Lemma 15.104.23. Let $A \to B$ be a ring homomorphism. Assume
$A$ is a henselian local ring,
$A \to B$ is integral,
$B$ is a domain.
Then $B$ is a henselian local ring and $A \to B$ is a local homomorphism. If $A$ is strictly henselian, then $B$ is a strictly henselian local ring and the extension $\kappa (\mathfrak m_ B)/\kappa (\mathfrak m_ A)$ of residue fields is purely inseparable.
Proof. Write $B$ as a filtered colimit $B = \mathop{\mathrm{colim}}\nolimits B_ i$ of finite $A$-sub algebras. If we prove the results for each $B_ i$, then the result follows for $B$. See Algebra, Lemma 10.154.8. If $A \to B$ is finite, then $B$ is a product of local henselian rings by Algebra, Lemma 10.153.4. Since $B$ is a domain we see that $B$ is a local ring. The maximal ideal of $B$ lies over the maximal ideal of $A$ by going up for $A \to B$ (Algebra, Lemma 10.36.22). If $A$ is strictly henselian, then the field extension $\kappa (\mathfrak m_ B)/\kappa (\mathfrak m_ A)$ being algebraic, has to be purely inseparable. Of course, then $\kappa (\mathfrak m_ B)$ is separably algebraically closed and $B$ is strictly henselian. $\square$
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