Closed vs Rational Points on Schemes

Question

Background: When Ueno builds the fully faithful functor from Var/k to Sch/k he mentions that the variety $V$ can be identified with the rational points of $t(V)$ over $k$. I know how to prove this on affine everything and will work out the general case at some future time.

The question that this got me thinking about was if $X$ is a $k$-scheme where $k$ is algebraically closed, then are the $k$-rational points of $X$ just the closed points? This is probably extremely well known, but I can't find it explicitly stated nor can I find a counterexample.

For $k$ not algebraically closed, I can come up with examples where this is not true. So in general is there some relation between the closed points and rational points on schemes (everything over $k$)?

This would give a bit more insight into what this functor does. It takes the variety and makes all the points into closed points of a scheme, then adds the generic points necessary to actually make it a legitimate scheme. General tangential thoughts on this are welcome as well.

Answer 1

If $k$ is algebraically closed and $X$ is a $k$-scheme locally of finite type, then the $k$-rational points are precisely the closed points. (See EGA 1971, Ch. I, Corollaire 6.5.3).

More generally: if $k$ is a field and $X$ is a $k$-scheme locally of finite type, then $X$ is a Jacobson scheme (i.e. it is quasi-isomorphic to its underlying ultrascheme) and the closed points are precisely the points $x \in X$ such that $\kappa(x)|k$ is a finite extension.

You should also confer the appendix of EGA 1971. There it is shown that for any field $k$ the category of $k$-schemes locally of finite type with morphisms locally of finite type is equivalent to the category of $k$-ultraschemes (a $k$-ultrascheme is locally the maximal spectrum of a $k$-algebra).

Answer 2

The following result deals with the case of finite type affine schemes over an arbitrary field $k$.

Theorem: Let $A$ be a finitely generated algebra over a field $k$. Let $\iota: A \rightarrow \overline{A} = A \otimes_k \overline{k}$.
a) For every maximal ideal $\mathfrak{m}$ of $A$, the set $\mathcal{M}(\mathfrak{m})$ of maximal ideals $\mathcal{M}$ of $\overline{A}$ lying over $\mathfrak{m}$ is finite and nonempty.
b) The natural action of $G = \operatorname{Aut}(\overline{k}/k)$ on $\mathcal{M}(\mathfrak{m})$ is transitive. Thus $\operatorname{MaxSpec}(A) = G \backslash \operatorname{MaxSpec}(\overline{A})$.
c) If $k$ is perfect, the size of the $G$-orbit on $\mathfrak{m} \in \operatorname{MaxSpec}(A)$ is equal to the degree of the field extension of $k$ generated by the coordinates in $\overline{k}^n$ of any $\mathcal{M}$ lying over $\mathfrak{m}$.

In brief, the closed points correspond to the Galois orbits of the geometric points.

This is Theorem 8 in http://alpha.math.uga.edu/~pete/8320notes3.pdf.

The proof is left as an exercise, with some suggestions.

Exactly where this result came from, I cannot now remember. The text for the course that these notes accompany was Qing Liu's Algebraic Geometry and Arithmetic Curves (+1!), so it's a good shot that there is at least some cognate result in there.

Answer 3

It is certainly true for schemes of finite type over $k$ (algebraically closed) that the closed points are exactly the $k$-points. To see this, notice that if $x \in X$ is any point, then the closure $\overline{\{x\}}$, equipped with its reduced subscheme structure, is integral and has dimension equal to the transcendence degree of its function field over $k$ (Hartshorne, exercise 3.20 in chapter 2). I hope that's clear enough?

For $k$-scheme which are not (locally) of finite type, this doesn't work, as Martin shows below.