Lemma 28.5.9. Any nonempty locally Noetherian scheme has a closed point. Any nonempty closed subset of a locally Noetherian scheme has a closed point. Equivalently, any point of a locally Noetherian scheme specializes to a closed point.

Proof. The second assertion follows from the first (using Schemes, Lemma 26.12.4 and Lemma 28.5.6). Consider any nonempty affine open $U \subset X$. Let $x \in U$ be a closed point. If $x$ is a closed point of $X$ then we are done. If not, let $X_0 \subset X$ be the reduced induced closed subscheme structure on $\overline{\{ x\} }$. Then $U_0 = U \cap X_0$ is an affine open of $X_0$ by Schemes, Lemma 26.10.1 and $U_0 = \{ x\}$. Let $y \in X_0$, $y \not= x$ be a specialization of $x$. Consider the local ring $R = \mathcal{O}_{X_0, y}$. This is a Noetherian local ring as $X_0$ is Noetherian by Lemma 28.5.6. Denote $V \subset \mathop{\mathrm{Spec}}(R)$ the inverse image of $U_0$ in $\mathop{\mathrm{Spec}}(R)$ by the canonical morphism $\mathop{\mathrm{Spec}}(R) \to X_0$ (see Schemes, Section 26.13.) By construction $V$ is a singleton with unique point corresponding to $x$ (use Schemes, Lemma 26.13.2). By Algebra, Lemma 10.61.1 we see that $\dim (R) = 1$. In other words, we see that $y$ is an immediate specialization of $x$ (see Topology, Definition 5.20.1). In other words, any point $y \not= x$ such that $x \leadsto y$ is an immediate specialization of $x$. Clearly each of these points is a closed point as desired. $\square$

Comment #447 by Qing Liu on

Hi, I think there is some flaw in the proof (that $V$ is a singleton). If we identify Spec($R$) with its image in $X$, then $V$ is the set of points of $U$ which specializes to $y$. In general $V$ is bigger than $\{ x \}$, thus is not a singleton (example: $X$ equal to the affine line over a DVR, $U$ its generic fiber, $x$ a closed point of $U$ specializing to the origin of the closed fiber; then $V$ is a kind of open unit disk). However, the proof can be easily repaired by replacing $X$ with $X_0:=\overline{\{ x\}}$ (say with the reduced structure) and $U$ with $U\cap X_0$. In the new situation, $X_0$ is an integral scheme, its generic point $x$ is open. Then the same argument shows that $O_{X_0, y}$ has dimension $1$.

Comment #449 by on

Oops! Yes. What a bizarre error. Thanks! If you come by before I run out of them, I'll give you a Stacks project mug. Fixed here.

Comment #1073 by Antoine Chambert-Loir on

Second statement has to be corrected: every non-empty closed subset... contains a closed point. (Sorry...)

Comment #1077 by on

Yes, indeed! Also, it seems your name hadn't been added to the list of contributors yet and I've done so this time. Thanks! The change is here.

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