The Stacks project

Lemma 35.23.23. The property $\mathcal{P}(f) =$“$f$ is finite” is fpqc local on the base.

Proof. An finite morphism is the same thing as an integral morphism which is locally of finite type. See Morphisms, Lemma 29.44.4. Hence the lemma follows on combining Lemmas 35.23.10 and 35.23.22. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 35.23: Properties of morphisms local in the fpqc topology on the target

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Your email address will not be published.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



View Lemma 35.23.23 as pdf