Varieties where every non-zero effective divisor is ample

The following question seems very intuitive, but I haven't been able to find any proof (or counterexample).

Let $X$ be a non-singular projective variety of $\dim X\ge 2$ and let $NS^1(X)$ be its Neron-Severi group. If every non-zero effective divisor on $X$ is ample, does it follow that $X$ has Picard number one, i.e., $\rho=$ rank $NS^1(X)=1$?

Motivation:

1) In the case of Fano varieties the result is true (the proof is an easy application of Riemann-Roch). In fact this result was a key ingredient in Mori's proof of Hartshorne's conjecture for projective 3-space (i.e., any 3-fold with ample tangent bundle is isomorphic to $\mathbb{P}^3$). See Mori's original article for the details.

2) In this Mathoverflow question Charles Staats asks for a surface with the property that any two curves on the surface have nontrivial intersection. In his comment, BCnrd considered a K3 surface with Picard number one, which satisfies the condition precisely because any effective divisor is ample. A natural question is whether any such surface has Picard number one.

I am mostly interested in the case where $X$ is a complex projective variety. In the case the result does not hold, I'd also be interested in seeing a concrete counterexample and other examples of varieties where the result holds.

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For algebraic surfaces this result follows from the hodge index theorem. If the picard number is bigger than 1, then the intersection pairing on the orthogonal completement of any ample divisor is negative definite. Then any effective divisor with negative self-intersection is not ample. Is there is higher dimensional analogue of this? –  Daniel Loughran Dec 1 '10 at 12:45
J.C: I asked a similar question here mathoverflow.net/questions/41619/… –  Hailong Dao Dec 1 '10 at 13:24
@Daniel: how do you show that an effective divisor with negative slf intersection exists? –  rita Dec 1 '10 at 13:24
@ J.C: In fact, I just looked at my question again and Francesco's answer seems to be a counter-example to yours question. –  Hailong Dao Dec 1 '10 at 13:54
Oops, I just read Francesco's answer in the linked question. He says essentially the same thing I do below, much more succinctly. I guess I'll leave mine there, since it gives a few more details. –  user5117 Dec 1 '10 at 13:57

Answer: no. Example: take a simple abelian surface X with real multiplication by Q($\sqrt{d}$) (where d is a square-free positive number). X has Picard number 2, and the intersection form on N^1(X) diagonalises over Q to diag(a,-b) where b/a=d. The nef cone is just the cone of classes x in N^1(X) with x^2 >= 0 (more precisely, the part of this cone which also satisfies x.h >=0 for a chosen ample class h), and a simple computation shows that the boundary rays of this cone are irrational (by square-freeness of d). Now for any abelian variety the effective cone is equal to the nef effective cone, which equals the union of the ample cone with the rational boundary faces of the nef cone. In our example the nef cone has no rational boundary faces, so the ample cone and the effective cone (minus zero) coincide.

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Good answer. It nicely expands (in dimension 2) the one I gave to Hailong's question. +1 –  Francesco Polizzi Dec 1 '10 at 14:45
This is a very nice example! It's interesting that the result fails in the case $K_X=0$, wheras it holds in the case $K_X<0$. After this example however, I guess one could also come up with counterexamples when $X$ is of general type. –  J.C. Ottem Dec 1 '10 at 15:36
e.g., by taking a hypersurface of an abelian variety with the above property. –  J.C. Ottem Dec 1 '10 at 16:03

It is a general fact that on any simple abelian variety, an effective divisor is ample. The following result underlies the usual algebraic proof of the projectivity of abelian varieties (defined initially only as complete group varieties). It is therefore rather standard; the first reference which comes to mind is Lemma 8.5.6 on page 253 in the abelian varieties chapter of the book Heights in diophantine geometry by Bombieri and Gubler, from which I quote literally.

Let A be an abelian variety and $D$ an effective divisor such that the subgroup $$Z_D : \hspace{3cm} \{ a \in A \mid a + D = D \}$$ is finite. Then $D$ is ample on $A$.

If the abelian variety $A$ is simple, and $D$ is non-zero, then the $Z_D$ is a fortiori finite, since it is a proper algebraic subgroup of $A$; and it follows from the quoted Lemma 8.5.6 that $D$ is ample.

An application. A surjective morphism $f: A \to X$ from a simple abelian variety onto a positive-dimensional projective variety $X$ is finite.

Proof. Choose $H \subset X$ an ample divisor. The divisor $f^*H$ is effective on $A$, hence it is ample. This is equivalent to $f$ being finite.

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