Question: If $X$ is a projective surface and $U$ is an open affine subset of $X$, then is it true that $X \setminus U$ is the support of an (effective) ample divisor on $X$?

Background: I was reading Goodman's paper "Affine open subsets of algebraic varieties and ample divisors" which considers this same question for general varieties. Here is what I understand so far:

  1. If $\dim X = 1$, then the answer to the question is always affirmative. Indeed, if $X$ is a complete curve and $S$ is any finite set of points on $X$, then there is an effective ample Cartier divisor on $X$ which has support $S$ (this is Proposition 5 of the paper, and a straightforward application of the Nakai-Moishezon criterion of ampleness).

  2. For $\dim X = 2$, Theorem 2 of the paper states that the answer is positive if each point of $X\setminus U$ is factorial (i.e. its local ring is a UFD). Actually he proves it only assuming that $X$ is complete (i.e. a priori not necessarily projective) and as a corollary he proves Zariski's theorem that if all the singularities of a complete surface $X$ are contained in an affine open subset, then $X$ is projective.

  3. He presents two examples (of Hironaka and Zariski) where $X$ is a non-singular projective $3$-folds, but $X\setminus U$ is not the support of any ample divisor.

  4. In general he proves (in Theorem 1) that if $X$ is complete then a Zariski open subset $U$ of $X$ is affine iff the complement of (the isomorphic image of $U$) in a blow-up $X'$ of $X$ along a closed subscheme $F$ not meeting $U$ is the support of an effective ample Cartier divisor on $X'$.

  5. For $\dim X \geq 3$, he gives a criterion (in Theorem 3) for when the answer to the question is positive.

As far as I can see, he does not mention anything about the status of the question (i.e. whether if there is a counter-example or not) for general projective surfaces. Therefore I ask it here. I would expect the answer to be negative, but can not think of any examples. For me particularly interesting would be the case when $X$ is normal.

Edit: As the example of Jason Starr in the comment shows: The answer is negative even for normal surfaces (see my comments for an attempt of proof). I wonder what happens if $X$ is rational. In any event, I would gladly accept Jason's answer if he writes one. (And I would also greatly appreciate any answer/remark about the rational case.)

  • 5
    $\begingroup$ It is not true, even for normal surfaces. Consider the projective cone over a smooth, plane cubic. Take a point on the plane cubic whose difference with any flex point is non-torsion in the group of linear equivalence classes of 0-divisors. The line of that point gives a counterexample. $\endgroup$ – Jason Starr Oct 17 '11 at 11:00
  • 2
    $\begingroup$ @Jason: Why is the complement in your example affine? $\endgroup$ – J.C. Ottem Oct 17 '11 at 11:59
  • 2
    $\begingroup$ In LNM #156, p.64, Hartshorne shows, following Goodman, that on a complete integral surface, the complement of the support of an effective divisor with no base points is affine. Thus a counterexample to your question arises from any such divisor which is not ample. $\endgroup$ – roy smith Oct 18 '11 at 3:34
  • 2
    $\begingroup$ apparently such examples do not exist for non singular surfaces. $\endgroup$ – roy smith Oct 18 '11 at 3:36
  • 2
    $\begingroup$ @roysmith: Roy, what you are saying couldn't possibly be true as stated. If you take a projective surface fibred over a curve and choose your divisor to be the pull-back of a very ample divisor from the curve, then it is effective with no base points, yet its complement contains proper curves (the other fibers) so can't be affine. You'd need an additional assumption, something like "it doesn't contain proper curves". Cheers! $\endgroup$ – Sándor Kovács Jul 1 '17 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.