# $A$ abelian variety. Is the multiplication by $n_A$ surjective?

According with Mumford the answer is yes, but there are some obscure points in the proof.

We know that there exists a very ample line bundle on $A$ since every abelian variety is projective. Hence, via some corollaries of the Theorem of the Cube, we get an ample line bundle whose restriction to the kernel of $n_A$ is both the trivial bundle and very ample. This should imply that the dimension of $\ker n_A$ is $0$.

Furthermore, Mumford says that the fact that the dimension of the kernel is $0$ implies the surjectivity of $n_A$.

Is anybody able to clarify these two points to me?

If $Z$ is a projective scheme and the trivial bundle is ample, then $\dim Z=0$. This is because, multiple of the trivial bundle is still trivial and thus you may assume that the trivial bundle is very ample. This means we can embed $Z$ in a projective space and the hyperplane bundle restricts to the trivial bundle on $Z$. Thus, a hyperplane does not intersect $Z$, which easily implies that $\dim Z=0$.
The second part, hypothesis implies the fiber over any point $n_A^{-1}(x)$ is a translate of $\mathrm{Ker}\, n_A$, since $n_A$ is a group homomorphism and so its dimension is zero. Then use the formula $\dim A=\dim \mathrm{Im}\, A +\dim n_A^{-1}(x)$ for a general $x$ and thus the image has the same dimension as that of $A$ and being irreducible, must be all of $A$.
• Don't we need that $n_A$ is flat? – Symòn May 19 '16 at 10:03