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
A is local and \mathfrak a = \mathfrak m is the maximal ideal,
A has a dualizing complex,
I = (f) is a principal ideal for a nonzerodivisor f \in \mathfrak m,
\mathcal{F}_ n is a finite locally free \mathcal{O}_ U/f^ n\mathcal{O}_ U-module,
if \mathfrak p \in V(f) \setminus \{ \mathfrak m\} , then \text{depth}((A/f)_\mathfrak p) + \dim (A/\mathfrak p) > 1, and
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 Cohomology, Lemma 20.36.2.
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 Cohomology, Lemma 20.36.5. 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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