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?


1 Answer 1


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 $X$.

  • $\begingroup$ Ok, nice. Now how would this argument go in higher dimensions? We would want to intersect with curves to use Kleiman's Theorem, but I don't see how we can generalize what you wrote. If $(L\cdot C)<0$, we could take the 1-class $-(L\cdot C)L_0^{n-1}+(L_0^{n-1}\cdot L)C$ and this intersected with $L$ would be 0. But I'm not sure what else could be done... $\endgroup$
    – rfauffar
    Commented Aug 28, 2013 at 18:57
  • $\begingroup$ In higher dimension, since $L_0$ is ample, you can take a surface $Y \subset X$ through $C$ in the linear equivalence class $L_0^{g-2} \in \mathrm{CH}^2(X)$ (i.e., linearly equivalent to a multiple of a $2$-plane in the appropriate projective embedding). Since $L_0^{g-2}.L^2 > 0$ we still have $L_{|Y}^2 > 0$, and the same argument shows that $L.C < 0$ contradicts the Hodge index theorem on the surface $Y$. $\endgroup$ Commented Aug 28, 2013 at 23:31
  • $\begingroup$ Awesome answer! Thank you! It's really clear to me now. $\endgroup$
    – rfauffar
    Commented Aug 29, 2013 at 0:54
  • $\begingroup$ One last question: Why can we assume that $C$ is contained in $L_0^{g-2}$? This doesn't seem obvious to me... $\endgroup$
    – rfauffar
    Commented Aug 29, 2013 at 13:53
  • 1
    $\begingroup$ I'm sorry, I didn't write it cleanly in my last comment. I meant to say that $C$ lies on a surface which is linearly equivalent to a multiple of $c_1(L_0)^{g-2}$ in the Chow group $\mathrm{CH}_2$ of $2$-cycles mod rational equivalence. To fix ideas, say $g = \dim{X} = 3$. Because $L_0$ is ample, the sheaf $L_0^{\otimes n}(-C)$ has non-zero sections for $n \gg 0$, whose zero loci give surfaces through $C$ linearly equivalent to $nL_0$. $\endgroup$ Commented Aug 29, 2013 at 14:16

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