Let $k$ be a field of characteristic 0 (not necessarily algebraically closed), let $G$ be a connected split reductive group over $k$ and let $\mathfrak{g}$ be the Lie algebra of $G$. 


Let $X \in \mathfrak{g}$ be a nilpotent element. Does there exist a unipotent subgroup $U$ of $G$ such that $X$ is contained in the Lie algebra of $U$ ?


If $k$ is algebraically closed this is Theorem 5.1 of <http://virtualmath1.stanford.edu/~conrad/249BW16Page/handouts/applgr.pdf>.


Here is a rough idea for a proof in the general case but I can't make the details work.


By the result over an algebraically closed field there exists a unipotent subgroup $U\_{\overline{k}}$ of $G\_{\overline{k}}$ such that $X$ is in the Lie algebra of $U\_{\overline{k}}$. Via the expoential/logarithm there is an isomorphism between $U\_{\overline{k}}$ and it's Lie algebra thus there exists $x \in U\_{\overline{k}}(\overline{k})$ whose exponential is $X$.


Since $X$ is defined over $k$ I would expect it to be also true for $x$ (as an element of $G$). Then by Theorem 3.6 (very easy in characteristic 0) of the Conrad's notes, $x$ will be an element of a unipotent subgroup $U$ of $G$ and we would deduce that $X$ is contained in the Lie algebra of $U$.


Since I work in characteristic 0, I imagine there might be a much simpler way. Also I don't think the split reductive hypothesis is necessary (at least I don't use it in my "proof idea").



Nilpotent elements of Lie algebra and unipotent groups
Indeed there's a much simpler way in characteristic zero, with $G$ an arbitrary $k$-defined linear algebraic group.


Let $X$ be nilpotent. Fix a faithful $k$-defined linear representation $\rho$ of $G$ and let $\rho'$ be the corresponding representation of $\mathfrak{g}$.


Now $\rho'(X)$ being nilpotent, it preserves a complete flag (defined over $k$) in the linear space $\mathfrak{g}$. Let $H$ be the subgroup of $G$ preserving this flag (for the representation $\rho$), and acting as identity on all successive 1-dimensional quotients. Then, since $\rho$ is faithful, $H$ is a $k$-defined unipotent subgroup of $G$, whose Lie algebra $\mathfrak{h}$ is the subalgebra of $\mathfrak{g}$ of those element preserving the flag (for $\rho'$), and acting as zero on all successive 1-dimensional quotients. Hence $X\in\mathfrak{h}$.


