# 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)

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.

• 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
• 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
• 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
• 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
• @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