Thomas Bauer shows in http://arxiv.org/pdf/alggeom/9712019v1.pdf that for a complex abelian variety a nef line bundle is numerically equivalent to an effective divisor (this is shown in Lemma 1.1). It seems to me (by comments in other papers) that this is known over a general algebraically closed field, but I have yet to find a proof. Since Bauer's proof deals with the eigenvalues of the Hermitian form associated to the (polarized) abelian variety, I'm not quite sure how a proof would proceed over an arbitrary algebraically closed field. Any suggestions or references?
1 Answer
Here is a sketch of a purely algebraic proof based on the theory developed in Chapter 3 of Mumford's "Abelian Varieties".
Let $L$ be a nef line bundle on the abelian variety $A$ of dimension $g$. If $K(L)$ is finite, then the index of $L$ is a well defined integer between $0$ and $g$ and we need to show that $g=0$. By the theorem on p. 155, the index is the number of positive roots of the polynomial $P(n) = \chi(L \otimes M^n)$, where $M$ is any ample line bundle. Using RiemannRoch and the nefness of $L$ one sees that all the coefficients of the polynomial are nonnegative so it has no positive roots.
We now reduce the general case to the one above. We may assume, by replacing $L$ by $L \otimes (1)^*L$, that $L$ is symmetric. If $K(L)$ is not finite, i.e. $T_x^*L \cong L$ for $x$ in a positive dimensional subvariety $B$ of $A$ we claim that $L^{2}$ descends to a line bundle $L'$ on $A/B$. This follows as we can get descent data for $L^2$ by choosing symmetric isomorphisms $T_x^*L \cong L$ for all $x \in B$. (Note that one cannot always descend $L$ as shown by the example of a line bundle of order $2$ in $Pic(A)$.)
Since $L$ is nef it follows that $L'$ is also nef, so it is numerically equivalent to an effective line bundle by induction.
(In the sketch above, the class of $L$ modulo numerical equivalence is replaced with a multiple. One can perhaps avoid this by an additional argument.)

$\begingroup$ Where can I read about descent, in this sense? Why does the data you gave assure me that $L^2$ descends to a line bundle on $A/B$? $\endgroup$– rfauffarJul 11, 2013 at 21:16

1$\begingroup$ This is a special case of flat descent; see, for example, Theorem 4 in Chapter 6 of the book Neron Models by Bosch, Lutkebohmert and Raynaud. (The situation here is simpler though since the morphism $A \to A/B$ is a smooth homomorphism so you should be able to rewrite the condition for the existence of descent datum). After fixing an isomorphism of $L$ with $(1)^*L$, there is a symmetric isomorphism $T_x^*L \cong L$ which is well defined up to sign. The sign ambiguity disappears if you square these isomorphisms. $\endgroup$– nafJul 12, 2013 at 13:11

$\begingroup$ Ok great, I'm going to take a look at that book. Thanks a lot! $\endgroup$– rfauffarJul 12, 2013 at 15:11

$\begingroup$ After reading a bit, I see that one of the conditions necessary for $L$ to descend is that the commutator map $e^{L^2}:K(L^2)\times K(L^2)\to k^\times$ be trivial on $K(L)\times K(L)$. I don't see why this should happen by only asking for $L$ to be symmetric... $\endgroup$– rfauffarJul 18, 2013 at 20:53

1$\begingroup$ I don't see why you need this condition. It suffices to descend to $A/B$ where $B$ is the identity component of $K(L)$; the pullback of the theta group to $B$ is then commutative and what you need to show is that the corresponding extension splits. Assuming $L$ symmetric allows one to define a symmetric isomorphism $T_x^*L \cong L$. By this I mean that this isomorphism is preserved by $(1)^*$ (after using the fixed identification of $L$ with $(1)^*L$ and translating by $x$). This is well defined after squaring and should give the required splitting. $\endgroup$– nafJul 21, 2013 at 5:59