Considering a complex algebraic group G defined over the reals, one knows from an article of Borel and Harish-Chandra (Arithmetic subgroups of algebraic groups, Annals of Mathematics 75 (1962)) that G is reductive (as a complex group) if and only if G(R), the subgroup of its real points, is reductive (as a real group). One natural question is whether the same is true replacing the reals by an arbitrary field (say of characteristic 0) and the complexes by its algebraic closure?


Clarifying:
If $G$ is an affine reductive algebraic group defined over $k$ we know that $G$ can be seen as a subgroup of $Gl(n,\bar{k})$, with $\bar{k}$ the algebraic closure of k. Let $G(k)=G\cap Gl(n,k)$. My question is if $G(k)$ is also reductive? Meaning reductive when the unipotent radical of the group is trivial.



Reductive groups question
I'm a little hesitant to say anything in the face of all the comments above, but I think Ana is asking for an answer to the following question: If $G$ is a reductive algebraic $k$-group, and we choose:


1. an algebraic closure $\overline{k}$ of $k$,
2. a $\overline{k}$-group embedding $G\_{\overline{k}} \hookrightarrow GL\_n$,
3. a set-theoretic normal subgroup $U \trianglelefteq G(k)$ such that any element $g \in U$ has unipotent image in $GL\_n(\overline{k})$ under the composition $\operatorname{Spec} \overline{k} \to \operatorname{Spec} k \overset{g}{\to} G \to GL\_n$,


then is $U$ necessarily trivial?


I think the answer is yes, but I don't know a complete proof. When $k$ is infinite, the Zariski closure of $U$ in $G$ is a normal unipotent subgroup (at least, it sounds plausible). When $k$ is finite, you may have to eliminate small order cases by hand.


