Lemma 52.20.4. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Assume

1. $A$ is local and $\mathfrak a = \mathfrak m$ is the maximal ideal,

2. $A$ has a dualizing complex,

3. $I = (f)$ is a principal ideal,

4. $(\mathcal{F}_ n)$ satisfies the $(2, 3)$-inequalities.

Then $(\mathcal{F}_ n)$ extends to $X$. In particular, if $A$ is $I$-adically complete, then $(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ U$-module.

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 Cohomology, Lemma 20.36.2.

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

$0 \to M_ n \to H^0(U, \mathcal{F}_ n) \to H^1_\mathfrak m(M_ n) \to 0$

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

$H^ i_\mathfrak m(M_ n) \to \mathop{\mathrm{Hom}}\nolimits _ A(N_ i, E), \quad i = 1, 2$

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

$Ob = \mathop{\mathrm{lim}}\nolimits H^1(U, \mathcal{F}_ n) = \mathop{\mathrm{lim}}\nolimits H^2_\mathfrak m(M_ n)$

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

$0 \to K \to Ob \to \mathop{\mathrm{Hom}}\nolimits _ A(N_2, E)$

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

$\delta _ n : H^0(U, \mathcal{F}_ n) \longrightarrow \mathop{\mathrm{lim}}\nolimits _ N H^1(U, f^ n\mathcal{F}_ N) \xrightarrow {f^{-n}} Ob$

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

$0 \to f^ n\mathcal{F}_ N \to \mathcal{F}_ N \to \mathcal{F}_ n \to 0$

For each $n$ set

$P_ n = \mathop{\mathrm{Im}}(H^0(U, \mathcal{F}_{n + m}) \to H^0(U, \mathcal{F}_ n))$

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

$\bigoplus \nolimits _{n \geq 0} \mathop{\mathrm{Im}}(P_ n \to Ob)$

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. $\square$

[1] Choose homogeneous generators of the form $\delta _{n_ j}(p_ j)$ for the displayed module. Then if $k = \max (n_ j)$ we find that for $n \geq k$ and any $p \in P_ n$ we can find $a_ j \in A$ such that $p - \sum a_ j f^{n - n_ j} p_ j$ is in the kernel of $\delta _ n$ and hence in the image of $P_{n'}$ for all $n' \geq n$. Thus $\mathop{\mathrm{Im}}(P_ n \to P_{n - k}) = \mathop{\mathrm{Im}}(P_{n'} \to P_{n - k})$ for all $n' \geq n$.

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