Lemma 10.126.6. Let $\varphi : R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Assume

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

$\varphi $ induces an isomorphism $R_\mathfrak p \cong S_\mathfrak q$.

Then there exist $f \in R$, $f \not\in \mathfrak p$ and an $R_ f$-algebra $C$ such that $S_ f \cong R_ f \times C$ as $R_ f$-algebras.

**Proof.**
Write $S = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. Let $a_ i \in R_\mathfrak p$ be an element mapping to the image of $x_ i$ in $S_\mathfrak q$. Write $a_ i = b_ i/f$ for some $f \in R$, $f \not\in \mathfrak p$. After replacing $R$ by $R_ f$ and $x_ i$ by $x_ i - a_ i$ we may assume that $S = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$ such that $x_ i$ maps to zero in $S_\mathfrak q$. Then if $c_ j$ denotes the constant term of $g_ j$ we conclude that $c_ j$ maps to zero in $R_\mathfrak p$. After another replacement of $R$ we may assume that the constant coefficients $c_ j$ of the $g_ j$ are zero. Thus we obtain an $R$-algebra map $S \to R$, $x_ i \mapsto 0$ whose kernel is the ideal $(x_1, \ldots , x_ n)$.

We have the isomorphisms $R_\mathfrak p \to S_\mathfrak q \to R_\mathfrak p$ and $S \to R$ sends $x_ i$ to zero. Thus we must have $S_\mathfrak q = R_\mathfrak p[x_1, \ldots , x_ n]/(x_1, \ldots , x_ n)$ and a fortiori $S_\mathfrak q = S_\mathfrak p/(x_1, \ldots , x_ n)S_\mathfrak p$. This means that the finitely generated ideal $(x_1, \ldots , x_ n)S_\mathfrak p$ is pure in $S_\mathfrak p$, see Definition 10.108.1. Hence $(x_1, \ldots , x_ n)S_\mathfrak p$ is generated by an idempotent $e$ in $S_\mathfrak p$ by Lemma 10.108.5. After replacing $R \to S$ by $R_ f \to S_ f$ for some $f \in R$, $f \not\in \mathfrak p$ we can find an idempotent $e' \in S$ mapping to $e$. Then $e'S$ and $(x_1, \ldots , x_ n)S$ are finitely generated ideals which become equal in $S_\mathfrak p$. Hence after replacing $R \to S$ by $R_ f \to S_ f$ for some $f \in R$, $f \not\in \mathfrak p$ we may assume $e'S = (x_1, \ldots , x_ n)S$. Setting $C = e'S$ finishes the proof.
$\square$

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