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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?

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If $X$ is of finite type over a field $k$, then this is true. Indeed a closed point $P$ then coincides with a point of residue field a finite extension of $k$ and this property is unchanged if $X$ is replaced by an open subset containing $P$. Now ${p}$ is constructible if and only if ${p}$ is closed in some open subset of $X$ (see Hartshorne, Alg. Geom., II, Ex. 3.17 & 3.18). Also, notice that a counterexample is given by the generic point of a discrete valuation ring (which is constructible but not closed). This is written in a rush. I hope I didn't make a silly mistake. –  Damian Rössler Sep 28 '11 at 17:54
    
Finitely-generated over a field should do the trick. Is that weak enough? –  Will Sawin Sep 28 '11 at 17:55
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up vote 10 down vote accepted

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.

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Great, this is exactly what I was looking for. Thanks! –  Ian A Sep 28 '11 at 19:27
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