are base change and restriction of scalars "inverses" in this case? Let $l/k$ be a finite extension of fields. Let $G$ be an affine $k$-groups scheme. Let $G_l = G \times_k l$. Is it true that $\mathfrak{R}_{l/k}(G_l) = G$, where $\mathfrak{R}_{l/k}(-)$ is the restriction of scalars?
This appears to be false. So my new question is:
Is it true that $\mathfrak{R}_{l/k}(G_l) = G^{[l:k]}$?
 A: What does "=" mean here, and do you have any examples of group schemes where this holds for some nontrivial extension?
If $k = \mathbb{R}$, $l=\mathbb{C}$, and $G=\operatorname{Spec} \mathbb{C}$, then certainly something goes wrong, as $G\times_k l$ is isomorphic to two copies of $G$.  This is a counterexample to every interpretation I can think of to the problem, but it's possible that I'm overlooking something.
A: They are not inverses, but adjoints.  For all affine $k$-group schemes $G$ and $l$-group schemes $H$, there is a bijection
$$\operatorname{Hom}_{\textrm{$l$-grp}}(G \times_k l,H) = \operatorname{Hom}_{\textrm{$k$-grp}}(G, \mathfrak R_{l/k}(H))$$
which is natural in $G$ and $H$.  In particular, if we fix $G$ and let $H = G \times_k l$, then we get a bijection
$$\operatorname{Hom}_{\textrm{$l$-grp}}(G \times_k l,G \times_k l) = \operatorname{Hom}_{\textrm{$k$-grp}}(G, \mathfrak R_{l/k}(G \times_k l))$$
and therefore a special homomorphism $G \rightarrow \mathfrak R_{l/k}(G \times_k l)$ over $k$ coming from the identity map on $G \times_k l$.  But it is not an isomorphism.
For example, let $H$ be the additive group over $l$.  To make life easy, assume $l/k$ is quadratic.  Let $\Omega$ be a separable closure of $k$ containing $l$, and $\sigma$ is an element of $\operatorname{Gal}(\Omega/k)$ whose restriction to $l$ is not trivial .  If we identify $\mathscr R_{l/k}(H)$ with its $\Omega$-points, it is the group $\Omega \times \Omega$, on which $\sigma$ has the effect $\sigma.(x,y) = (\sigma(y),\sigma(x))$.  If $\sigma$ does restrict to the identity on $l$, then $\sigma.(x,y) = (y,x)$.
Given a homomorphism $f: G \times_k l \rightarrow H$ over $l$, the corresponding $k$-group homomorphism $G \rightarrow \mathscr R_{l/k}(H)$ is defined on $\Omega$-points by $x \mapsto (f(x),f(x))$.
In particular, taking $G = \mathbf G_{a,k}$, so that $H = \mathbb G_{a,l} = G \times_k l$, the canonical map $G \rightarrow \mathfrak R_{l/k}(H)$ is the diagonal map $x \mapsto (x,x)$, which is not an isomorphism.
