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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?

Thanks in advance!

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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 $A$.

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  • $\begingroup$ Thanks,you have been very clear. However, do you have any references for the "dimension formula"? I would like to see a proof of it. $\endgroup$
    – Symòn
    May 14, 2016 at 15:50
  • $\begingroup$ All standard texts (Shafarevich, Hartshorne) will have it. $\endgroup$
    – Mohan
    May 14, 2016 at 15:54
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    $\begingroup$ Don't we need that $n_A$ is flat? $\endgroup$
    – Symòn
    May 19, 2016 at 10:03
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    $\begingroup$ Finite maps between two smooth varieties of same dimension is always flat. $\endgroup$
    – Mohan
    May 19, 2016 at 12:06

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