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Let $A$ be an abelian variety over $\mathbb{C}$. If $A$ has an effective non-big divisor, then $A$ is not simple. (In a simple abelian variety, every non-zero effective divisor is ample.)

What can we say about the structure of effective non-big divisors in $A$?

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    $\begingroup$ On A we should have effective=>nef. Then the big cone equals the ample cone. And effective non big is the same as strictly nef. Any such divisor (at least if it’s a $\mathbb{Q}$-divisor) should be numerically equivalent to the pull back of an ample divisor from some quotient map of abelian varieties $A\rightarrow B$. $\endgroup$
    – Yosemite Stan
    Commented Mar 23, 2020 at 17:40
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    $\begingroup$ I think when you say "strictly nef" you really mean "nef but not ample". "strictly nef" is actually a term of its own: it means $D \cdot C > 0$ for all curves $C$. $\endgroup$
    – Mark
    Commented Mar 25, 2020 at 14:54
  • $\begingroup$ Can you expand on what you would like to know about "the structure of effective non-big divisors"? Yosemite Stan's comment gives one possible answer; is that the kind of thing you were looking for? $\endgroup$
    – Pop
    Commented Mar 25, 2020 at 14:59
  • $\begingroup$ I was looking for what @YosemiteStan explained. Basically, from what I understand now, non-big divisors are pull-backs of ample divisors along some quotient map $A\to B$. Can you make this into an answer? $\endgroup$
    – Pat
    Commented Apr 1, 2020 at 18:48

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