The Stacks project

Smooth over a field implies regular

Lemma 33.25.3. Let $X \to \mathop{\mathrm{Spec}}(k)$ be a smooth morphism where $k$ is a field. Then $X$ is a regular scheme.

Proof. (See also Lemma 33.12.6.) By Algebra, Lemma 10.139.3 every local ring $\mathcal{O}_{X, x}$ is regular. And because $X$ is locally of finite type over $k$ it is locally Noetherian. Hence $X$ is regular by Properties, Lemma 28.9.2. $\square$


Comments (1)

Comment #1049 by Lenny Taelman on

Suggested slogan: Smooth over a field implies regular


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Your email address will not be published.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 056S. Beware of the difference between the letter 'O' and the digit '0'.