Lemma 15.116.3. Let $A' \subset A$ be an extension of rings. Let $f \in A'$. Assume that (a) $A$ is finite over $A'$, (b) $f$ is a nonzerodivisor on $A$, and (c) $A'_ f = A_ f$. Then there exists an integer $n_0 > 0$ such that for all $n \geq n_0$ the following is true: given a ring $B'$, a nonzerodivisor $g \in B'$, and an isomorphism $\varphi ' : A'/f^ n A' \to B'/g^ n B'$ with $\varphi '(f) \equiv g$, there is a finite extension $B' \subset B$ and an isomorphism $\varphi : A/fA \to B/gB$ compatible with $\varphi '$.

Proof. Since $A$ is finite over $A'$ and since $A'_ f = A_ f$ we can Cchoose $t > 0$ such that $f^ t A \subset A'$. Set $n_0 = 2t$. Given $n, B', g, \varphi '$ as in the statement of the lemma, denote $N \subset B'$ the set of elements $b \in B'$ such that $b \bmod g^ nB' \in \varphi '(f^ tA)$. Set $B = g^{-t}N$. As $f^ tA' \subset f^ tA$ and $\varphi '$ sends $f$ to $g$ we have $g^ tB' \subset N$, hence $B' \subset B$. Since $f^ tA \cdot f^ tA \subset f^ t \cdot f^ tA$ and $\varphi '$ sends $f$ to $g$, we see that $N \cdot N \subset g^ t N$. Hence we obtain a multiplication on $B$ extending the multiplication of $B'$. We have an isomorphism of $A'/f^ nA'$-modules

$A/f^ tA' \xrightarrow {f^ t} f^ tA/f^ nA' \xrightarrow {\varphi '} g^ tB/g^ nB' \xrightarrow {g^{-t}} B/g^ tB'$

where the module structures on the right are defined using $\varphi '$. Since $A/f^ tA'$ is a finite $A'$-module, we conclude that $B/g^ tB'$ is a finite $B'$-module and hence we see that $B' \to B$ is finite. Finally, we leave it to the reader to see that the displayed isomorphism of modules sends $fA$ into $gB$ and induces an isomorphism of rings $\varphi : A/fA \to B/gB$ compatible with $\varphi '$ (it even induces an isomorphism $A/f^ tA \to B/g^ tB$ but we don't need this). $\square$

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