$\DeclareMathOperator{\Hom}{Hom}$
$\DeclareMathOperator{\Res}{Res}$
Let $G$ be an algebraic group over $K$ and let $L/K$ be a finite separable extension. Consider the Weil restriction $G' := \Res^L_K(G \times_K L)$ of the base change $G \times_K L$.
Is it true that $G'$ is isomorphic to $G$ as algebraic groups over $K$? Is it at least true if $G$ is an elliptic curve (1-dimensional projective algebraic group)? If not, what is the relation between $G$ and $G'$?
I believe that $G' \times_K L$ is isomorphic to $G^{[L:K]}$ (at least if $L/K$ is Galois). But I am not sure.
Best Answer
No, they need not be isomorphic as schemes.
It might be helpful to see an example. Let $\mathbb{C}/\mathbb{R}$ be an extension of fields.
Consider, some algebraic group over $\mathbb{R}$. I will take the multiplicative group $G=\text{Spec}\mathbb{R}[x,y]/(xy-1)$ as an example and its base change is $G\times_\mathbb{R}\mathbb{C}=\text{Spec}\mathbb{C}[x,y]/(xy-1)$ If I want to look at the Weil restriction to $\mathbb{R}$, this becomes, $G'=\mathbb{R}[x_1,x_2,y_1,y_2]/(x_1y_1-x_2y_2-1, x_2y_1-x_1y_2)$.
How did I get these relations? Basically, we are saying that $\Re(xy-1)=0$ and $\Im (xy-1)=0$ and we wrote $x=x_1+ix_2$ and $y=y_1+iy_2$.
I am not sure what meaningful things we can say apart from the obvious statements. i.e. Look at the functor of points, $G'(K)=(G\times_K L)(L)$.