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.



Polarization of an abelian variety made by the sum of two divisors
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 


