Lemma 29.35.16 (02GG)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 29: Morphisms of Schemes
Section 29.35: Unramified morphisms
Lemma 29.35.16 (cite)

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Lemma 29.35.16. Let $f : X \to Y$ be a morphism of schemes over $S$ .

If $X$ is unramified over $S$ , then $f$ is unramified.
If $X$ is G-unramified over $S$ and $Y$ is locally of finite type over $S$ , then $f$ is G-unramified.

Proof. Assume that $X$ is unramified over $S$ . By Lemma 29.15.8 we see that $f$ is locally of finite type. By assumption we have $\Omega _{X/S} = 0$ . Hence $\Omega _{X/Y} = 0$ by Lemma 29.32.9. Thus $f$ is unramified. If $X$ is G-unramified over $S$ and $Y$ is locally of finite type over $S$ , then by Lemma 29.21.11 we see that $f$ is locally of finite presentation and we conclude that $f$ is G-unramified. $\square$

Comments (3)

Comment #7363 by Alex Ivanov on May 16, 2022 at 11:14

In (2), it should in fact suffice to assume that $Y$ is locally of finite type over $S$ . (This is also consistent with Lemma 02FW, to which the proof refers).

Comment #7385 by Zhiyu Z on May 25, 2022 at 15:49

@7363 I agree with you, and "Y of finite type" shall be "Y is of finite type".

Comment #7573 by Stacks Project on August 12, 2022 at 14:35

This was fixed a while back here.

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