TITLE: Galois Group of $x^p - 2$, $p$ an odd prime
QUESTION [28 upvotes]: Question is to determine the Galois group of $x^p-2$ for an odd prime $p$.
For finding the Galois group, we look for the splitting field of $x^p-2$ which can be seen as $\mathbb{Q}(\sqrt[p]{2},\zeta)$ where $\zeta$ is a primitive $p^{th}$ root of unity.
Consider $\mathbb{Q}\subset \mathbb{Q}(\zeta) \subset \mathbb{Q}(\sqrt[p]{2},\zeta)$.we know that $\mathbb{Q}(\sqrt[p]{2},\zeta)$ is Galois over $\mathbb{Q}(\zeta)$,  we find Corresponding Galois Group say $G_1$.
Consider $\mathbb{Q}\subset \mathbb{Q}(\sqrt[p]{2}) \subset \mathbb{Q}(\sqrt[p]{2},\zeta)$.
we know that  $\mathbb{Q}(\sqrt[p]{2},\zeta)$ is galois over $ \mathbb{Q}(\sqrt[p]{2})$, we find Corresponding Galois Group say $G_2$.
Then Galois Group of $\mathbb{Q}(\sqrt[p]{2},\zeta)$ would possibly be Product of these two subgroups $G_1$ and $G_2$ with some relation between the generators.
For $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\zeta))$, consider $\tau: \mathbb{Q}(\sqrt[p]{2},\zeta) \rightarrow \mathbb{Q}(\sqrt[p]{2},\zeta)$ fixing $\zeta$ and sending $\sqrt[p]{2} \rightarrow \sqrt[p]{2}\zeta$.
$\tau(\sqrt[p]{2})=\sqrt[p]{2}\zeta$,
$\tau^2(\sqrt[p]{2})=\tau(\tau(\sqrt[p]{2}))=\tau(\sqrt[p]{2}\zeta)=\tau(\sqrt[p]{2})\tau(\zeta)=\sqrt[p]{2}\zeta^2$,
For similar Reasons, $\tau^{p}(\sqrt[p]{2})=\sqrt[p]{2}\zeta^p=\sqrt[p]{2}$.
No power of $\tau$ less than $p$ gives identity as no power of $\zeta$ less than $p$ gives identity.
So, $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\zeta)) \cong \mathbb{Z}_p \cong \big< \tau \big>$.
For $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\sqrt[p]{2}))$, consider $\sigma : \mathbb{Q}(\sqrt[p]{2},\zeta) \rightarrow \mathbb{Q}(\sqrt[p]{2},\zeta)$ fixing $\sqrt[p]{2}$ and sending $\zeta \rightarrow \zeta^2$
$\sigma(\zeta)=\zeta^2$
$\sigma^2(\zeta)=\sigma(\sigma(\zeta))=\sigma(\zeta^2)=\zeta^{(2^2)}$
For similar reasons, $\sigma^{p-1}(\zeta)=\zeta^{(2^{p-1})}$, as for every $a\in \mathbb{F}_p^\times$, we have $a^{p-1}=1$ we have in particular $2^{p-1} \equiv~1~mod~p$.
So, $\sigma^{p-1}(\zeta)=\zeta^{(2^{p-1})}=\zeta$ (as $\zeta$ is a $p^{th}$ root of unity).
No power of $\sigma$ less than $p-1$ gives identity as $2\in \mathbb{F}_p$ generates Multiplicative group, no power of $2$ less than $p-1$ can be equal to $1~mod~p$.
So, $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\sqrt[p]{2})) \cong \mathbb{Z}_{p-1} \cong \big< \sigma\big>$.
As $[\mathbb{Q}(\sqrt[p]{2},\zeta):\mathbb{Q}]=p(p-1)$ and $|\sigma|=p-1$ and $|\tau|=p$ i strongly feel Galois group should be possibly generated by $\sigma$ and $\tau$ with "Some extra related conditions"But not very sure to confirm this.
I am not able to go any further, I can see that $\sigma$ and $\tau$ do not commute with each other. I am unable to produce a know group which contain isomorphic copies of $\mathbb{Z}_{p-1}$ and $\mathbb{Z}_p$ as subgroups.
I would be thankful if some one can help me out in this case.
Thank You.

REPLY [11 votes]: I thought I'd add a realization of this Galois group that can be analysed easily with direct computation, since it turns out that the Galois group is actually quite easy to describe.
Let $G = Gal(X^p-2)$ (or indeed $Gal(X^p-q)$ for any prime $q$). Then, as observed, the splitting field is $L = \mathbb{Q}(\zeta,\sqrt[p]{2})$. So any element $\varphi \in G$ is determined by $\varphi(\zeta)$ and $\varphi(\sqrt[p]{2})$. But $\varphi(\zeta) = \zeta^a$ for some $1 \leq i \leq p-1$ and $\varphi(\sqrt[p]{2}) = \zeta^b \sqrt[p]{2}$ for some $0 \leq b \leq p-1$ We thus get a map:
$$\chi:G \rightarrow G' :=\bigg\{\begin{pmatrix}
 a & b \\
  0 & 1\\ 
\end{pmatrix} : a \in \mathbb{F}_p^{\times}, b \in \mathbb{F}_p\bigg\}  \subset \mathrm{GL}(2, \mathbb{F}_p)
$$
$$\chi(\varphi)= \begin{pmatrix}
 a & b \\
  0 & 1\\ 
\end{pmatrix}$$
This is quickly seen to be both bijective and a homomorphism, so an isomorphism, explicitly identifying $G$ with the easy to understand group $G'$.
The nicest thing about this isomorphism, I think, is that it is extremly explicit. You can see the subgroups generated by $\sigma$ and $\tau$, you can easily tell which subgroups are normal and you can quickly compute the composition of given elements. This also gives you for free that the size of the extension is $p(p-1)$, although that was actually easy to compute.