Lemma 10.153.8. Let $(R, \mathfrak m, \kappa )$ be a strictly henselian local ring. Let $R \to S$ be an unramified ring map. Then

$S = A_1 \times \ldots \times A_ n \times B$

with each $R \to A_ i$ surjective and no prime of $B$ lying over $\mathfrak m$.

Proof. First write $S = A_1 \times \ldots \times A_ n \times B$ as in Lemma 10.153.5. Now we see that $R \to A_ i$ is finite unramified and $A_ i$ local. Hence the maximal ideal of $A_ i$ is $\mathfrak mA_ i$ and its residue field $A_ i / \mathfrak m A_ i$ is a finite separable extension of $\kappa$, see Lemma 10.151.5. However, the condition that $R$ is strictly henselian means that $\kappa$ is separably algebraically closed, so $\kappa = A_ i / \mathfrak m A_ i$. By Nakayama's Lemma 10.20.1 we conclude that $R \to A_ i$ is surjective as desired. $\square$

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