Lemma 10.126.10. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \to S'$ be an $R$-algebra map. Let $IS \subset \mathfrak q \subset S$ be a prime ideal. Assume that

$S \to S'$ is surjective,

$S_\mathfrak q/IS_\mathfrak q \to S'_\mathfrak q/IS'_\mathfrak q$ is an isomorphism,

$S$ is of finite type over $R$,

$S'$ of finite presentation over $R$, and

$S'_\mathfrak q$ is flat over $R$.

Then $S_ g \to S'_ g$ is an isomorphism for some $g \in S$, $g \not\in \mathfrak q$.

**Proof.**
Let $J = \mathop{\mathrm{Ker}}(S \to S')$. By Lemma 10.6.2 $J$ is a finitely generated ideal. Since $S'_\mathfrak q$ is flat over $R$ we see that $J_\mathfrak q/IJ_\mathfrak q \subset S_\mathfrak q/IS_{\mathfrak q}$ (apply Lemma 10.39.12 to $0 \to J \to S \to S' \to 0$). By assumption (2) we see that $J_\mathfrak q/IJ_\mathfrak q$ is zero. By Nakayama's lemma (Lemma 10.20.1) we see that there exists a $g \in S$, $g \not\in \mathfrak q$ such that $J_ g = 0$. Hence $S_ g \cong S'_ g$ as desired.
$\square$

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