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It is well known that for any polarization ( that is, ample line bundle) $L$ on an abelian variety $A$, there is an isogeny $\phi\colon A \to B$ to another abelian variety with a principal polarization $M$ such that $\phi^*M=L$.

If instead I take a nef but not ample divisor $D$ on $A$, is it always true that $D$ is the pullback of some ample divisor on a quotient of $A$?

Thanks for help!

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    $\begingroup$ No, take a generic degree 0 divisor on an elliptic curve. $\endgroup$
    – Ennio Mori cone
    Commented Feb 21, 2020 at 15:19
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    $\begingroup$ If you work in the complex setting I think the Appel-Humbert theorem should give you essentially what you want (en.wikipedia.org/wiki/Appell%E2%80%93Humbert_theorem). The only caveat, is that you might have to work "up to translation." Nefness I think corresponds to positive semidefiniteness of the Hermitian form H in the Appell-Humbert theorem. Then the kernel of this form should define a sublattice corresponding to a subcomplex torus. If you quotient by this subtorus, I think it should be straightforward to check that the Hermitian form "descends" and is ample on that quotient torus. $\endgroup$
    – Yosemite Stan
    Commented Feb 21, 2020 at 18:23
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    $\begingroup$ ... A minor annoyance is that one might not be able to descend the $\alpha$ in the Appell-Humbert theorem (e.g. in @wnx's answer - indeed this is true for "most" strictly nef line bundles with a given $c_1$). However, if you only want up to topological equivalence, i.e. to say that there is a quotient $A\rightarrow B$ such that $D$ is the image of some ample divisor under $\mathrm{NS}(B)\rightarrow \mathrm{NS}(A)$ then you should be okay I think. $\endgroup$
    – Yosemite Stan
    Commented Feb 21, 2020 at 18:30
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    $\begingroup$ Thinking more algebraically, it seems that the quotient should be taken by (the connected component of) $T(L) = \{ x\in A \,:\, T_x^* L \cong L\}$. Indeed, the restriction of $L$ to $T(L)^\circ$ lies in its $\mathrm{Pic}^0$, so by twisting by an element of $\mathrm{Pic}^0(A)$ one should be able to ensure $L$ trivial on $T(L)^\circ$. Then it descends to the quotient and is ample there (I think finiteness of $T(L)$ will ensure this). $\endgroup$
    – Piotr Achinger
    Commented Feb 21, 2020 at 20:22
  • $\begingroup$ Thanks to YosemiteStan and PiotrAchinger for their useful comments. Can you write them as an answer, so I can accept them? $\endgroup$
    – TartagliaTriangle
    Commented Feb 23, 2020 at 17:26

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