Let $X$ and $Y$ be complete curves over a field $k$ of characteristic zero. Let $S = X \times_k Y$. Assume that Y has a $k$-rational point and use this point to consider $X$ as a divisor (also denoted by $X$) on $S$. Is this divisor ample?
If yes, for which $m_0$ do we have that the higher cohomology of $mX$ vanishes for all $m>m_0$?
I'm mostly interested in the case where $X$ and $Y$ are of genus at least $2$. The case where both curves are of genus zero being, of course, very easy.
By definition you have a projection $f:X\times Y\rightarrow Y$ whose fiber is the divisor $X$. Since $X$ is the fiber of a fibration $X^2 = 0$ and $X$ can not be ample (an ample divisor intersect positively any effective curve). On the other hand $X\subset S$ is nef. Furthermore, since $X$ is nef, $X^2 = 0$ implies that $X$ is not big.
The easiest example is $S=\mathbb{P}^1\times\mathbb{P}^1$. Then $Pic(S)$ is generated by the classes of the two $\mathbb{P}^1$'s. Let call them $L,R$. You have $L^2=R^2 = 0$ and $L\cdot R = 1$. Therefore both $L,R$ are nef but neither big nor ample.
The fact that the divisor "$X\,$" is not ample can also be seen directly from the definition, and does not depend on the characteristic.
As usual we may assume $k$ is algebraically closed, so that
$Y$ has infinitely many points $p$.
Choose some $p$ other than the image of the divisor "$X\,$".
Then $X \times p$ is a complete curve, and
the restriction to $X \times p$ of any section of $mX$
has no poles and is therefore constant. Therefore there is no $m$
for which the linear series $\Gamma(mX)$ separates the points of $X \times Y$.
(In fact the map from $X \times Y$ to projective space
associated with $\Gamma(mX)$ always factors through the projection
$X \times Y \rightarrow Y$.)
P.S. How did this question from 2012 make it to the top page?
