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.