The Stacks project

Lemma 5.23.4. Let $X$ be a spectral space. Let $E \subset X$ be closed in the constructible topology (for example constructible or closed). Then $E$ with the induced topology is a spectral space.

Proof. Let $Z \subset E$ be a closed irreducible subset. Let $\eta $ be the generic point of the closure $\overline{Z}$ of $Z$ in $X$. To prove that $E$ is sober, we show that $\eta \in E$. If not, then since $E$ is closed in the constructible topology, there exists a constructible subset $F \subset X$ such that $\eta \in F$ and $F \cap E = \emptyset $. By Lemma 5.15.15 this implies $F \cap \overline{Z}$ contains a nonempty open subset of $\overline{Z}$. But this is impossible as $\overline{Z}$ is the closure of $Z$ and $Z \cap F = \emptyset $.

Since $E$ is closed in the constructible topology, it is quasi-compact in the constructible topology (Lemmas 5.12.3 and 5.23.2). Hence a fortiori it is quasi-compact in the topology coming from $X$. If $U \subset X$ is a quasi-compact open, then $E \cap U$ is closed in the constructible topology, hence quasi-compact (as seen above). It follows that the quasi-compact open subsets of $E$ are the intersections $E \cap U$ with $U$ quasi-compact open in $X$. These form a basis for the topology. Finally, given two $U, U' \subset X$ quasi-compact opens, the intersection $(E \cap U) \cap (E \cap U') = E \cap (U \cap U')$ and $U \cap U'$ is quasi-compact as $X$ is spectral. This finishes the proof. $\square$


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Your email address will not be published.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 0902. Beware of the difference between the letter 'O' and the digit '0'.