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
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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