and I should say that for a proper divisor $d|n$ $\#C(\mathbb{F}_{q^d})=q^d+1+(error term)$ so weak estimates imply that $C(\mathbb{F}_{q^n})-\bigcup_{d|n,d\neq n}C(\mathbb{F}_{q^d})$ is larger than 1.

Well now that I think it is obvious. If $C$ is the corresponding curve then we have an exact formula for $\#C(\mathbb{F}_{q^n})$. It is $q^n+1+(error term)$. Nous using weak Weil estimates we are done.

Yes you are right, I use the fact that the fixed field of $Aut(C)$ is $Q$. The fact that $Aut(R)={Id}$ is not a problem since you may work in a suitable algebraic closure and as you know $C$ is an algebraic closure of $R$. I guess that in general if you have a field $k\subseteq K$ then you want to show the existence of a field $L$ which contains $K$ such that $Aut_k(L)=k$. Once you have that the proof works.

Thanks a lot Neil for this nice argument. I'm still wondering if it is possible to use this $C^{\infty}$-structure on the topological group $G$ and prove very cheaply that $K$ is a deformation retract of $G$.

Yes, I did not check it but the proof might carry over to power series ring in many variables.

Thanks a lot for the reference. (Actually the proof was taken from Bourbaki Algebre IV)

Yes Gjergji, I think that you may rephrase the problem in these terms