The Weil restriction of a simple algebraic group Let $F$ be a number field, $G$ an $F$-simple affine algebraic group.
Then is the Weil restriction $\operatorname{Res}_{F/\mathbb{Q}} G$ $\mathbb{Q}$-simple?
I couldn’t find any references.
 A: As noted, the answer is yes. Let $G$ be a simple algebraic group (meaning $F$-simple) over $F$.
Assume first that $F$ is Galois over $\mathbb{Q}$. Then $(\mathrm{Res}_{F/\mathbb{Q}}G)_{F}\simeq G_{1}\times\cdots\times G_{r}$, where the $G_{i}$ are the conjugates of $G$ under $\mathrm{Gal}(F/\mathbb{Q})$. If $H$ is a smooth connected normal algebraic subgroup of $\mathrm{Res}_{F/\mathbb{Q}}G$, then $H_{F}$ is a product of a certain number of the $G_{i}$ (Milne, Algebraic Groups, 21.51), which is stable under the Galois action. But $\mathrm{Gal}% (F/\mathbb{Q})$ acts transitively on the set of $G_{i}$, and so $H=\mathrm{Res}_{F/\mathbb{Q}}G$.
In the general case, the same argument works if $G$ is geometrically simple. If not, then we may suppose that $G=\mathrm{Res}_{F^{\prime}/F}G^{\prime}$ with $G^{\prime}$ geometrically simple (ibid. 24.3) and apply the argument to $G^{\prime}$.
This argument fails for algebraic groups like $\mathbb{G}_{a}^{n}$ because for them the decomposition into simple objects is not unique.
Note that the proof only uses standard properties of semisimple groups, for which I gave a modern reference. I don't understand the comment of LSpice.
