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.

isogenousto 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! $\endgroup$2more comments