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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.

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    $\begingroup$ No, because of (the easy direction of) the Nakai-Moishezon criterion. If $X'$ is another copy of $X$ obtained by taking a different point in $Y$ then $X.X'= 0$, as they don't meet. $\endgroup$ Nov 6, 2012 at 20:46
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    $\begingroup$ In particular, the genus zero case is easily false. $Pic(S) = {\mathbb Z}\times {\mathbb Z}$, your divisor is $(1,0)$, and ample divisors are $(m,n)$ with $m,n>0$. $\endgroup$ Nov 6, 2012 at 23:00

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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.

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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?

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    $\begingroup$ Probably a Community user action. meta.stackexchange.com/questions/48578/… Questions with positive score, no activity for a month, one answer with 0 score, and no positive score answer have a chance to get bumped to the front page. $\endgroup$ Jul 22, 2014 at 14:39
  • $\begingroup$ Thanks. I see that this was in fact a "Community" action. This question's score was zero, not positive, but the page you linked to says "non-negatively scored" so a score of zero still fits that criterion. $\endgroup$ Jul 22, 2014 at 16:46

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