# The Stacks Project

## Tag: 01IS

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The corresponding content:

Lemma 22.11.1. 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 5.7.4.

Proof. Let $Z \subset X$ be an irreducible closed subset. For every affine open $U \subset X$, $U = \mathop{\rm Spec}(R)$ we know that $Z \cap U = V(I)$ for a unique radical ideal $I \subset R$. Note that $Z \cap I$ 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 $\xi' = \xi$. $\square$

\begin{lemma}
\label{lemma-scheme-sober}
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 \ref{topology-definition-generic-point}.
\end{lemma}

\begin{proof}
Let $Z \subset X$ be an irreducible closed subset.
For every affine open $U \subset X$, $U = \Spec(R)$
we know that $Z \cap U = V(I)$ for a unique
radical ideal $I \subset R$. Note that $Z \cap I$ 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 $\xi' = \xi$.
\end{proof}


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