Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
5  
Your argument is nearly complete, but the hypothesis you need is that X is not isogenous to a product of elliptic curves; that's necessary since abelian surfaces that are isogenous to a product also contain non-ample curves, e.g., the image of an elliptic curve under the isogeny. Now to complete your argument: by the Poincare irreducibility theorem, if X contains a 1-dimensional subgroup, it is isogenous to product of elliptic curves. That does it! –  Bjorn Poonen May 25 '10 at 0:44
1  
P.S. If I wanted to be annoying, I would complain about the title of your question: the zero divisor is effective but not ample! –  Bjorn Poonen May 25 '10 at 0:46
3  
Alternate pf in any dimension, assuming $X$ abs. simple: if $D$ is effective divisor and $L$ is associated line bundle (inverse of ideal sheaf of $D$) then $L$ is ample if and only if $\phi_L:X \rightarrow X^{\rm{t}}$ has finite kernel (proved in Mumford's book on A.V.; uses crucially that $D$ is effective). The reduced scheme of geometric fiber of identity component of $\ker \phi_L$ is trivial or $X$ (as it is an abelian subvariety, $X$ abs. simple). Thus, $L$ is ample or $\phi_L = 0$. Since $\phi_L = 0$ iff $L$ is alg. equiv. to 0, & over $\mathbf{C}$ iff $[D] = 0$ in ${\rm{H}}_1$, QED –  BCnrd May 25 '10 at 2:01
1  
Dear Professor Poonen and BCnrd - Thanks a lot for all the help! It was really nice to learn about the Poincare irreducibility theorem. Here is a link to a proof of it. Thanks again! books.google.co.uk/… –  James O May 25 '10 at 19:16
1  
@James O: when you learn the algebraic theory, you'll be glad to see that the Poincare reducibility theorem is valid over any field $k$ (with appropriate notion of $k$-simplicity); in particular, $k$ need not be algebraically closed (e.g., finite, rationals, imperfect, whatever). –  BCnrd May 25 '10 at 20:00

1 Answer 1

up vote 3 down vote accepted

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.