Let $k$ be a field of characteristic zero (I'm only interested in number fields), and let $\mathbb{G}\_{/k}$ be a linear algebraic group defined over $k$ which is almost $k$-simple (all normal subgroups defined over $k$ are finite). This group need not be absolutely almost simple, and one way this situation can arise is when $\mathbb{G}$ is the Weil restriction of scalars of an almost-simple $K$-group defined over a finite extension $K/k$. Is that the only possibility?


In other words, suppose $\mathbb{G}$ is $k$-almost simple. Is there a finite extension $K$ and an absolutely almost simple group $\mathbb{H}\_{/K}$ such that $R^K\_k\mathbb{H}$ is $k$-isogenous to $\mathbb{G}$?



Converse to Weil Restriction of Scalars
No. Algebraic groups geometrically isomorphic to $\mathbb G\_m^n$ are classified by homomorphisms from the Galois group to $\operatorname{Aut}( \mathbb G\_m^n) = GL\_n(\mathbb Z)$. Almost simple ones correspond to irreducible representations. You are asking whether the irreducible representations are induced representations of trivial representations - in fact induced representations are never irreducible, so the answer is no.


All other cases are fine, though:


Consider the group $G$ over an algebraically closed field. It has no proper canonical normal subgroups - ones definable using just the structure of the group, which would be definable over $k$. Hence it is either reductive or unipotent.


If unipotent, it must be abelian, hence of the form $\mathbb G\_a^n$. Then it must be of the form $\mathbb G\_a^n$ over $k$ by Hilbert 90, hence $k=1$ and we are done.


If reductive, it must be either a torus or semisimple. If semisimple it is isogenous to a product of copies of simple groups. The semisimple case is fine - there is a transitive Galois action on the set of simple factors, which you can use to write it as a Weil restriction from the field corresponding to the stabilizer of one of the factors. The torus case, as we saw earlier, is problematic. 


