Tag 05WP
Chapter 10: Commutative Algebra > Section 10.150: Henselization and strict henselization
Lemma 10.150.9. 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 10.150.6 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 $\mathfrak q$.
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 10.121.6. Hence the localization $S'$ of $R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$ is henselian, see Lemma 10.148.4. As a localization $S'$ is a filtered colimit of étale $R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$-algebras. By Lemma 10.150.8 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 10.149.6. $\square$
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\begin{lemma}
\label{lemma-quasi-finite-henselization}
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 \ref{lemma-henselian-functorial}
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 $\mathfrak q$.
\end{lemma}
\begin{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 \ref{lemma-four-rings}. Hence the localization $S'$ of
$R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$ is henselian, see
Lemma \ref{lemma-finite-over-henselian}. As a localization $S'$ is a filtered
colimit of \'etale
$R_{\mathfrak p}^h \otimes_{R_{\mathfrak p}} S_{\mathfrak q}$-algebras.
By Lemma \ref{lemma-henselian-functorial-improve} 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 \ref{lemma-uniqueness-henselian}.
\end{proof}Comments (0)
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