}_ n)$ stabilize for $N$ large enough. By Lemmas 52.17.1 and 52.16.3 in order to prove the lemma we may replace $(\mathcal{F}_ n)$ by the image of such a map. Thus we may assume $f$ is injective on $(\mathcal{F}_ n)$. After this replacement the equivalent conditions of Lemma 52.3.1 hold for the inverse system $(\mathcal{F}_ n)$ on $U$. We will use this without further mention in the rest of the proof.

We will check hypotheses (a), (b), and (c) of Lemma 52.16.10. Hypothesis (b) holds by Lemma 52.3.2.

Pick a inverse system of modules $\{ M_ n\} $ as in Lemma 52.16.2. We may assume $H^0_\mathfrak m(M_ n) = 0$ by replacing $M_ n$ by $M_ n/H^0_\mathfrak m(M_ n)$ if necessary. Then we obtain short exact sequences

for all $n$. Let $E$ be an injective hull of the residue field of $A$. By Lemma 52.20.3 and our current assumption (4) we can choose, an integer $m \geq 0$, finite $A$-modules $N_1$ and $N_2$ annihilated by $f^ c$ for some $c \geq 0$ and compatible systems of maps

for $n \geq m$ with the properties stated in the lemma.

We know that $M = \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}_ n)$ is an $A$-module whose limit topology is the $f$-adic topology. Thus, given $n$, the module $M/f^ nM$ is a subquotient of $H^0(U, \mathcal{F}_ N)$ for some $N \gg n$. Looking at the information obtained above we see that $f^ cM/f^ nM$ is a finite $A$-module. Since $f$ is a nonzerodivisor on $M$ we conclude that $M/f^{n - c}M$ is a finite $A$-module. In this way we see that hypothesis (c) of Lemma 52.16.10 holds.

Next, we study the module

For $n \geq m$ let $K_ n$ be the kernel of the map $H^2_\mathfrak m(M_ n) \to \mathop{\mathrm{Hom}}\nolimits _ A(N_2, E)$. Set $K = \mathop{\mathrm{lim}}\nolimits K_ n$. We obtain an exact sequence

By the above the limit topology on $Ob = \mathop{\mathrm{lim}}\nolimits H^2_\mathfrak m(M_ n)$ is the $f$-adic topology. Since $N_2$ is annihilated by $f^ c$ we conclude the same is true for the limit topology on $K = \mathop{\mathrm{lim}}\nolimits K_ n$. Thus $K/fK$ is a subquotient of $K_ n$ for $n \gg 1$. However, since $\{ K_ n\} $ is pro-isomorphic to a inverse system of finite length $A$-modules (by the conclusion of Lemma 52.20.3) we conclude that $K/fK$ is a subquotient of a finite length $A$-module. It follows that $K$ is a finite $A$-module, see Algebra, Lemma 10.96.12. (In fact, we even see that $\dim (\text{Supp}(K)) = 1$ but we will not need this.)

Given $n \geq 1$ consider the boundary map

(the second map is an isomorphism) coming from the short exact sequences

For each $n$ set

where $m$ is as above. Observe that $\{ P_ n\} $ is an inverse system and that the map $f : \mathcal{F}_ n \to \mathcal{F}_{n + 1}$ on global sections maps $P_ n$ into $P_{n + 1}$. If $p \in P_ n$, then $\delta _ n(p) \in K \subset Ob$ because $\delta _ n(p)$ maps to zero in $H^1(U, f^ n\mathcal{F}_{n + m}) = H^2_\mathfrak m(M_ m)$ and the composition of $\delta _ n$ and $Ob \to \mathop{\mathrm{Hom}}\nolimits _ A(N_2, E)$ factors through $H^2_\mathfrak m(M_ m)$ by our choice of $m$. Hence

is a finite graded $A[T]$-module where $T$ acts via multiplication by $f$. Namely, it is a graded submodule of $K[T]$ and $K$ is finite over $A$. Arguing as in the proof of Cohomology, Lemma 20.35.1¹ we find that the inverse system $\{ P_ n\} $ satisfies ML. Since $\{ P_ n\} $ is pro-isomorphic to $\{ H^0(U, \mathcal{F}_ n)\} $ we conclude that $\{ H^0(U, \mathcal{F}_ n)\} $ has ML. Thus hypothesis (a) of Lemma 52.16.10 holds and the proof is complete. $\square$