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Let $k$ be a local field of char $0$ (which is the case I concern). Let $V$ be a variety defined over $k$ and let $f: \mathbb A^n\to V$ be a surjective map (over the algebraic closure of $k$) defined over $k$. Is it true that the restriction of $f$ to $k$ rational points $k^n\to V(k)$ surjective?

After it is answered I realized that I simplified what I want to know too much. Please see the comments for the answer for more information.

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No it is not ! Let $f:\mathbb{A}^1\to\mathbb{A}^1$ be defined by $f(x)=x^2$. It is not surjective at the level of $k$-points, because there are elements of $k$ that are not squares.

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  • $\begingroup$ Thanks. In fact what I want to know is the following: Assume in addition that $V$ is a linear algebraic group. Is it true that the group generated by $f(k^n)$ has finite index in $V$? $\endgroup$
    – ronggang
    Commented Jun 28, 2012 at 8:20
  • $\begingroup$ Sorry, finite index in $V(k)$. $\endgroup$
    – ronggang
    Commented Jun 28, 2012 at 8:24
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    $\begingroup$ @ronggang -- What precisely do you mean by "index"? Why should the image of f be a subgroup of the group of $k$-points? For instance, the image on $k$-points of the map $f(x) = x^2 + 1$ is not a subgroup of the additive group. $\endgroup$
    – Jason Starr
    Commented Jun 28, 2012 at 12:35
  • $\begingroup$ I mean the group generated by the set $f(k^n)$ has finite index in the group $V(k)$. Here is the main example in my mind. Suppose $V$ is generated by connected unipotent groups defined over $k$. Then we get such a map $f$ according to I Porp. 2.2 of Borel's algebraic groups. Then the question is whether the group generated by $k$ points of these unipotent groups has finite index in $V(k)$. If $V$ is semisimple then the question is whether the group they generate contains $V^+$ in the sense of 1973 paper of Borel and Tits( Homomorphismes "Abstraits" de Groupes Algebriques Simples). $\endgroup$
    – ronggang
    Commented Jun 28, 2012 at 14:33
  • $\begingroup$ If you're thinking of those specific maps, then it should be surjective. The reason is that the map from $\mathbb A^n$ to a connected unipotent group is not just surjective, it is an isomorphism. The algebraic inverse comes from the formal power series for $\log$. Isomorphisms are surjective on rational points. You might run into additional troubles if the decomposition into unipotents is not canonical, though. $\endgroup$
    – Will Sawin
    Commented Jun 28, 2012 at 16:02

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