# why is multiplication by n surjective on an abelian variety [duplicate]

I've just read the proof of $\S4$ (iv) in Mumford's "Abelian Varieties", on page 42-43, which says that multiplication by $n$ not divisible by $p=\text{char}(k)$ on an abelian variety $X$ is surjective.

I can understand that it works (as in, I can follow the proof line by line), yet I don't have intuition for it. In particular, I don't understand why does passing to tangent spaces work. It may be my poor understanding of tangent spaces and differentials.

It seems to say something along the lines of "surjectivity of multiplication by $n$" can be checked on first order linear approximations". Does this make any sense? Would this make sense for more general isogenies? A reference where this is discussed in more detail would also be welcome.

• In the smooth situation, a group endomorphism of a connected compact Lie group whose differential at the identity is surjective is itself surjective, since you can use the inverse function theorem to show that the image contains an open subset. – Mariano Suárez-Álvarez Mar 12 '18 at 1:58
• Here is another way to think about this: The map $[n]:X \to X$ is proper and the image is a closed, irreducible subvariety $Z$ of $X$. Since $X$ is irreducible, we know that either $Z=X$ or that $Z$ has dimension strictly smaller than $X$. But the fibers of $[n]$ are finite so the dimension of $Z$ has to be the same as the dimension of $X$. – user45878 Mar 13 '18 at 8:45