The Stacks project

Lemma 66.46.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

  1. $f$ is finite locally free,

  2. $f$ is finite, flat, and locally of finite presentation.

If $Y$ is locally Noetherian these are also equivalent to

  1. $f$ is finite and flat.

Proof. In each of the three cases the morphism is representable and you can check the property after base change by a surjective ├ętale morphism $V \to Y$, see Lemmas 66.45.3, 66.46.3, 66.30.5, and 66.28.4. If $Y$ is locally Noetherian, then $V$ is locally Noetherian. Hence the result follows from the corresponding result in the schemes case, see Morphisms, Lemma 29.48.2. $\square$


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Your email address will not be published.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 0416. Beware of the difference between the letter 'O' and the digit '0'.