Lemma 65.28.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is of finite presentation and $Y$ is Noetherian, then $X$ is Noetherian.
Proof. Assume $f$ is of finite presentation and $Y$ Noetherian. By Lemmas 65.28.5 and 65.23.5 we see that $X$ is locally Noetherian. As $f$ is quasi-compact and $Y$ is quasi-compact we see that $X$ is quasi-compact. As $f$ is of finite presentation it is quasi-separated (see Definition 65.28.1) and as $Y$ is Noetherian it is quasi-separated (see Properties of Spaces, Definition 64.24.1). Hence $X$ is quasi-separated by Lemma 65.4.9. Hence we have checked all three conditions of Properties of Spaces, Definition 64.24.1 and we win. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.