Lemma 66.23.5. Let $S$ be a scheme. Let $X$ be a decent locally Noetherian algebraic space over $S$. Let $x \in |X|$. Then

$W = \{ x' \in |X| : x' \leadsto x,\ x' \not= x\}$

is a Noetherian, spectral, sober, Jacobson topological space.

Proof. We may replace by any open subspace containing $x$. Thus we may assume that $X$ is quasi-compact. Then $|X|$ is a Noetherian topological space (Properties of Spaces, Lemma 64.24.2). Thus $W$ is a Noetherian topological space (Topology, Lemma 5.9.2).

Combining Lemma 66.14.1 with Properties of Spaces, Lemma 64.15.2 we see that $|X|$ is a spectral toplogical space. By Topology, Lemma 5.24.7 we see that $W \cup \{ x\}$ is a spectral topological space. Now $W$ is a quasi-compact open of $W \cup \{ x\}$ and hence $W$ is spectral by Topology, Lemma 5.23.5.

Let $E \subset W$ be an irreducible closed subset. Then if $Z \subset |X|$ is the closure of $E$ we see that $x \in Z$. There is a unique generic point $\eta \in Z$ by Proposition 66.12.4. Of course $\eta \in W$ and hence $\eta \in E$. We conclude that $E$ has a unique generic point, i.e., $W$ is sober.

Let $x' \in W$ be a point such that $\{ x'\}$ is locally closed in $W$. To finish the proof we have to show that $x'$ is a closed point of $W$. If not, then there exists a nontrivial specialization $x' \leadsto x'_1$ in $W$. Let $U$ be an affine scheme, $u \in U$ a point, and let $U \to X$ be an étale morphism mapping $u$ to $x$. By Lemma 66.12.2 we can choose specializations $u' \leadsto u'_1 \leadsto u$ mapping to $x' \leadsto x'_1 \leadsto x$. Let $\mathfrak p' \subset \mathcal{O}_{U, u}$ be the prime ideal corresponding to $u'$. The existence of the specializations implies that $\dim (\mathcal{O}_{U, u}/\mathfrak p') \geq 2$. Hence every nonempty open of $\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u}/\mathfrak p')$ is infinite by Algebra, Lemma 10.60.1. By Lemma 66.12.1 we obtain a continuous map

$\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u}/\mathfrak p') \setminus \{ \mathfrak m_ u/\mathfrak p'\} \longrightarrow W$

Since the generic point of the LHS maps to $x'$ the image is contained in $\overline{\{ x'\} }$. We conclude the inverse image of $\{ x'\}$ under the displayed arrow is nonempty open hence infinite. However, the fibres of $U \to X$ are finite as $X$ is decent and we conclude that $\{ x'\}$ is infinite. This contradiction finishes the proof. $\square$

## 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Your email address will not be published.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.

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 0ED2. Beware of the difference between the letter 'O' and the digit '0'.