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).

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    $\begingroup$ If we take the square of an elliptic curve (over C) doesn't its automorphism group contains SL(2,Z)? Isn't necessary to take non-torsion points to rigidify it? $\endgroup$ Oct 23, 2009 at 2:49

3 Answers 3


I get that the answer is "no" for an abelian variety over the algebraic closure of Fp 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 Fp 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(Fq) for some q=p^n. Since X(Fq) is finite, some power of f will act trivially on X(Fq). 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?

  • $\begingroup$ I am commenting for completely vain reasons: I am kind of surprised that no one has either voted this up or pointed out an error in the argument. I'm not trying to beg for votes, but if people think I got this wrong I'd like to know. $\endgroup$ Oct 22, 2009 at 16:58
  • $\begingroup$ Yes, another nice example (you need to choose Fq large enough so that the variety, the points, and the automorphism are defined over it). So I was wrong in my claim for abelian varieties --- the result I was thinking of applies to polarized abelian varieties. $\endgroup$
    – JS Milne
    Oct 23, 2009 at 2:43

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 $\mathbb C$.

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    $\begingroup$ Nice. But you should fix N+1 points; think about the case N=0. $\endgroup$
    – BCnrd
    Feb 25, 2010 at 15:31

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 (xi, yi), there is a polynomial f(x) which vanishes at each xi. 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.


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