Lemma 37.60.11. 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. $\square$

Comment #4295 by comment_bot on

It would be useful to strengthen this statement: it suffices to assume that $X$ is a local complete intersection (with $Y$ 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 37.60.12 and 37.59.6. 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 37.60: Local complete intersection morphisms

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).