Can algebraic varieties be rigidified by finite sets of points?

Question

For an algebraic variety X over an algebraically closed field, does there always exist a finite set of (closed) points on X such that the only automorphism of X fixing each of the points is the identity map? If Aut(X) is finite, the answer is obviously yes (so yes for varieties of logarithmic general type in characteristic zero by Iitaka, Algebraic Geometry, 11.12, p340). For abelian varieties, one can take the set of points of order 3 [added: not so, only for polarized abelian varieties]. For P^1 one can take 3 points. Beyond that, I have no idea.

The reason I ask is that, for such varieties, descent theory becomes very easy (see Chapter 16 of the notes on algebraic geometry on my website).

Answer 1

I get that the answer is "no" for an abelian variety over the algebraic closure of F_p with complex multiplication by a ring with a unit of infinite order. Since you say you have already thought through the abelian variety case, I wonder whether I am missing something.

More generally, let X be any variety over the algebraic closure of F_p with an automorphism f of infinite order. A concrete example is to take X an abelian variety with CM by a number ring that contains units other than roots of unity. Any finite collection of closed points of X will lie in X(F_q) for some q=p^n. Since X(F_q) is finite, some power of f will act trivially on X(F_q). Thus, any finite set of closed points is fixed by some power of f.

As I understand the applications to descent theory, this is still uninteresting. For that purpose, we really only need to kill all automorphisms of finite order, right?

Answer 2

If the variety $X$ is defined over $\mathbb C$, is compact and smooth, then it should be possible to make it rigid. Indeed, the variety $Aut(X)$ has at most countable number of irreducible components and all these components should be of bounded dimension $N$.

Let us prove it will be sufficient to fix $N+1$ points (where was $N$ here before the comment of Brian). For every component Comp of $Aut(X)$ let us consider the subvariety of $X$ that is moved by at least one element of this component Comp. It is clear that moved point will be an open subset of $X$, so all of them will have an intersection (a countable intersection of open subsets is non-empty). So there will be a point on $X$ that is moved by at least on element from every component of $Aut(X)$. Let us mark this point, call it $P$. Note now that the components of $Aut(X,P)$ have dimensions at most $N-1$. So we proceed by induction.

In fact we did not really used smoothness of $X$ but we surely used that $X$ is defined over $\mathbb C$.

Answer 3

If you don't ask for the variety to be proper, the answer is no.

Consider the affine plane A^2. Given any finite set of points (x_i, y_i), there is a polynomial f(x) which vanishes at each x_i. Then (x,y) --> (x, y+f(x)) is an automorphism fixing those points.

From your stated motivation, I'm sure you want to say the variety is proper. That's an interesting question; I'll think about it.