Assume $X_1$ and $X_2$ are two smooth projective curves and let $M_i$ be an ample line bundle on $X_i$, for $i=1,2$. Further, denote the natural projection map on the $i$-th factor by $$\pi_i:X_1\times X_2 \to X_i \,.$$

I need to show that the line bundle $\pi_1^* M_1\otimes\pi_2^* M_2$ over $X_1\times X_2$ is ample.

I know that, by Nakai's criterion, if $\pi_i$ were a finite morphism then the line bundle $\pi_i^* M_i$ would be ample. Nevertheless this is not the case here, as $\pi_i$ is not finite. I think $\pi_i^* M_i$ alone is not ample on the product space, but $\pi_1^* M_1\otimes\pi_2^* M_2$ actually is.

What could I do to see this? Any help is appreciated.


The bundles $\pi^*_i M_i$ are definitely not ample: Nakai's criterion is an if and only if, but $\pi^*_i M_i$ is trivial on curves of the form $\pi_i^*(p)$ for $p$ a point, so it cannot be ample.

On the other hand, you can just apply Nakai's criterion to the bundle you actually want. Call it $L$. Since each $M_i$ is ample, some multiple of each of them is very ample, so replacing $L$ by a positive multiple if necessary (which doesn't change the fact of its ampleness or otherwise), we can assume $L$ has a section that is of the form $\sum_i F^i_1 + \sum_j F^j_2$ where the $F^i_1$ are fibres of $\pi_1$ and the $F^j_2$ are fibres of $\pi_2$.

Now to calculate $L \cdot C$ for $C$ a curve in $X_1 \times X_2$, we are free to choose the $F^i_1$ and $F^j_2$ to be any fibres we like of the respective projections, since these fibres are all agebraically equivalent. So choose them all to intersect $C$ properly. Moreover, at least one of these curves must actually intersect $C$ in a nonzero number of points, so we get $L \cdot C >0$.

This argument applies to any effective divisor on the surface, including $L$ itself, so we also get $L^2>0$. Hence by Nakai, $L$ is an ample line bundle.


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