For $\ell/k$ a finite separable extension of fields and an affine variety $X\_{/\ell}$, the Weil restriction of scalars $R\_{\ell/k}(X\_{/\ell})$ represents the functor $R\_{\ell/k}(X\_{/\ell})(S) := X\_{/\ell}(S \times\_k \ell)$, for any $k$-scheme $S$. One can show that if $X$ is a smooth affine algebraic group over $\ell$, then so is $R\_{\ell/k}(X\_{/\ell})$, but over $k$. Moreover, for any affine smooth algebraic group, the Weil restriction of scalars preserves finiteness, commutativity, reductivity, connectivity, unipotency, and possibly other properties. Now my question:



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> Does the Weil restriction of scalars preserve solubility of a smooth affine algebraic group?
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I can't find any references on this. Also, I've tried showing that the Weil restriction commutes with the formation of derived subgroups, but with no success (due to my lack of ability, probably). 


For completeness, an algebraic group is soluble if its derived series becomes trivial in a finite number of steps. 



Solubility of an algebraic group and Weil restriction
Let me try to deduce a positive answer from standard properties of the Weil restriction and the results presented in this paper (found by Google): <http://math.stanford.edu/~conrad/papers/appbnew.pdf>. This is probably an overkill, especially if you are only interested in perfect ground fields.


Weil restriction preserves left exactness; by Corollary A.5.4 (1) of "Pseudo-reductive groups" by Conrad, Gabber, Prasad, it also preserves smooth surjections between affine groups. Thus, all you need to prove is that your smooth affine $\ell$-group $X$ has a composition series whose subquotients are all affine smooth and commutative. In the sequel, we argue the existence of such a filtration. For this, as a first reduction, we may assume that $X$ is connected (its component group is etale and inherits solvability).


Since tori are affine smooth and commutative, Theorem 5.4 in the linked pdf reduces us to assuming that $X$ is unipotent and either $k$-split or $k$-wound. The split case is immediate from Definition 1.3. In the $k$-wound case, the filtration by cckp-kernels provided by Corollary 3.3 gives the claim.


EDIT: The original question did not have the assumption that $\ell/k$ be separable.


