Lemma 61.8.5. Let $A$ be a ring such that every faithfully flat étale ring map $A \to B$ has a section. Then every local ring of $A$ at a maximal ideal is strictly henselian.

Proof. Let $\mathfrak m$ be a maximal ideal of $A$. Let $A \to B$ be an étale ring map and let $\mathfrak q \subset B$ be a prime lying over $\mathfrak m$. By the description of the strict henselization $A_\mathfrak m^{sh}$ in Algebra, Lemma 10.155.11 it suffices to show that $A_\mathfrak m = B_\mathfrak q$. Note that there are finitely many primes $\mathfrak q = \mathfrak q_1, \mathfrak q_2, \ldots , \mathfrak q_ n$ lying over $\mathfrak m$ and there are no specializations between them as an étale ring map is quasi-finite, see Algebra, Lemma 10.143.6. Thus $\mathfrak q_ i$ is a maximal ideal and we can find $g \in \mathfrak q_2 \cap \ldots \cap \mathfrak q_ n$, $g \not\in \mathfrak q$ (Algebra, Lemma 10.15.2). After replacing $B$ by $B_ g$ we see that $\mathfrak q$ is the only prime of $B$ lying over $\mathfrak m$. The image $U \subset \mathop{\mathrm{Spec}}(A)$ of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is open (Algebra, Proposition 10.41.8). Thus the complement $\mathop{\mathrm{Spec}}(A) \setminus U$ is closed and we can find $f \in A$, $f \not\in \mathfrak p$ such that $\mathop{\mathrm{Spec}}(A) = U \cup D(f)$. The ring map $A \to B \times A_ f$ is faithfully flat and étale, hence has a section $\sigma : B \times A_ f \to A$ by assumption on $A$. Observe that $\sigma$ is étale, hence flat as a map between étale $A$-algebras (Algebra, Lemma 10.143.8). Since $\mathfrak q$ is the only prime of $B \times A_ f$ lying over $A$ we find that $A_\mathfrak p \to B_\mathfrak q$ has a section which is also flat. Thus $A_\mathfrak p \to B_\mathfrak q \to A_\mathfrak p$ are flat local ring maps whose composition is the identity. Since a flat local homomorphism of local rings is injective we conclude these maps are isomorphisms as desired. $\square$

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