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Consider a number field $K/\mathbb{Q}$ and the embedding of $K^* \hookrightarrow GL_n(\mathbb{Q})$. This is the set of rational points of a $\mathbb{Q}$-algebraic group $G \subseteq GL_n(\mathbb{C})$. Then is it true that any $\mathbb{Q}$-characters of $G$ will look like $g \mapsto \det(g)^k$ for some $k \in \mathbb{Z}$. That is, on $G_\mathbb{Q} = K^*$ the character will look like $x \mapsto N_\mathbb{Q}^K(x^k) $.

I heard someone make this remark in a discussion that all the rational characters on a number field are powers of norm. I have not been able to find a reference (or other non-norm characters!).

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If $H$ denotes the multiplicative group defined over $K$, then $G=\mathrm{Res}_{K/\mathbb{Q}}H$. By Section 2.61 of Milne's "Algebraic Groups - The Theory of Group Schemes of Finite Type over a Field", the group $G_{\overline{\mathbb{Q}}}$ obtained from $G$ by extension of scalars is isomorphic to the product of $H_\sigma$, where $\sigma$ runs through the embeddings $K\hookrightarrow\overline{\mathbb{Q}}$. It follows that the characters of $G$ defined over $\overline{\mathbb{Q}}$ are the maps $\prod_\sigma f_\sigma^\sigma$, where each $f_\sigma$ is a character of $H$ defined over $\overline{\mathbb{Q}}$. The characters of $G$ defined over $\mathbb{Q}$ are those maps $\prod_\sigma f_\sigma^\sigma$, which are fixed by $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. The condition "fixed by $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$" means that $f_\sigma$ is independent of $\sigma$, and it is defined over $K$. In other words, $f_\sigma$ is the map $x\mapsto x^k$ on $K^\times$ with some $k\in\mathbb{Z}$ independent of $\sigma$, which means that $$\prod_\sigma f_\sigma^\sigma(x)=\prod_\sigma(x^k)^\sigma=N_{K/\mathbb{Q}}(x^k).$$

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