Definition of unipotent linear algebraic groups over non algebraically closed fields Suppose we have a field $F$ with $\text{char}\ F=0$ and $F$ is not necessarily algebraically closed.

What is the definition of a unipotent linear algebraic group over $F$?

I'd really appreciate any definition having to do with the additive group $\mathbb{G}_{a,F}$.
 A: I haven't checked Milne's updated version (I'm having trouble opening the PDF), but here are some (hopefully helpful!) definitions.
Since $\text{char }F=0$, we know that the unipotent $U/F$ will be smooth. This gives a quick and dirty way to define unipotence. Namely, $U$ is unipotent if for all $g\in U(\bar{k})$ $g=g_u$. Here $g=g_u g_{ss}$ is the (unique!) Jordan decomposition of $g\in U(\bar{k})$. Recall this comes from choosing some embedding $U\hookrightarrow \text{GL}(V)$ and pulling back the usual Jordan decomposition in $\text{GL}_n(V_{\bar{k}})$.
Not shockingly (by appealing to the Lie-Kolchin theorem!) you can prove equivalently that $U/k$ is unipotent iff for all embeddings $U\hookrightarrow \text{GL}_n$ you can conjugate the embedding by some $g\in \text{GL}_n(k)$ to have the embedding land inside of the standard unipotent $U_n\subseteq\text{GL}_n$ given by:
$$U_n=\left\{\begin{pmatrix}1 & \cdots & \ast\\ \vdots & \ddots & \vdots\\ 0 & \cdots & 1\end{pmatrix}\right\}$$
of upper triangular matrices with $1$ along the diagonal.
One nice theorem that almost says that unipotents are the opposite of tori is the following. If $G/k$ is smooth and connected, and not unipotent, then it contains a nontrivial torus $T\subseteq G$.
As for definitions involving $\mathbf{G}_a$, I'm not too sure. It is true (for any perfect field, in particular for characteristic $0$, that all unipotents are 'generalized extensions' of $\mathbf{G}_a$. By this I just mean that $U/k$ has a composition series whose quotients are $\mathbf{G}_a$:
$$1=G_0\unlhd G_1\unlhd\cdots\unlhd G_{n-1}\unlhd G_n=U$$
with $G_i/G_{i-1}\cong\mathbf{G}_{a}$. This is not true for non-perfect base fields though. Restricting scalars $\mathsf{Res}_{K'/K}\mathbf{G}_a$ for a purely inseparable field extension $K'/K$ gives an example of a unipotent group (since you can check unipotence by base changing to $K'$!), but contains no copy of $\mathbf{G}_a$.
If you want to learn more, I'd suggest looking at the book Pseudo-reductive Groups by Conrad and Prasad. I'm sure the full story is contained in there.
