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Let $T$ be an algebraic group of multiplicative type over a field $K$. Let $$X^*(T)=\operatorname{Hom}_{\overline{K}}(T_{\overline{K}},(G_m)_{\overline{K}}) = \operatorname{Hom}_{\overline{K}}(\overline{K}[X,X^{-1}],O_T \otimes_K \overline{K})$$ be its character group.

How is defined the action of $\operatorname{Gal}(\overline{K}/K)$ on $X^*(T)$ ?

I am reading Milne's notes (lemma 5.2 and theorem 5.3, p.223) about the fact that $G \mapsto X^*(G)$ is an equivalence of categories between algebraic groups of multiplicative type and Galois modules. He is talking about 'the canonical action of $\operatorname{Gal}$ on $X^*$'. It may be simple but I can't figure out what it is.

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What do you mean by multiplicative type? –  Tobias Kildetoft Feb 6 '13 at 15:54
    
It means that $T_{\overline{K}}$ is diagonalizable. –  user10676 Feb 6 '13 at 15:56
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up vote 3 down vote accepted

For any algebraic variety $Z$ defined over $K$, the absolute Galois group $G$ of $K$ acts canonically on $Z_{\overline{K}}$ through its action on $\overline{K}$. If you prefer, it acts on the points $Z(\overline{K})$ through its action on the coordinates.

For any $\sigma\in G$, let us denote by $\sigma_Z$ the automorphism of $Z_{\overline{K}}$ defined by the action of $\sigma$. Then $G$ acts canonically on $X^*(T)$ by
$$ \sigma * \chi = \sigma_{\mathbb G_m} \circ \chi \circ (\sigma_{T})^{-1}, \quad \sigma\in G, \chi\in X^*(T).$$ You see that $\sigma *\chi=\chi$ if and only if $\chi$ is already defined over $K$.

For example, if $T=\mathbb G_m^d$ is a split torus, then all characters of $T$ are defined over $K$ and $X^*(T)=\mathbb Z^d$ with the trivial Galois action. Conversely, if $X^*(T)$ has trivial Galois action, then all characters of $T$ are defined over $K$. If we choose a basis of $X^*(T)$, they define an isomorphism $T_{\overline{K}}\to (\mathbb G_m)^d_{\overline{K}}$ which is invariant by Galois, hence is defined over $K$. This implies that $T$ is split over $K$.

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