Lemma 26.11.1. Let $X$ be a scheme. Any irreducible closed subset of $X$ has a unique generic point. In other words, $X$ is a sober topological space, see Topology, Definition 5.8.6.

Proof. Let $Z \subset X$ be an irreducible closed subset. For every affine open $U \subset X$, $U = \mathop{\mathrm{Spec}}(R)$ we know that $Z \cap U = V(I)$ for a unique radical ideal $I \subset R$. Note that $Z \cap U$ is either empty or irreducible. In the second case (which occurs for at least one $U$) we see that $I = \mathfrak p$ is a prime ideal, which is a generic point $\xi$ of $Z \cap U$. It follows that $Z = \overline{\{ \xi \} }$, in other words $\xi$ is a generic point of $Z$. If $\xi '$ was a second generic point, then $\xi ' \in Z \cap U$ and it follows immediately that $\xi ' = \xi$. $\square$

Comment #935 by correction_bot on

Typo: replace $Z \cap I$ with $Z \cap U$.

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