Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers

up vote 12 down vote accepted

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).

share|improve this answer
add comment

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://www.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.

share|improve this answer
2  
To the downvoter: if there is something wrong with my answer, I would appreciate learning what it is. –  Pete L. Clark Feb 3 '10 at 11:12
    
It is true that Liu mentions quite some things about the action about the absolute Galois group of the base field. Some of these things are left as exercises (usually in a more general setting), so be prepared to look at these exercises as well. I also remember that Mumford adresses this in his second chapter of "The red book of varieties and schemes". I think the right paragraph is called "fields of definition" (not entirely sure though). –  Wanderer Feb 3 '10 at 11:13
    
Your answer really adds something to the discussion, so I am upvoting it :) –  Wanderer Feb 3 '10 at 11:16
2  
Thanks, AS. It seems that the downvote was removed. Again, I had no problem with it per se, but I'm quoting a result from a course that I taught last year. If the result is not correct, please tell me why! –  Pete L. Clark Feb 3 '10 at 11:21
add comment

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.

share|improve this answer
    
Thanks. If I was accepting the fact $V$ can be identified with rational points of $t(V)$, then I should have conjectured this considering Hartshorne 4.10 tells us this is the image of the functor. –  Matt Feb 3 '10 at 23:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.