The Stacks project

Lemma 10.154.15. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p$ in $R$. Let $\kappa (\mathfrak q) \subset \kappa ^{sep}$ be a separable algebraic closure. Assume $R \to S$ is quasi-finite at $\mathfrak q$. The commutative diagram

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

of Lemma 10.154.12 identifies $S_{\mathfrak q}^{sh}$ with a localization of $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$.

Proof. The residue field of $R_{\mathfrak p}^{sh}$ is the separable algebraic closure of $\kappa (\mathfrak p)$ in $\kappa ^{sep}$. Note that $R_{\mathfrak p}^{sh} \otimes _ R S$ is quasi-finite over $R_{\mathfrak p}^{sh}$ at the prime ideal corresponding to $\mathfrak q$, see Lemma 10.121.6. Hence the localization $S'$ of $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$ is henselian, see Lemma 10.152.4. Note that the residue field of $S'$ is $\kappa ^{sep}$ since it contains both the separable algebraic closure of $\kappa (\mathfrak p)$ and $\kappa (\mathfrak q)$. Furthermore, as a localization $S'$ is a filtered colimit of ├ętale $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$-algebras. By Lemma 10.154.14 we see that $S_{\mathfrak q}^{sh}$ is a strict henselization of $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$. Thus $S' = S_\mathfrak q^{sh}$ by the uniqueness result of Lemma 10.153.6. $\square$

Comments (2)

Comment #4679 by Peng DU on

In the last formula in the Proof, it should be .

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  • 6 comment(s) on Section 10.154: Henselization and strict henselization

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