Lemma 10.156.1. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p$ in $R$. Assume $R \to S$ is quasi-finite at $\mathfrak q$. The commutative diagram

\[ \xymatrix{ R_{\mathfrak p}^ h \ar[r] & S_{\mathfrak q}^ h \\ R_{\mathfrak p} \ar[u] \ar[r] & S_{\mathfrak q} \ar[u] } \]

of Lemma 10.155.6 identifies $S_{\mathfrak q}^ h$ with the localization of $R_{\mathfrak p}^ h \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$ at the prime generated by $\mathfrak q$. Moreover, the ring map $R_{\mathfrak p}^ h \to S_{\mathfrak q}^ h$ is finite.

**Proof.**
Note that $R_{\mathfrak p}^ h \otimes _ R S$ is quasi-finite over $R_{\mathfrak p}^ h$ at the prime ideal corresponding to $\mathfrak q$, see Lemma 10.122.6. Hence the localization $S'$ of $R_{\mathfrak p}^ h \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$ is henselian and finite over $R_{\mathfrak p}^ h$, see Lemma 10.153.4. As a localization $S'$ is a filtered colimit of étale $R_{\mathfrak p}^ h \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$-algebras. By Lemma 10.155.8 we see that $S_\mathfrak q^ h$ is the henselization of $R_{\mathfrak p}^ h \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$. Thus $S' = S_\mathfrak q^ h$ by the uniqueness result of Lemma 10.154.7.
$\square$

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