Description of the Functor of Points of Commutators of Algebraic Groups

Question

Let $k$ be a field and consider an algebraic group $G/k$ (i.e. an affine $k$-group scheme of finite type). Furthermore, suppose we are given two algebraic subgroups $H_1, H_2 \subseteq G$. Then the commutator group $[H_1, H_2]$ is defined to be the smallest algebraic subgroup $K \subseteq G$ such that $[H_1(R), H_2(R)] \subseteq K(R)$ for all $k$-algebras $R$ (compare for example with J.S. Milne's book 'Algebraic Groups').

My question is now if this is the same thing as taking the commutator 'in the category of fppf-sheaves over $k$'. More precisely: Given $g \in [H_1, H_2](R) \subseteq G(R)$ for some $k$-algebra $R$, does there always exist an fppf map of $k$-algebras $R \to R'$ such that the image of $g$ in $G(R')$ is actually contained in $[H_1(R'), H_2(R')]$?

Milne remarks in his book that this indeed holds true, but I don't see why. In their book 'Groupes Algebriques, Tome 1', Demazure and Gabriel show it when $H_1$ and $H_2$ are both smooth and one of them is connected (see II.5.4.9). There is also an article by Battiston (see https://arxiv.org/abs/1803.06965) that gives a similar result in a slightly orthogonal situation.

Answer 1

Okay I think I found a counterexample that answers my question:

Assume that $k$ is of characteristic $0$. Consider the semidirect product $G = \mathbf{G}_a \rtimes \mathbf{Z}/2 \mathbf{Z}$ where $1 \in (\mathbf{Z}/2\mathbf{Z})(k)$ acts on $\mathbf{G}_a$ via $x \mapsto -x$. Also, let $a \in k \setminus \{ 0 \}$ and consider the algebraic subgroups $H_1 = \{ (0, 0), (0, 1) \}$ and $H_2 = \{ (0, 0), (a, 1) \}$ of $G$, both isomorphic to $\mathbf{Z}/2\mathbf{Z}$. Then the sheaf theoretic commutator of $H_1$ and $H_2$ is given by $\{ (2na, 0) \mid n \in \mathbf{Z} \} \cong \mathbf{Z}$ and thus is not an algebraic group. The commutator $[H_1, H_2]$ in the world of algebraic groups is given by $\{ (x, 0) \mid x \in \mathbf{G}_a \} \cong \mathbf{G}_a$.

Answer 2

In Milne (6.24), $[H_{1},H_{2}]$ is defined in two cases: (a) $G$ is affine; (b) $H_{1}$ and $H_{2}$ are smooth.

In case (a), the claim follows from a general statement about the algebraic subgroup generated by a map $\varphi\colon X\rightarrow G$ with the property that $\varphi(X(R))$ is closed under inversion for all $R$, namely, that the algebraic subgroup is defined by a certain ideal $I$ (Milne 2.46).

In case (b), D&G (II, Section 5, 4.9) state that "one shows in the same manner" that, when one of the groups $H_{1}$ and $H_{2}$ is connected, then the obvious functor is represented by a connected, smooth, closed subgroup $[H_1,H_2]$. I suspect that the connectedness hypothesis is only required to show that $[H_{1},H_{2}]$ is connected, but since D&G don't write out their proof, one can't be sure.