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Let $a,m$ be two positive integers $\geq 2$. Consider the following chain of extensions of $\mathbb{Q}$:

$$\mathbb{Q} \subset \mathbb{Q}(\zeta_m) \subset \mathbb{Q}(\sqrt[m]{a}, \zeta_m),$$

where as usual $\sqrt[m]{a}$ is a real number such that to the $m$-power equals $a$, and $\zeta_m$ is a $m$-th root of 1. So the latter field is the decomposition field of the polynomial $x^m-a$.

My question is the following: is always true that the extension $\mathbb{Q}(\sqrt[m]{a}, \zeta_m)/ \mathbb{Q}(\zeta_m)$ is abelian? Do we know to which group the Galois group is isomorphic?

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My attempt: We know that $\mathbb{Q}(\sqrt[m]{a}, \zeta_m)/\mathbb{Q}$ is a Galois extension, which is not always abelian; consider for example $a=2,m=3$; here the full extension is not abelian, as the Galois group is $S_3$, however, the extension that I am interested is, because the Galois group is $\mathbb{Z}/3\mathbb{Z}$.

We also know that if $\mathbb{Q}(\sqrt[m]{a}) \cap \mathbb{Q}(\zeta_m) = \mathbb{Q}$ then the full Galois group is isomorphic to a semi-direct product, namely

$$\mbox{Gal}(\mathbb{Q}(\sqrt[m]{a}, \zeta_m)/\mathbb{Q}) \cong \mathbb{Z}/m\mathbb{Z} \rtimes \left(\mathbb{Z}/m\mathbb{Z}\right)^{\times},$$

however I don't know if this is useful or not for my problem. I am guessing that the group that I am interested is actually isomorphic to $\mathbb{Z}/m\mathbb{Z}$, but I don't even believe that, because asking this group to be cyclic is too much, isn't it?

El.Gon.Zalo
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1 Answers1

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For each $\sigma\in G:=\operatorname{Gal}(\mathbb Q(\sqrt[m]{a},\zeta_m)/\mathbb Q(\zeta_m))$ there exists a unique $n(\sigma)\in \mathbb Z/m\mathbb Z$ such that $\sigma(\sqrt[m]{a}) = \zeta_m^{n(\sigma)}\cdot \sqrt[m]{a}$. This gives you an injective group homomorphism $G\to \mathbb Z/m\mathbb Z$, $\sigma\mapsto n(\sigma)$. So in particular, $G$ is cyclic.

Claudius
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