# Multiplication by $n$ on commutative algebraic groups

Let $K$ be a field of characteristic zero. Let $G/K$ be a group scheme of finite type. Assume that $G$ is commutative and connected. For a natural number $n$ denote by $n_G: G\to G$ the multiplication by $n$ morphism. Is it true that $n_G$ is surjective with finite kernel?

(I know that the answer is yes provided $G$ is an abelian variety. But the proof of this fact makes use of a very ample invertible sheaf on $G$, so it does not carry over to the general case directly.)

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My approach would be something like the following. The statement is clearly true over $K=\mathbb C$, by the theory of Lie groups. So check that it suffices to assume that $K$ is algebraically closed, and then use one of those model-theoretic statements along the lines of "any purely-algebraic theorem true in one algebraically closed field of characteristic zero is true in another". – Theo Johnson-Freyd Oct 19 '10 at 1:05

Yes, this is true. The derivative of $n_G$ at the identity element $e$ is multiplication by $n$ on the Lie algebra (tangent space at $e$) which gives that the kernel is finite and the image of $n_G$ has the same dimension as $G$ which implies that it is equal to $G$ as $G$ is connected.
Addendum: Jim may be right: The statements may be checked over an algebraic closure of $K$ so we may assume $K$ algebraically closed. As $dn_G$ is an isomorphism the kernel is finite and the dimension of the image is equal to that of $G$. As $G$ is of finite type the image is constructible by Chevalley's theorem and hence contains an open subset of $G$. As it is also a subgroup it is open and as $G$ is connected (and the cosets are also open) we get that the image equals $G$.