Lemma 37.62.11. Let $f : X \to Y$ be a morphism of schemes. The following are equivalent

1. $f$ is weakly étale, and

2. for $x \in X$ the local ring map $\mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$ induces an isomorphism on strict henselizations.

Proof. Let $x \in X$ be a point with image $y = f(x)$ in $Y$. Choose a separable algebraic closure $\kappa ^{sep}$ of $\kappa (x)$. Let $\mathcal{O}_{X, x}^{sh}$ be the strict henselization corresponding to $\kappa ^{sep}$ and $\mathcal{O}_{Y, y}^{sh}$ the strict henselization relative to the separable algebraic closure of $\kappa (y)$ in $\kappa ^{sep}$. Consider the commutative diagram

$\xymatrix{ \mathcal{O}_{X, x} \ar[r] & \mathcal{O}_{X, x}^{sh} \\ \mathcal{O}_{Y, y} \ar[u] \ar[r] & \mathcal{O}_{Y, y}^{sh} \ar[u] }$

local homomorphisms of local rings, see Algebra, Lemma 10.155.10. Since the strict henselization is a filtered colimit of étale ring maps, More on Algebra, Lemma 15.104.14 shows the horizontal maps are weakly étale. Moreover, the horizontal maps are faithfully flat by More on Algebra, Lemma 15.45.1.

Assume $f$ weakly étale. By Lemma 37.62.2 the left vertical arrow is weakly étale. By More on Algebra, Lemmas 15.104.9 and 15.104.11 the right vertical arrow is weakly étale. By More on Algebra, Theorem 15.104.24 we conclude the right vertical map is an isomorphism.

Assume $\mathcal{O}_{Y, y}^{sh} \to \mathcal{O}_{X, x}^{sh}$ is an isomorphism. Then $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}^{sh}$ is weakly étale. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{X, x}^{sh}$ is faithfully flat we conclude that $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is weakly étale by More on Algebra, Lemma 15.104.10. Thus (2) implies (1) by Lemma 37.62.2. $\square$

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