Nakai-Moishezon theorem for abelian varieties In Birkenhake and Lange's book, they prove a version of the Nakai-Moishezon theorem for complex abelian varieties that says that if $L_0$ is an ample line bundle on a complex abelian variety $X$ of dimension $g$, then a line bundle $L$ is ample if and only if $(L^\nu\cdot L_0^{g-\nu})>0$ for $\nu=1,\ldots,g$ (Corollary 4.3.3 on page 77 of the second edition). 
I would love for this to be true in arbitrary characteristic. The problem is that the proof over $\mathbb{C}$ explicitly uses the Hermitian forms associated to the line bundles, and I don't see how this could be adapted to arbitrary characteristic.
Does anyone know if this is true in arbitrary characteristic? Are there any references or at least some idea for a proof?
 A: On an abelian variety (regardless of the characteristic), an effective divisor with positive self-intersection is ample. To be more precise, it suffices here to recall that on any simple abelian variety, all non-zero effective divisors are ample; apply this to the factors in a decomposition (mod isogenies) of an arbitrary abelian variety into irreducibles.
It is therefore enough to show that a multiple of $L$ is effective. By Riemann-Roch and the theorem on page 32 (Example 1.2.31: Higher cohomology of nef divisors, I) in Lazarsfeld's "Positivity in alg. geom.," this will follow if we show that $L$ is nef, and Kleiman's theorem thus reduces the problem to checking $L.C \geq 0$ on every curve $C$ in $X$. Let me restrict to the case $\dim{X} = 2$ "for simplicity;" this should given an idea for the general case. Curves in abelian varieties are in any case nef, and if $a := -(L.C) > 0$, the divisor $H := aL_0 + (L_0.L)C$ would be ample. But $L$ has a positive self-intersection and is orthogonal to $H$, contradicting the Hodge index theorem on the surface $X$.
