TITLE: Density question in algebraic group
QUESTION [12 upvotes]: Suppose G is an algebraic group defined over F, the algebraic closure of F is K. Consider the Zariski topology on G(K), is G(F) Zariski dense in G(K)?

REPLY [5 votes]: A (smooth, connected) unipotent group $U$ is said to be $k$-split if there is a filtration by $k$-subgroups for which the successive quotients are isomorphic to $\mathbf{G}_{a/k}$.
The examples mentioned in comments (e.g. the subgroup of $\mathbf{G}_a^2$ defined by $y^p - y = tx^p$) are non-split unipotent groups. Any $k$-split unipotent
group $U$ is even a rational variety (in fact, $k$-isomorphic as a variety to $\mathbf{A}^n$) so it is clear that $U(k)$ is Zariski dense in $U(k_{alg})$ when $k$ is infinite.
More generally, let $G$ be a (smooth) linear algebraic group over $k$ and assume that the unipotent radical of $G$ is defined and split over $k$ (both of these conditions can fail). Then as a $k$-variety, $G$ is just the product of its reductive quotient $G_{red}$ and its unipotent radical (result of Rosenlicht). In particular, $G$ is unirational and if $k$ is infinite, $G(k)$ is dense in $G(k_{alg})$.
Of course, this observation isn't that interesting -- in some sense, it just "identifies" the problem.