Lemma 37.62.11 (0E9K)—The Stacks project

Table of contents
Part 2: Schemes
Chapter 37: More on Morphisms
Section 37.62: Local complete intersection morphisms
Lemma 37.62.11 (cite)

Lemma 37.62.11. 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 July 06, 2019 at 05:42

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 Johan on September 01, 2019 at 16:34

The statement you want is a combination of Lemmas 37.62.12 and 37.61.6. 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).

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