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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.)

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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. –  Jason Starr Oct 17 '11 at 11:00
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@Jason: Why is the complement in your example affine? –  J.C. Ottem Oct 17 '11 at 11:59
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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. –  roy smith Oct 18 '11 at 3:34
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apparently such examples do not exist for non singular surfaces. –  roy smith Oct 18 '11 at 3:36
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Ok, I think I can see the proof for Jason Starr's claim at least in the case that the base field is $\mathbb{C}$: 1. The line $L$ of Jason Starr's comment is not $\mathbb{Q}$-Cartier, so $L$ cannot be the support of an ample divisor. 2. The (minimal) desingularization $X'$ of $X$ replaces the vertex of the cone of $X$ by a copy $E$ of the original smooth plane cubic $C$ (it is a "cylinder" over $C$ near $E$). 3. To show $U := X\setminus L$ is affine, it suffices to show that there is an ample divisor on $X'$ with support $L' \cup E$, $L'$ being the strict transform of $L$. (contd. ...) –  auniket Oct 18 '11 at 19:26
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