Polarization of an abelian variety made by the sum of two divisors Let $X$ be an abelian variety of dimension $n$, and let $L$ be a polarization, that is, an ample line bundle on $X$, with $\chi(L)=3$.
In my specific case, I have that $L=\mathcal{O}_X(\Theta + D)$, where $\Theta$ is an ample divisor with $\chi(\Theta)=1$ and $D$ is an effective Cartier divisor.
I want to show that $(D^2)=0$ (self-intersection of $D$), or equivalently that $(\Theta^{n-2}.D^2)=0$.
For $n=2$, $X$ is a surface, and using Riemann-Roch I have that $2\chi(L)=6=(\Theta^2)+2(\Theta.D)+(D^2)$, where the first two intersection numbers are strictly positive because of ampleness of $\Theta$ (in particular $(\Theta^2)=2$). If I suppose $(D^2)\ne 0$, then $(D^2)=2$ and so $(\Theta.D)$ must be 1. But this is impossible by the index theorem, because we have $4=(\Theta^2)(D^2)\le (\Theta.D)^2$.
But for dimension $n>2$, I don't know how to procede, because in Riemann-Roch formula $n!$ increases too fast, so it seems impossible to make the same argument.
Thanks for help!
Note: I have already posted this question on Math StackExchange, but maybe it is better to post it here.
 A: For any $P\in Pic^0(A)$ consider the map $|\Theta +P|\times |D-P|\to |\Theta +D|\cong \mathbb P ^2$. Since $D$ is effective, there is an abelian subvariety $T\subset Pic ^0(A)$ such that $|D-P'|\ne \emptyset$ for any $P'\in P+T$ and if $t=\dim T$ the $D^t\ne 0$ but $D^{t+1}=0$. If $t\geq 3$, then a general element in the image of the above map may be written as $\Theta _{P'}+D_{P'}$ in for infinitely many $P'\in P+T\subset Pic ^0(A)$. Thus $\Theta _{P'}+D_{P'}=\Theta _{P''}+D_{P''}$, but then $D_{P''}\geq \Theta _{P'}$ (as $\Theta _{P'}\in |\Theta +P'|$ is unique and different from $\Theta _{P''}$) and hence $\chi (\Theta +D)\geq \chi (2\Theta )>3$. Finally, if $t=2$, the above argument shows that any element $G\in |\Theta +D|$ can be written as a sum of elements $\Theta _{P'}\in |\Theta +P'|$ and $D_{P'}\in |D-P'|$ and $\dim |D-P'|=0$. Thus, the corresponding rational map $T\to \mathbb P ^2$ is generically finite, and of degree $>1$ (as $T$ is not rational). But then, for general $G\in |\Theta +D|$, we have $G=\Theta _{P'}+D_{P'}=\Theta _{P''}+D_{P''}$ which implies $\Theta _{P'}=D_{P''}$ and hence $\chi (L)=4$.
NB I originally misread the question so I have edited the answer appropriately 
