Morphism of algebraic groups defined over $\overline{\mathbb{Q}}$

Question

My question is about the general theory of algebraic groups, on which I'm still far from comfortable. I want to understand a sentence in Milne's "Shimura varieties" http://www.jmilne.org/math/xnotes/svi.pdf , bottom of page 60.

We consider a reductive group $G$ defined over $\mathbb{Q}$, for which Milne defined a weight homomorphsm (cf. page 54 for a definition) $w_X:\mathbb{G}_m\rightarrow \mathbb{G}_{\mathbb{R}}$. The sentence I'm not getting is: "The weight homomorphism $w_X$ is a homomorphism $\mathbb{G}_m\rightarrow \mathbb{G}_{\mathbb{R}}$ over $\mathbb{R}$ of algebraic groups that are defined over $\mathbb{Q}$. It is therefore defined over $\overline{\mathbb{Q}}$"

Why can we reduce to $\overline{\mathbb{Q}}$? I guess it is because one works here only with algebraic tori (the image of $\mathbb{G}_m$ lies in the center $Z$ of $G$). Because for a general morphism between algebraic groups, this doesn't seem to be true: take $\mathbb{G}_{a,\mathbb{R}}=\mathrm{Spec}$ $\mathbb{R}[T]$, the additive group over $\mathbb{R}$, and the morphism $\mathbb{G}_{a,\mathbb{R}}\rightarrow \mathbb{G}_{a,\mathbb{R}}$ to be given by $T\mapsto \pi T$, which doesn't seem to be defined over $\overline{\mathbb{Q}}$.

I thank you for your help !

Yoël.

Answer 1

Yes, this is a statement about tori (or more generally groups of multiplicative type). The homomorphisms from $\mathbb{G}_m$ to itself are the maps $t\mapsto t^n$, $n\in\mathbb{Z}$, and this doesn't change when you extend the field. Therefore, for split tori, you don't get more homomorphisms when you extend the field. As every torus splits over a finite algebraic extension, it follows that for tori defined over $\mathbb{Q}$, every homomorphism over $\mathbb{C}$ is defined over the algebraic closure of $\mathbb{Q}$.