The Stacks project

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

$$X({\bar{y}})\cong\coprod_{a\in X_{\bar{y}}}{\rm Spec}(\mathscr{O}_{X({\bar{y}}), a})=\coprod_{x\in X_y}\coprod_{a\in(X_{\bar{y}}\to X_y)^{-1}(x)}{\rm Spec}(\mathscr{O}_{X({\bar{y}}), a}).$$ 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)

$$X({\bar{y}})\cong\coprod_{x\in X_y}{\rm Spec}((\mathscr{O}_{X, \bar{x}})^{[\kappa_x: \kappa_y]_s})=\coprod_{x\in X_y}\coprod_{[\kappa_x: \kappa_y]_s}{\rm Spec}(\mathscr{O}_{X, \bar{x}}).$$

Thanks and fixed here.