Exercise 6.4.15

Proof. Let $L=\mathbb{Q}({\zeta }_p,\sqrt[p]{2})$, the splitting field of $f=x^p-2$ over $\mathbb{Q}$. Then $\mathrm{Gal}(L∕\mathbb{Q})\simeq \mathrm{AGL}(1,{𝔽}_p)$.

We show that $x^p-2$ is irreducible over $\mathbb{Q}({\zeta }_p)$.

${\Phi }_p=1+x+\cdots +x^{p-1}$ is irreducible over $\mathbb{Q}$, thus $[\mathbb{Q}({\zeta }_p):\mathbb{Q}]=p-1$.

$[L:\mathbb{Q}]=p(p-1)$ by Section 6.4. We deduce of $[L:\mathbb{Q}]=[L:\mathbb{Q}({\zeta }_p)][\mathbb{Q}({\zeta }_p):\mathbb{Q}]$ that

(If $g$ is the minimal polynomial of $\sqrt[p]{2}$ over $\mathbb{Q}({\zeta }_p)$, then $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=[\mathbb{Q}({\zeta }_p,\sqrt[p]{2}):\mathbb{Q}({\zeta }_p)]=p$. Moreover $\sqrt[p]{2}$ is a root of $f=x^p-2\in \mathbb{Q}[x]\subset \mathbb{Q}({\zeta }_p)[x]$, thus $g\mid f$ in $\mathbb{Q}({\zeta }_p)[x]$, where $f,g$ are of the same degree $p$ and monic, thus $g=f=x^p-2$. Therefore $x^p-2$ is irreducible over $\mathbb{Q}({\zeta }_p)$.)

Following the wording, we note that $\sigma ={\sigma }_{1,1}\in \mathrm{Gal}(L∕\mathbb{Q}({\zeta }_p))$ defined by

is of order $p$, and corresponds to the $p$-cycle $\tilde{\sigma }=(1,2,\cdots ,p)$, if we number the roots by $z_1=\sqrt[p]{2},\dots ,z_p={\zeta }_p^{p-1}\sqrt[p]{2}$. Since ${\tilde{\sigma }}^{k-1}(1)=k,k=1,\dots ,p-1$, the subgroup $⟨\tilde{\sigma }⟩\subset S_n$ is transitive, and so is $⟨\sigma ⟩$. Since $\mathrm{Gal}(L∕\mathbb{Q}({\zeta }_p))\supset ⟨\sigma ⟩$, $\mathrm{Gal}(L∕\mathbb{Q}({\zeta }_p))$ acts transitively on the roots of $x^p-2$. So, by Proposition 6.3.7, $x^p-2$ is irreducible over $\mathbb{Q}({\zeta }_p)$. □