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I am having trouble computing the Galois Group of the splitting field $E$ of $x^p-a$ (where $p$ and $a$ are prime) over $\mathbb{Q}$.

Let $w$ be a $p^\text{th}$ root of unity, and $\alpha$ a root of $x^p-a$. $E$ should be isomorphic to $\mathbb{Q}(w,\alpha)$. The extension $\mathbb{Q}(w,\alpha)/\mathbb{Q}(\alpha)$ has basis $\{w,\ldots, w^p\}$ and $\mathbb{Q}(\alpha)/\mathbb{Q}$ has basis $\{1,\alpha,\ldots, \alpha^{p-1}\} \implies [E:\mathbb{Q}] = p(p-1)$.

But I am not sure what the Galois Group should be. Is it $C_p \times C_{p-1}$, and how do I justify this? There can only be one (cyclic) group of order $p$, but there could be other groups of order $p-1$ that are not cyclic.

Viktor Vaughn
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user138798
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  • @Quasicoherent Why is E isomorphic to $\mathbb Q(w,\alpha)$? – P-S.D Jan 07 '18 at 10:06
  • @P-S.D The roots of $x^p - a$ are $\alpha, \omega \alpha, \ldots, \omega^{p-1} \alpha$. So $\mathbb{Q}(\omega, \alpha)$ contains all the roots of $x^p - a$ and conversely, since $\alpha, \omega \alpha \in E$, then $\omega = \frac{\omega \alpha}{\alpha} \in E$, so $\mathbb{Q}(\omega, \alpha) \subseteq E$. – Viktor Vaughn Jan 07 '18 at 18:42

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You should not expect the Galois group to be abelian, even though the correspondence gives you two subgroups $\operatorname{Gal}(\Bbb Q(\omega, \sqrt[n](\alpha))/\Bbb Q(\omega))$ and $\operatorname{Gal}(\Bbb Q(\omega, \sqrt[n](\alpha))/\Bbb Q(\sqrt[n]{a}))$ that are cyclic of order $p$ and $p-1$, respectively. These two subgroups generate the whole group, and the one of order $p$ must be normal, but this only implies that $G$ is a semidirect product of these groups. It remains to determine the action of $C_{p-1}$ on $C_p$.

Note that this action will in general not be trivial. An example is the splitting field for $x^3-2$. This extension is degree six, but we also know that the Galois group is a subgroup of $S_3$. Therefore, it is the nonabelian group $S_3$. In any case, since we know that $\Bbb Q(\sqrt[n]{a})$ is not a normal extension of $\Bbb Q$, that implies that the corresponding subgroup $C_{p-1}$ will not be a normal supgroup.

My advice, if you are having trouble determining the action, is to write down generators for your two cyclic subgroups explicitly as field automorphisms, and compute the conjugation action by hand.

Rolf Hoyer
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  • is there a way to show what is the automorphism which has order $p - 1$? I know how to find this automorphism when $p$ is given, for example $p = 3$ for a polynomial $x^3 - 7$, it's just write all automorphisms and see what them have order $2$, but I'm trying give an explicit automorphism to the Galois group of $x^p - 7$ and I don't know how determine the element of the group which has order $p - 1$, I just could find the element which is isomorphic to $C_p$ which will be $\sigma: \alpha \mapsto \alpha \omega$ and $\omega \mapsto \omega$. – Math enthusiast Nov 24 '18 at 14:38
  • @Mathenthusiast This is equivalent to finding an element $i\in \Bbb F_p^\times$ of order $p-1$. Such primitive roots are known to exist (as $\Bbb F_p^\times$ is known to be cyclic) but I don't think there's an easy way to generate them other than brute force. If you find a suitable $i$ then the corresponding Galois automorphism would be $\alpha \mapsto \alpha, \omega \mapsto \omega^i$. – Rolf Hoyer Nov 24 '18 at 15:46
  • I found a way to show what is the automorphism which has order $p - 1$ for the case when $p$ is an odd prime. I will put the link here if someone would like to know how to find this automorphism, it's just read what the author of this OP did. – Math enthusiast Jan 06 '19 at 14:07