The Stacks project

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 ├ętale hence smooth $R$-algebras, the result follows from Algebra, Lemmas 10.163.9 and 10.37.17. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06DI. Beware of the difference between the letter 'O' and the digit '0'.