Lemma 66.23.1. Let $S$ be a scheme. Let $X$ be a Jacobson algebraic space over $S$. Any algebraic space locally of finite type over $X$ is Jacobson.
Proof. Let $U \to X$ be a surjective étale morphism where $U$ is a scheme. Then $U$ is Jacobson (by definition) and for a morphism of schemes $V \to U$ which is locally of finite type we see that $V$ is Jacobson by the corresponding result for schemes (Morphisms, Lemma 29.16.9). Thus if $Y \to X$ is a morphism of algebraic spaces which is locally of finite type, then setting $V = U \times _ X Y$ we see that $Y$ is Jacobson by definition. $\square$
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Your email address will not be published.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.