Lemma 26.11.1 (01IS)—The Stacks project

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Part 2: Schemes
Chapter 26: Schemes
Section 26.11: Zariski topology of schemes
Lemma 26.11.1 (cite)

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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 $. $\square$

Comments (3)

Comment #935 by correction_bot on August 22, 2014 at 19:46

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

Comment #9572 by Lucas Henrique on August 21, 2024 at 18:39

There's a slight hypothesis missing: any nonempty irreducible closed subset. Otherwise, the proof fails when it claims that $Z\cap U$ is nonempty for at least one $U$ .

Comment #9994 by Joe Lamond on February 18, 2025 at 13:42

Lucas, in the Stacks Project an irreducible topological space is nonempty by definition (see https://stacks.math.columbia.edu/tag/004U#:~:text=Let%20X%20be%20a%20topological,maximal%20irreducible%20subset%20of%20X.). One reason this convention makes sense is that an irreducible topological space can be defined as a topological space which cannot be expressed as a finite union of proper closed subspaces. But the empty space equals the empty union, hence it is not irreducible. A similar line of reasoning explains why the empty topological space is not connected.

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