Lemma 15.45.6. Let $R$ be a local ring. The following are equivalent: $R$ is a normal domain, the henselization $R^ h$ of $R$ is a normal domain, and the strict henselization $R^{sh}$ of $R$ is a normal domain.
Proof. A preliminary remark is that a local ring is normal if and only if it is a normal domain (see Algebra, Definition 10.37.11). The ring maps $R \to R^ h \to R^{sh}$ are faithfully flat. Hence one direction of the implications follows from Algebra, Lemma 10.164.3. Conversely, assume $R$ is normal. Since $R^ h$ and $R^{sh}$ are filtered colimits of étale hence smooth $R$-algebras, the result follows from Algebra, Lemmas 10.163.9 and 10.37.17. $\square$
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