Lemma 5.23.5. 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. $\square$

Comment #5059 by Laurent Moret-Bailly on

Obvious converse (from previous lemma): if $E$ is a spectral subspace (with spectral inclusion map) then $E$ is constructibly closed.

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