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
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