0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $.

$\endgroup$
  • $\begingroup$ Oh right, of course. Thanks $\endgroup$ – Auclair May 11 '16 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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