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I have some trouble while reading a proof of a lemma in the book

Conrad, Brian; Gabber, Ofer; Prasad, Gopal, Pseudo-reductive groups., New Mathematical Monographs 26. Cambridge: Cambridge University Press (ISBN 978-1-107-08723-1/hbk; 978-1-316-09243-9/ebook). xxiv, 665 p. (2015). ZBL1314.20037.

Here is the lemma:

Lemma C.4.4. Let $G$ be a group scheme locally of finite type over a field $k$, $T$ is a maximal $k$-torus in $G$ and $K/k$ an extension field. If $G$ is smooth or commutative, or if $K/k$ is a separable extension, then $T_{K}$ is a maximal $K$-torus in $G_{K}$.

In the proof, the authors wrote:

Suppose $G$ is connected and commutative. Any $k_{s}$-torus in $G_{k_{s}}$ has a $\text{Gal}(k_{s}/k)$-orbit consisting of finitely many $k_{s}$-tori, so by commutativity the collectively generate a $\text{Gal}(k_{s}/k)$-stable $k_{s}$-subtorus.

I don't know why a Galois orbit of a $k_{s}$-torus has only finite tori, can anybody explain this for me? Thanks a lot.

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    $\begingroup$ I guess you can take an affine neighborhood $U$ of $1$ in your $k_s$-torus, affine means it is $Spec(A_s)$, locally finite type means $U$ can be chosen such that $A_s$ is a finitely generated $k_s$ algebra thus defined by finitely many algebraic equations whose coefficients are in a finite extension of $k$ so that $Gal(k_s/k)$ sends $A_s$ to finitely many different copies $\endgroup$
    – reuns
    Commented Sep 28, 2019 at 14:55
  • $\begingroup$ @reuns I'm sorry if this question is silly but why those equations have coefficients in a finite extension of $k$? Is it related to the fact that every torus splits over a finite Galois extension or something else? $\endgroup$
    – vutuanhien
    Commented Sep 28, 2019 at 16:41
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    $\begingroup$ They have coefficients in $k_s\subset k^{alg}$ and finitely many algebraic numbers generate a finite extension. $Gal(k_s/k)$ corresponds to infinitely many isomorphisms $A_s\to \sigma(A_s)$ but there are only finitely many $\sigma(A_s)$ $\endgroup$
    – reuns
    Commented Sep 28, 2019 at 17:22
  • $\begingroup$ Thanks a lot. So now there can only be finitely many different $\sigma(U)$, right? $\endgroup$
    – vutuanhien
    Commented Sep 29, 2019 at 2:24

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