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Suppose that $G$ is an affine algebraic group defined over $\mathbb{Q}$. Then we can take its group $G(\mathbb{Q})$ of $\mathbb{Q}$-points, and abelianisation followed by rationalisation provides a homomorphism

$$\varphi\colon G(\mathbb{Q}) \longrightarrow G(\mathbb{Q})^\mathrm{ab} \otimes_{\mathbb{Z}} \mathbb{Q}$$

Since the right side is isomorphic to a direct sum of copies of $\mathbb{Q}$, it is also isomorphic to the $\mathbb{Q}$-points of an affine algebraic group. My question is whether this isomorphism can be chosen such that $\varphi$ is algebraic:

Is there a morphism $f \colon G \to H$ of affine algebraic groups defined over $\mathbb{Q}$ and an isomorphism $\lambda \colon H(\mathbb{Q}) \to G(\mathbb{Q})^\mathrm{ab} \otimes_{\mathbb{Z}} \mathbb{Q}$ such that $\lambda \circ f_\mathbb{Q} = \varphi$?

I naively expect the answer to be "no", and if this is the case a counterexample would be helpful.

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    $\begingroup$ If $G =\mathbb G_m$ then the rationalization is infinite-dimensional over $\mathbb Q$ and so doesn't arise from any affine algebraic group. $\endgroup$
    – Will Sawin
    Commented Dec 3, 2022 at 2:01
  • $\begingroup$ I think the issue (notably for $G=\mathbb{G}_m$) is not infinite-dimensionality, but rather the fact that the abelian group $H(\mathbf{Q})$ is not torsion-free divisible. $\endgroup$
    – YCor
    Commented Dec 3, 2022 at 10:16
  • $\begingroup$ @WillSawin You are of course right. Thanks! $\endgroup$
    – skupers
    Commented Dec 3, 2022 at 17:31
  • $\begingroup$ @YCor It depends on what $H$ is. What I said was a problem for $H$ an additive group, which seems to be what skupers has in mind, while your objection is a problem for $H$ a multiplicative group. $\endgroup$
    – Will Sawin
    Commented Dec 3, 2022 at 19:35

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