The Stacks project

Lemma 37.59.11. Let $f : X \to Y$ be a morphism of schemes. If $f$ is locally of finite type and $X$ and $Y$ are regular, then $f$ is a local complete intersection morphism.

Proof. We may assume there is a factorization $X \to \mathbf{A}^ n_ Y \to Y$ where the first arrow is an immersion. As $Y$ is regular also $\mathbf{A}^ n_ Y$ is regular by Algebra, Lemma 10.163.10. Hence $X \to \mathbf{A}^ n_ Y$ is a regular immersion by Divisors, Lemma 31.21.12. $\square$


Comments (2)

Comment #4295 by comment_bot on

It would be useful to strengthen this statement: it suffices to assume that is a local complete intersection (with still regular). A similar comment applies to https://stacks.math.columbia.edu/tag/0E9J

Comment #4460 by on

The statement you want is a combination of Lemmas 37.59.12 and 37.58.6. I have added it as a lemma in this commit. Hope you will use it. The corresponding analogue of Lemma 31.21.12 immediately follows from this new lemma (as a local complete intersection morphism of locally Noetherian schemes which is also an immersion is a regular immersion essentially by definition).

There are also:

  • 4 comment(s) on Section 37.59: Local complete intersection morphisms

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Your email address will not be published.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 0E9K. Beware of the difference between the letter 'O' and the digit '0'.