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.


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.

  • $\begingroup$ Thanks, this is great. What do you mean by $[k_\sigma]_p$? $\endgroup$ – Auclair Jun 7 '16 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 Jun 7 '16 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 Jun 7 '16 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 Jun 7 '16 at 12:46
  • 1
    $\begingroup$ @Auclair : you're welcome. Good luck for your exam! :-) $\endgroup$ – Watson Jun 7 '16 at 12:50

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.