# Show that Gal$(E/\mathbb{Q})$ is abelian, where $E$ is the splitting field of $f(x)=x^{14} - 1$

Let $E$ denote the splitting field of $f(x)=x^{14}-1$. I want to show that the Galois group is abelian. Here's my attempt:

The different 14'th roots of unity are given by $w=e^{i \pi n/7}$ where $n = 1, 2,..., 14$. The primitive 14'th roots of unity are given by $n = 1,3,5,9,11,13$.

We have $E = \mathbb{Q}(w)$ where $w=e^{i\pi/7}$ and the basis for $E$ over $Q$ is $B=\{1,w,w^2,...,w^{12},w^{13}\}$ and $[E:\mathbb{Q}] = 14$. Therefore the Galois group Gal$(E/\mathbb{Q})$ is isomorphic to either $\mathbb{Z}_{14}$ or the dihedral group $D_{14}$. I know it's isomorphic to $\mathbb{Z}_{14}$ since it's supposed to be abelian, but how can I determine which it is? Which automorphism in the Galois group will generate it?

• – sqtrat May 11 '16 at 10:44
• @sqtrat Great, thanks! – Auclair May 11 '16 at 10:47

Actually, the degree $[E:Q] = 6 = \phi(14)$, where $\phi$ is the Euler totient function. For $m$ odd, the $m$-th cyclotomic field is the same as the $2m$-th cyclotomic field. Therefore, $E = Q( \xi)$, where $\xi$ is a primitive $7$-th root of unity. The Galois group is the cyclic group of order $6$ (and hence, Abelian) generated by the automorphism, $\sigma$, $\sigma (\xi ) = \xi^3$.