Verifying that $\operatorname{Res}_{E/F} \mathbb G_a$ is a split unipotent group Let $G$ be a unipotent connected linear algebraic group over a field $F$. Then $G$ is called split if there is a series of closed subgroup schemes $1 = G_0 \subset G_1 \subset \cdots \subset G_t  =G$ with each $G_i$ normal in $G_{i+1}$ and $G_i/G_{i+1} \cong_F \mathbb G_a$.
I have read that for $F$ perfect, every such $G$ is split, provided it is connected.  And $F$ of characteristic zero, every such $G$ is moreover automatically connected.
I was trying to verify this in the case of $G = \operatorname{Res}_{E/F} \mathbb G_a$, where $E/F$ is a quadratic extension.  Does $G$ really have such a composition series?
I tried considering the diagonal embedding $\Delta$ of $\mathbb G_a$ into $G$ and looking at the quotient $G/\Delta$.  It should be the case that $G/\Delta \cong \mathbb G_a$, but I'm not able to verify this.  The natural map $G/\Delta \rightarrow \mathbb G_a$ given by $(x,y)\Delta \mapsto x- y $ is not defined over $F$.  
 A: It's not that complicated. I believe you have all the needed knowledge, but just have to unpack the definitions.
By fixing a basis of $E/F$, we know that $E$ is isomorphic to $F^2$ as $F$-vector spaces, hence for any $F$-algebra $R$, one has $\mathbb{G}_a(E\otimes_F R) = E\otimes_F R \simeq R^2 = \mathbb{G}_a^2(F)$, where the middle isomorphism is functorial in $R$.
Hence by the definition of restriction of scalars, we see that $\operatorname{Res}_{E/F}\mathbb{G}_a \simeq \mathbb{G}_a^2$.
A: Tricky business if you were using the definition of $G = \operatorname{Res}_{E/F}(\mathbb G_a)$ as a form of $\mathbb G_a \times \mathbb G_a$ and you didn't recollect Hilbert's Theorem 90...  
Worse, I actually did my PhD thesis on the restriction of scalars of a certain group and I didn't realize that $\operatorname{Res}_{E/F}(\mathbb G_a)$ is the same algebraic group as  $\mathbb G_a \times \mathbb G_a$.  It's obvious from the Yoneda lemma, but less obvious from the way I was thinking about restriction of scalars.  
Here I'm taking $G$ to be the group given on $\overline{F}$-points by $G(\overline{F})  = \overline{F} \times \overline{F}$, with the Galois action
$$\sigma.(x,y) = \begin{cases} (\sigma(x),\sigma(y)) & \textrm{ if } \sigma|_E =1 \\ (\sigma(y),\sigma(x)) & \textrm{ if } \sigma|_E \neq 1 \end{cases}$$
for $\sigma \in \operatorname{Gal}(\overline{F}/F)$.  Now choose $\beta \in \overline{F}$ such that $E = F(\sqrt{\beta})$, and define 
$$\phi: G \rightarrow \mathbb G_a \times \mathbb G_a$$
$$\phi(x,y) = (x+y, \sqrt{\beta}(y-x))$$
This can be checked to be defined over $F$.
