Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 4 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 $K$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.