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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. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 40922–40939 (see updates for more information).

    \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}

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