Is there an Abelian surface such that every effective divisor is ample? (Together with a boil down version to a question in Complex Lie group theory)

2010-05-25T00:15:57Z

The Nakai-Moishezon criterion states that a line bundle $L$ over a surface $X$ is ample iff $L \cdot L > 0$ and $L \cdot C > 0$ for every curve $C$. We can use this criterion to check that if $X$ is the product of two elliptic curves, then lots of divisors of $X$ are not ample. The fibers of the projection maps of $X$ to its factors have zero self intersection and hence cannot be ample.

Question: is there an Abelian surface such that everyone of its curves is ample?

This is what I attempted. I don't believe it leads anywhere, tough... Suppose $X$ is an Abelian surface that is not the product of two elliptic curves. Suppose that $C_1$ and $C_2$ are two curves in $X$ representing different homology classes. Then, they must intersect [fix an element $\theta \in X$ such that $\theta$ sends $C_1$ to a curve that intersects $C_2$...]. So, all that matters is to check that $C_1 \cdot C_1 > 0$.

We do it by contradiction. Assume that $C_1 \cdot C_1 = 0$. By acting with the inverse of a point of $C_1$ on $C_1$, we can assume that the identity element of $X$ is in $C_1$. Since $C_1 \cdot C_1 = 0$, $C_1$ is a subgroup of $X$, furthermore, it is smooth [just act on $C_1$ with $C_1$ itself]. So, we have a mapt $X \rightarrow X/C_1$, a elliptic fibration of $X$ with elliptic fibers. If this was a trivial HOLOMORPHIC bundle then we would get the contradiction we sought. But that is very unlikely to be the case.

Answer by David Speyer for Is there an Abelian surface such that every effective divisor is ample? (Together with a boil down version to a question in Complex Lie group theory)

2010-05-26T03:11:19Z

This answer exists simply to record that BCnrd and Bjorn Poonen gave excellent answers in the comments above. If someone votes up my answer, this will be removed from this list of unanswered questions. (And, as I have made this answer CW, I will not gain any reputation.)

Is there an Abelian surface such that every effective divisor is ample? (Together with a boil down version to a question in Complex Lie group theory) - MathOverflow

Is there an Abelian surface such that every effective divisor is ample? (Together with a boil down version to a question in Complex Lie group theory)

Answer by David Speyer for Is there an Abelian surface such that every effective divisor is ample? (Together with a boil down version to a question in Complex Lie group theory)