The Stacks project

Lemma 35.24.1. The property $\mathcal{P}(f) =$“$f$ is an immersion” is fppf local on the base.

Proof. The property of being an immersion is stable under base change, see Schemes, Lemma 26.18.2. The property of being an immersion is Zariski local on the base. Finally, let $\pi : S' \to S$ be a surjective morphism of affine schemes, which is flat and locally of finite presentation. Note that $\pi : S' \to S$ is open by Morphisms, Lemma 29.25.10. Let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is an immersion. In particular we see that $f'(X') = \pi ^{-1}(f(X))$ is locally closed. Hence by Topology, Lemma 5.6.4 we see that $f(X) \subset S$ is locally closed. Let $Z \subset S$ be the closed subset $Z = \overline{f(X)} \setminus f(X)$. By Topology, Lemma 5.6.4 again we see that $f'(X')$ is closed in $S' \setminus Z'$. Hence we may apply Lemma 35.23.19 to the fpqc covering $\{ S' \setminus Z' \to S \setminus Z\} $ and conclude that $f : X \to S \setminus Z$ is a closed immersion. In other words, $f$ is an immersion. Therefore Lemma 35.22.4 applies and we win. $\square$

Comments (0)

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Your email address will not be published.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 02YM. Beware of the difference between the letter 'O' and the digit '0'.