Theorem 15.104.24 (Olivier). Let $A \to B$ be a local homomorphism of local rings. If $A$ is strictly henselian and $A \to B$ is weakly étale, then $A = B$.

**Proof.**
We will show that for all $\mathfrak p \subset A$ there is a unique prime $\mathfrak q \subset B$ lying over $\mathfrak p$ and $\kappa (\mathfrak p) = \kappa (\mathfrak q)$. This implies that $B \otimes _ A B \to B$ is bijective on spectra as well as surjective and flat. Hence it is an isomorphism for example by the description of pure ideals in Algebra, Lemma 10.108.4. Hence $A \to B$ is a faithfully flat epimorphism of rings. We get $A = B$ by Algebra, Lemma 10.107.7.

Note that the fibre ring $B \otimes _ A \kappa (\mathfrak p)$ is a colimit of étale extensions of $\kappa (\mathfrak p)$ by Lemmas 15.104.7 and 15.104.16. Hence, if there exists more than one prime lying over $\mathfrak p$ or if $\kappa (\mathfrak p) \not= \kappa (\mathfrak q)$ for some $\mathfrak q$, then $B \otimes _ A L$ has a nontrivial idempotent for some (separable) algebraic field extension $L/\kappa (\mathfrak p)$.

Let $L/\kappa (\mathfrak p)$ be an algebraic field extension. Let $A' \subset L$ be the integral closure of $A/\mathfrak p$ in $L$. By Lemma 15.104.23 we see that $A'$ is a strictly henselian local ring whose residue field is a purely inseparable extension of the residue field of $A$. Thus $B \otimes _ A A'$ is a local ring by Algebra, Lemma 10.156.5. On the other hand, $B \otimes _ A A'$ is integrally closed in $B \otimes _ A L$ by Lemma 15.104.22. Since $B \otimes _ A A'$ is local, it follows that the ring $B \otimes _ A L$ does not have nontrivial idempotents which is what we wanted to prove. $\square$

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