Nef divisors on abelian varieties

The following question stems from a question I already asked on MO:

Nakai-Moishezon theorem for abelian varieties

I would like to prove that if $L_0$ is an ample line bundle on an abelian variety $A$ of dimension $n$ (defined over any algebraically closed field) and $L$ is any line bundle such that $(L^i\cdot L_0^{n-i})\geq0$ for all $i=1,\ldots,n$, then $L$ is nef. This is reasonably easy to prove over $\mathbb{C}$, using the Hermitian form associated to $L$.

The user that answered my previous question suggested taking a curve $C$ on $A$ such that $(L\cdot C)<0$ (hoping to arrive at a contradiction) and a smooth surface $S$ that gives the algebraic class of $L_0^{n-2}$ such that $C\subseteq S$. He then said that if we take $H=-(L\cdot C)L_0|_S+(L_0|_S\cdot L|_S)C$ then this is ample, but since $(L|_S^2)\geq 0$ and $(H\cdot L)=0$ we contradict the Hodge Index Theorem. The problem I see with this argument is that I don't see why $C$ should be nef on $S$, thus ensuring that $H$ is ample.

Is there any reason to see that $H$ is ample without $C$ being nef? Or is there another way to prove what I'm looking for?

• Yes you're right! Thanks so much for pointing that out! If you want you can copy what you wrote as an answer. – rfauffar Sep 12 '13 at 12:53
• I did that and deleted my comment. – naf Sep 12 '13 at 13:17

I think it is probably easier to prove that $L$ is ample when all the inequalities are strict:
The assumption for $i=n$ implies that $K(L)$ is finite, i.e., $L$ is non-degenerate, by the second statement in the Riemann-Roch theorem in Mumford's "Abelian Varieties" and then the index theorem together with Riemman-Roch again (here you use the assumption for the other $i$ as well) imply that $L$ is ample.