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So $p$ is prime, and we have $f = x^p -1 \in \mathbb{Q}[x]$ with splitting field $E$. I need to show that ${\rm Gal}(E/\mathbb{Q})$ is abelian and of order $p-1$.

The splitting field $E$ is $\mathbb{Q}(\zeta)$ for some primitive $p$th root of unity. Let $\zeta = e^{\pi i/p}$. From the theory of cyclotomic polynomials, the field $E$ is the splitting field of $f$, and also $[E:\mathbb{Q}] = \phi(p) = p-1 = \left|{\rm Gal}(E/\mathbb{Q})\right|$. Since any $\sigma$ in the Galois group moves $\zeta$ to another $p$th root of unity, every such $\sigma$ is determined by its image $\sigma(\zeta)$, correct? I don't really know how to proceed from here to show that the Galois group is abelian.

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Yes, you are correct when saying "every such $\sigma$ is determined by its image $\sigma(\zeta)$".

What do you know about $\sigma(\zeta)$ ? For instance, what is $\sigma(\zeta)^p$ ? Can you have $\sigma(\zeta)^k = 1$ if $1<k<p$ ? You can prove actually that $\sigma(\zeta)$ is a primitive root of the unity, so that $\sigma(\zeta) = \zeta^{k_{\sigma}}$ for some integer $k_{\sigma}$ which is coprime with $p$. Therefore, we can assume $1 \leq k_{\sigma} \leq p-1$.

Therefore you can consider the map $h : \text{Gal}(E/\Bbb Q) \longrightarrow (\Bbb Z/p\Bbb Z)^*$ defined by $\sigma \longmapsto [k_{\sigma}]_p$.

Then you can check that $h$ is a group isomorphism. In particular, the Galois group $\text{Gal}(E/\Bbb Q)$ of the Galois extension $E=\Bbb Q(\zeta_p)/\Bbb Q$ is abelian.


This can be generalized to compute the Galois group of $\Bbb Q(\zeta_n)/\Bbb Q$, where $n$ is composite. See for instance here for more details.

Moreover, if a Galois extension $F/\Bbb Q$ has abelian Galois group, then $F$ can be embedded in a cyclotomic extension $\Bbb Q(\zeta_n)$ ! This is known as the Kronecker-Weber theorem.

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  • $\begingroup$ Thanks, this is great. What do you mean by $[k_\sigma]_p$? $\endgroup$
    – Auclair
    Commented Jun 7, 2016 at 12:19
  • $\begingroup$ @Auclair: this is the equivalence class of the integer $k_{\sigma}$ in $(\Bbb Z/p\Bbb Z)^*$ (since $k_{\sigma}$ is coprime with $p$, its equivalence class is invertible in $\Bbb Z/p\Bbb Z$). $\endgroup$
    – Watson
    Commented Jun 7, 2016 at 12:21
  • $\begingroup$ Thanks for the edits. I haven't looked at the links you provided yet, but I will. By the way, the answer provided by Adrian in the other stackexchange post you also linked to seemed very simple and elegant. I tried to do it for my problem, and it seemed to work. I haven't taken into account anything about numbers being coprime or the map $h$ you gave. Is it necessary or will Adrians approach work here as well? $\endgroup$
    – Auclair
    Commented Jun 7, 2016 at 12:43
  • $\begingroup$ @Auclair : you're right. My arguments show that $\text{Gal}(E/\Bbb Q)$ is actually cyclic. This could be helpful to your later. But in order to prove that it is abelian, Adrian's approach is far more better. Sorry for that. $\endgroup$
    – Watson
    Commented Jun 7, 2016 at 12:46
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    $\begingroup$ @Auclair : you're welcome. Good luck for your exam! :-) $\endgroup$
    – Watson
    Commented Jun 7, 2016 at 12:50

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