Proof.
Recall that the local ring $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{U, u}$, see Properties of Spaces, Lemma 66.22.1. By Varieties, Lemma 33.6.2 we find that $U$ is geometrically reduced at $u$ if and only if $\mathcal{O}_{U, u}$ is geometrically reduced over $k$. Thus we have to show: if $A$ is a local $k$-algebra, then $A$ is geometrically reduced over $k$ if and only if $A^{sh}$ is geometrically reduced over $k$. We check this using the definition of geometrically reduced algebras (Algebra, Definition 10.43.1). Let $K/k$ be a field extension. Since $A \to A^{sh}$ is faithfully flat (More on Algebra, Lemma 15.45.1) we see that $A \otimes _ k K \to A^{sh} \otimes _ k K$ is faithfully flat (Algebra, Lemma 10.39.7). Hence if $A^{sh} \otimes _ k K$ is reduced, so is $A \otimes _ k K$ by Algebra, Lemma 10.164.2. Conversely, recall that $A^{sh}$ is a colimit of étale $A$-algebra, see Algebra, Lemma 10.155.2. Thus $A^{sh} \otimes _ k K$ is a filtered colimit of étale $A \otimes _ k K$-algebras. We conclude by Algebra, Lemma 10.163.7.
$\square$
Comments (0)