Lemma 10.155.12. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Choose separable algebraic closures $\kappa (\mathfrak p) \subset \kappa _1^{sep}$ and $\kappa (\mathfrak q) \subset \kappa _2^{sep}$. Let $R^{sh}$ and $S^{sh}$ be the corresponding strict henselizations of $R_\mathfrak p$ and $S_\mathfrak q$. Given any commutative diagram

$\xymatrix{ \kappa _1^{sep} \ar[r]_{\phi } & \kappa _2^{sep} \\ \kappa (\mathfrak p) \ar[r]^{\varphi } \ar[u] & \kappa (\mathfrak q) \ar[u] }$

The local ring map $R^{sh} \to S^{sh}$ of Lemma 10.155.10 identifies $S^{sh}$ with the strict henselization of $R^{sh} \otimes _ R S$ at a prime lying over $\mathfrak q$ and the maximal ideal $\mathfrak m^{sh} \subset R^{sh}$.

Proof. The proof is identical to the proof of Lemma 10.155.8 except that it uses Lemma 10.155.11 instead of Lemma 10.155.7. $\square$

Comment #4681 by Peng DU on

In the statement, it uses $\mathfrak{m}^{sh}$, which is not introduced.

BTW, this will answer the follow question https://mathoverflow.net/questions/345381/a-question-about-strict-henselian-local-rings in MathOverflow: Let $f: X\to Y$ be a finite morphism of schemes, let $y\in Y, Y(\bar{y}):={\rm Spec}(\mathscr{O}_{Y, \bar{y}}), X_{\bar{y}}:=X\times_Y{\rm Spec}(\kappa_y^s)=X_y\otimes_{\kappa_y}\kappa_y^s, X({\bar{y}}):=X\times_YY(\bar{y})$, here $\kappa_y^s$ is a separable closure of the residue field $\kappa_y$. By some results about strict henselian local rings one can see This result (tag/08HV) will give $\mathscr{O}_{X(\bar{y}), a}=\mathscr{O}_{X, \bar{x}}$ if $a\in(X_{\bar{y}}\to X_y)^{-1}(x)$.

This further gives a decomposition ($[-:-]_s$ stands for separable degree)

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