Exercise 7.3.15

Proof. Let $L=\mathbb{Q}({\zeta }_p,\sqrt[p]{2})$.

By the isomorphism $\psi :\mathrm{Gal}(L∕\mathbb{Q})\to \mathrm{AGL}(1,{𝔽}_p)$, ${\gamma }_{a,b}=\psi ({\sigma }_{a,b})$ corresponds to ${\sigma }_{a,b}$ uniquely determined by (see section 6.4)

(a)

$T^{\prime }$ is so the set of the ${\sigma }_{1,b},b\in {𝔽}_p$, where ${\sigma }_{1,b}({\zeta }_p)={\zeta }_p$. Therefore $$\mathbb{Q}({\zeta }_p)\subset L_{T^{\prime }}.$$

$T^{\prime }=\mathrm{Gal}(L∕L_{T^{\prime }})$, thus $p=|T^{\prime }|=[L:L_{T^{\prime }}]$.

Moreover, $[\mathbb{Q}({\zeta }_p,\sqrt[p]{2}):\mathbb{Q}({\zeta }_p)]=p$, since $p-1=[\mathbb{Q}({\zeta }_p):\mathbb{Q}]$ and $[\mathbb{Q}(\sqrt[p]{2}):\mathbb{Q}]=p$ are relatively prime.

Thus $[L:L_{T^{\prime }}]=[L:\mathbb{Q}({\zeta }_p)]$, so $[L_{T^{\prime }}:\mathbb{Q}]=[\mathbb{Q}({\zeta }_p):\mathbb{Q}]$, with $\mathbb{Q}({\zeta }_p)\subset L_{T^{\prime }}$, therefore

$$\mathbb{Q}({\zeta }_p)=L_{T^{\prime }}.$$

(b)

$D^{\prime }$ is the set of ${\sigma }_{a,0}$, where ${\sigma }_{a,0}(\sqrt[p]{2})=\sqrt[p]{2}$. Therefore $$\mathbb{Q}(\sqrt[p]{2})\subset L_{D^{\prime }}.$$

By Theorem 7.3.1(b), $[L:L_{D^{\prime }}]=|D^{\prime }|=p-1=[L:\mathbb{Q}(\sqrt[p]{2})]$, so we can conclude

$$\mathbb{Q}(\sqrt[p]{2})=L_{D^{\prime }}.$$

As ${\sigma }_{a,b}(\sqrt[p]{2})={\zeta }_p^b\sqrt[p]{2}$, the conjugate fields of $L_{D^{\prime }}$ are the fields

$$\mathbb{Q}({\zeta }_p^b\sqrt[p]{2}),b=0,\cdots ,p-1.$$