Lemma 52.20.1. 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 for a nonzerodivisor $f \in \mathfrak m$,

4. $\mathcal{F}_ n$ is a finite locally free $\mathcal{O}_ U/f^ n\mathcal{O}_ U$-module,

5. if $\mathfrak p \in V(f) \setminus \{ \mathfrak m\}$, then $\text{depth}((A/f)_\mathfrak p) + \dim (A/\mathfrak p) > 1$, and

6. if $\mathfrak p \not\in V(f)$ and $V(\mathfrak p) \cap V(f) \not= \{ \mathfrak m\}$, then $\text{depth}(A_\mathfrak p) + \dim (A/\mathfrak p) > 3$.

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

Proof. We will prove this by verifying hypotheses (a), (b), and (c) of Lemma 52.16.10.

Since $\mathcal{F}_ n$ is locally free over $\mathcal{O}_ U/f^ n\mathcal{O}_ U$ we see that we have short exact sequences $0 \to \mathcal{F}_ n \to \mathcal{F}_{n + 1} \to \mathcal{F}_1 \to 0$ for all $n$. Thus condition (b) holds by Lemma 52.3.2.

By induction on $n$ and the short exact sequences $0 \to A/f^ n \to A/f^{n + 1} \to A/f \to 0$ we see that the associated primes of $A/f^ nA$ agree with the associated primes of $A/fA$. Since the associated points of $\mathcal{F}_ n$ correspond to the associated primes of $A/f^ nA$ not equal to $\mathfrak m$ by assumption (3), we conclude that $M_ n = H^0(U, \mathcal{F}_ n)$ is a finite $A$-module by (5) and Local Cohomology, Proposition 51.8.7. Thus hypothesis (c) holds.

To finish the proof it suffices to show that there exists an $n > 1$ such that the image of

$H^1(U, \mathcal{F}_ n) \longrightarrow H^1(U, \mathcal{F}_1)$

has finite length as an $A$-module. Namely, this will imply hypothesis (a) by Lemma 52.3.5. The image is independent of $n$ for $n$ large enough by Lemma 52.5.2. Let $\omega _ A^\bullet$ be a normalized dualizing complex for $A$. By the local duality theorem and Matlis duality (Dualizing Complexes, Lemma 47.18.4 and Proposition 47.7.8) our claim is equivalent to: the image of

$\text{Ext}^{-2}_ A(M_1, \omega _ A^\bullet ) \to \text{Ext}^{-2}_ A(M_ n, \omega _ A^\bullet )$

has finite length for $n \gg 1$. The modules in question are finite $A$-modules supported at $V(f)$. Thus it suffices to show that this map is zero after localization at a prime $\mathfrak q$ containing $f$ and different from $\mathfrak m$. Let $\omega _{A_\mathfrak q}^\bullet$ be a normalized dualizing complex on $A_\mathfrak q$ and recall that $\omega _{A_\mathfrak q}^\bullet = (\omega _ A^\bullet )_\mathfrak q[\dim (A/\mathfrak q)]$ by Dualizing Complexes, Lemma 47.17.3. Using the local structure of $\mathcal{F}_ n$ given in (4) we find that it suffices to show the vanishing of

$\text{Ext}^{-2 + \dim (A/\mathfrak q)}_{A_\mathfrak q}( A_\mathfrak q/f, \omega _{A_\mathfrak q}^\bullet ) \to \text{Ext}^{-2 + \dim (A/\mathfrak q)}_{A_\mathfrak q}( A_\mathfrak q/f^ n, \omega _{A_\mathfrak q}^\bullet )$

for $n$ large enough. If $\dim (A/\mathfrak q) > 3$, then this is immediate from Local Cohomology, Lemma 51.9.4. For the other cases we will use the long exact sequence

$\ldots \xrightarrow {f^ n} H^{-1}(\omega _{A_\mathfrak q}^\bullet ) \to \text{Ext}^{-1}_{A_\mathfrak q}( A_\mathfrak q/f^ n, \omega _{A_\mathfrak q}^\bullet ) \to H^0(\omega _{A_\mathfrak q}^\bullet ) \xrightarrow {f^ n} H^0(\omega _{A_\mathfrak q}^\bullet ) \to \text{Ext}^0_{A_\mathfrak q}( A_\mathfrak q/f^ n, \omega _{A_\mathfrak q}^\bullet ) \to 0$

If $\dim (A/\mathfrak q) = 2$, then $H^0(\omega _{A_\mathfrak q}^\bullet ) = 0$ because $\text{depth}(A_\mathfrak q) \geq 1$ as $f$ is a nonzerodivisor. Thus the long exact sequence shows the condition is that

$f^{n - 1} : H^{-1}(\omega _{A_\mathfrak q}^\bullet )/f \to H^{-1}(\omega _{A_\mathfrak q}^\bullet )/f^ n$

is zero. Now $H^{-1}(\omega ^\bullet _\mathfrak q)$ is a finite module supported in the primes $\mathfrak p \subset A_\mathfrak q$ such that $\text{depth}(A_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \leq 1$. Since $\dim ((A/\mathfrak p)_\mathfrak q) = \dim (A/\mathfrak p) - 2$ condition (6) tells us these primes are contained in $V(f)$. Thus the desired vanishing for $n$ large enough. Finally, if $\dim (A/\mathfrak q) = 1$, then condition (5) combined with the fact that $f$ is a nonzerodivisor insures that $A_\mathfrak q$ has depth at least $2$. Hence $H^0(\omega _{A_\mathfrak q}^\bullet ) = H^{-1}(\omega _{A_\mathfrak q}^\bullet ) = 0$ and the long exact sequence shows the claim is equivalent to the vanishing of

$f^{n - 1} : H^{-2}(\omega _{A_\mathfrak q}^\bullet )/f \to H^{-2}(\omega _{A_\mathfrak q}^\bullet )/f^ n$

Now $H^{-2}(\omega ^\bullet _\mathfrak q)$ is a finite module supported in the primes $\mathfrak p \subset A_\mathfrak q$ such that $\text{depth}(A_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \leq 2$. By condition (6) all of these primes are contained in $V(f)$. Thus the desired vanishing for $n$ large enough. $\square$

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