Proof. Let $X$, $Y$ be spectral spaces. Denote $p : X \times Y \to X$ and $q : X \times Y \to Y$ the projections. Let $Z \subset X \times Y$ be a closed irreducible subset. Then $p(Z) \subset X$ is irreducible and $q(Z) \subset Y$ is irreducible. Let $x \in X$ be the generic point of the closure of $p(X)$ and let $y \in Y$ be the generic point of the closure of $q(Y)$. If $(x, y) \not\in Z$, then there exist opens $x \in U \subset X$, $y \in V \subset Y$ such that $Z \cap U \times V = \emptyset$. Hence $Z$ is contained in $(X \setminus U) \times Y \cup X \times (Y \setminus V)$. Since $Z$ is irreducible, we see that either $Z \subset (X \setminus U) \times Y$ or $Z \subset X \times (Y \setminus V)$. In the first case $p(Z) \subset (X \setminus U)$ and in the second case $q(Z) \subset (Y \setminus V)$. Both cases are absurd as $x$ is in the closure of $p(Z)$ and $y$ is in the closure of $q(Z)$. Thus we conclude that $(x, y) \in Z$, which means that $(x, y)$ is the generic point for $Z$.

A basis of the topology of $X \times Y$ are the opens of the form $U \times V$ with $U \subset X$ and $V \subset Y$ quasi-compact open (here we use that $X$ and $Y$ are spectral). Then $U \times V$ is quasi-compact as the product of quasi-compact spaces is quasi-compact. Moreover, any quasi-compact open of $X \times Y$ is a finite union of such quasi-compact rectangles $U \times V$. It follows that the intersection of two such is again quasi-compact (since $X$ and $Y$ are spectral). This concludes the proof. $\square$

Comment #435 by Kestutis Cesnavicius on

$V \subset T$ should be $V \subset Y$.

There are also:

• 4 comment(s) on Section 5.23: Spectral spaces

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).