Lemma 66.23.4. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Then $X$ is Jacobson if and only if $|X|$ is Jacobson.

**Proof.**
Assume $X$ is Jacobson and that $T \subset |X|$ is a closed subset. By Morphisms of Spaces, Lemma 65.25.6 we see that $T \cap X_{\text{ft-pts}}$ is dense in $T$. By Lemma 66.23.3 we see that $X_{\text{ft-pts}}$ are the closed points of $|X|$. Thus $|X|$ is indeed Jacobson.

Assume $|X|$ is Jacobson. Let $f : U \to X$ be an étale morphism with $U$ an affine scheme. We have to show that $U$ is Jacobson. If $x \in |X|$ is closed, then the fibre $F = f^{-1}(\{ x\} )$ is a finite (by definition of decent) closed (by construction of the topology on $|X|$) subset of $U$. Since there are no specializations between points of $F$ (Lemma 66.12.1) we conclude that every point of $F$ is closed in $U$. If $U$ is not Jacobson, then there exists a non-closed point $u \in U$ such that $\{ u\} $ is locally closed (Topology, Lemma 5.18.3). We will show that $f(u) \in |X|$ is closed; by the above $u$ is closed in $U$ which is a contradiction and finishes the proof. To prove this we may replace $U$ by an affine open neighbourhood of $u$. Thus we may assume that $\{ u\} $ is closed in $U$. Let $R = U \times _ X U$ with projections $s, t : R \to U$. Then $s^{-1}(\{ u\} ) = \{ r_1, \ldots , r_ m\} $ is finite (by definition of decent spaces). After replacing $U$ by a smaller affine open neighbourhood of $u$ we may assume that $t(r_ j) = u$ for $j = 1, \ldots , m$. It follows that $\{ u\} $ is an $R$-invariant closed subset of $U$. Hence $\{ f(u)\} $ is a locally closed subset of $X$ as it is closed in the open $|f|(|U|)$ of $|X|$. Since $|X|$ is Jacobson we conclude that $f(u)$ is closed in $|X|$ as desired. $\square$

## 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Your email address will not be published.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).

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.

## Comments (0)