When are constructible points closed?

Question

Let $X$ be a scheme. What technical hypotheses must be imposed on $X$ to assure that a point $p \in X$ is closed if and only if the 1-point set $\{p\}$ is constructible?

Answer 1

Let $X$ be locally noetherian. Then $\{x\}$ is constructible if and only if $\{x\}$ is locally closed (for non-noetherian schemes the notion of constructibility is more complicated and all kind of terrible things can happen, e.g. there exist closed points $x$ such that $\{x\}$ is not constructible). Moreover it is a nice exercise that this is the case if and only if the canonical morphism ${\rm Spec} \kappa(x) \to X$ is of finite type. On the other hand, $\{x\}$ is closed if and only if the canonical morphism ${\rm Spec}\kappa(x) \to X$ is finite (or, equivalently, a closed immersion). Thus we are looking for a scheme $X$ that has a covering by open affine subschemes $U = {\rm Spec}A$ such that every finitely generated $A$-algebra which is a field is already a finite $A$-algebra. Such rings are called Jacobson rings. They are also characterized by the property that every prime ideal is the intersection of (not necessarily finitely many) maximal ideals.

Upshot: For a locally noetherian scheme $X$ the following properties are equivalent.

(i) For all $x \in X$ the set $\{x\}$ is closed if and only if it is constructible.

(ii) $X$ is Jacobson.

Every scheme locally of finite type over a Jacobson ring is a Jacobson scheme. Examples for Jacobson rings are of course fields but also integral domains of dimension $1$ with infinitely many prime ideals. Thus every scheme locally of finite type over a field is Jacobson (as already remarked by Damian) but also every scheme locally of finite type over ${\mathbb Z}$ is Jacobson.