**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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