The Stacks project

Local rings of dimension zero are henselian.

Lemma 10.153.10. Let $(R, \mathfrak m)$ be a local ring of dimension $0$. Then $R$ is henselian.

Proof. Let $R \to S$ be a finite ring map. By Lemma 10.153.3 it suffices to show that $S$ is a product of local rings. By Lemma 10.36.21 $S$ has finitely many primes $\mathfrak m_1, \ldots , \mathfrak m_ r$ which all lie over $\mathfrak m$. There are no inclusions among these primes, see Lemma 10.36.20, hence they are all maximal. Every element of $\mathfrak m_1 \cap \ldots \cap \mathfrak m_ r$ is nilpotent by Lemma 10.17.2. It follows $S$ is the product of the localizations of $S$ at the primes $\mathfrak m_ i$ by Lemma 10.53.5. $\square$

Comments (1)

There are also:

  • 6 comment(s) on Section 10.153: Henselian local rings

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Your email address will not be published.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 06RS. Beware of the difference between the letter 'O' and the digit '0'.