Slick proof $\operatorname{Gal}(\mathbb Q[e^\frac{2\pi i}p, \sqrt[\leftroot{-2}\uproot{2}p]{2}]/\mathbb Q)\cong S_p$?

Question

After having seen a lengthy and painful calculation showing $\operatorname{Gal}(\mathbb Q[e^\frac{2\pi i}3, \sqrt[\leftroot{-2}\uproot{2}3]{2}]/\mathbb Q)\cong S_3$, I'm wondering whether there's a slick proof $\operatorname{Gal}(\mathbb Q[e^\frac{2\pi i}p, \sqrt[\leftroot{-2}\uproot{2}p]{2}]/\mathbb Q)\cong S_p$ for odd prime $p$, because these calculations are getting intractable fast.

What are some slick proofs of this fact (assuming it is indeed correct).

Correction: What IS $\operatorname{Gal}(\mathbb Q[e^\frac{2\pi i}p, \sqrt[\leftroot{-2}\uproot{2}p]{2}]/\mathbb Q)$ for prime $p$?

Answer 1

Your statement does not hold. Let $\zeta$ be some $p$-th root of unity. Remember that the order of the galois group $\text{Gal} \mathbb{Q}(\zeta, \sqrt[p]{2})$ is the degree of the extension $\mathbb{Q}(\zeta, \sqrt[p]{2})/ \mathbb{Q}$. Now $[\mathbb{Q}(\zeta) : \mathbb{Q}]=p-1$ and $[\mathbb{Q}(\sqrt[p]{2}) : \mathbb{Q}]=p$ because $X^p-2$ is irreducible by eisenstein. We have $[\mathbb{Q}( \zeta, \sqrt[p]{2}) : \mathbb{Q}( \sqrt{2} ) ] \leq p$ but $p \mid [ \mathbb{Q}(\zeta, \sqrt[p]{2}) : \mathbb{Q}]=[\mathbb{Q}(\zeta, \sqrt[p]{2}) : \mathbb{Q}(\zeta)][\mathbb{Q}(\zeta): \mathbb{Q}]= [\mathbb{Q}(\zeta, \sqrt[p]{2}) : \mathbb{Q}(\zeta)](p-1)$, by euclid's lemma we have $p \mid [\mathbb{Q}(\zeta, \sqrt[p]{2}) : \mathbb{Q}(\zeta)]$ because $ \gcd(p, p-1) =1 $. So $p=[\mathbb{Q}(\zeta, \sqrt[p]{2}) : \mathbb{Q}(\zeta)]$.

Conclusion: $[\mathbb{Q}(\zeta, \sqrt[p]{2}) : \mathbb{Q}]=p(p-1)$. Now if your statement would hold, then $p(p-1)=|S_p|=p!$. This is true for odd $p$ exactly when $p=3$. So for any other odd $p$ this is not true.

Answer 2

The field $\;K:=\Bbb Q\left(\zeta:=e^{2\pi i/p},\,\sqrt[p]2\right)\;$ is the splitting field of $\;f(x):=x^p-2\in\Bbb Q[x]\;$ , and since this is an irreducible polynomial (why?) then $\;G:=Gal(K/\Bbb Q)\;$ acts transitively over its roots, which are $\;\alpha_i:=\sqrt[p]2\,\zeta^k\;,\;\;k=0,1,2,...,p-1\;$ .

Now take the (Galois) subextension $\;E:=\Bbb Q(\zeta)/\Bbb Q\;$ . This is the cyclotomic extension of the rationals of order $\;\phi(p)=p-1\;$ and it, of course, is cyclic of that order since in fact $\;Gal(E/Q)\cong\left(\Bbb Z/p\Bbb Z\right)^*\;$ .

Likewise, the (non-Galois) subextension $\;F:=\Bbb Q(\sqrt[p]2\,\zeta)/\Bbb Q\;$ of order $\;p\;$ (Why is this extension not normal?) is of order $\;p\;$ and its automorphism group is cyclic of order $\;p\;$.

Finally, observe that $\;G=Gal(E/\Bbb Q)\cdot Aut(F/\Bbb Q)\;$ by orders considerations, and since $\;Gal(E/\Bbb Q)\lhd G\;$ we then have a semidirect product $\;\cong C_p\rtimes C_{p-1}\;$

Answer 3

Here is an 'easy' group it is isomorphic to:

$$ \left\{\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} : a, b \in \mathbb{F}_p, a \neq 0 \right\} $$ with the following isomorphism. If $\sigma \in \text{Gal}(\mathbb{Q}(\zeta,\sqrt[p]{2}))$ with $\sigma(\zeta)= \zeta^a$ and $ \sigma ( \sqrt{2} ) = \zeta^b \sqrt[p]{2}$, then send $\sigma$ to $$ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}. $$ This was actually an excercise in a Galois Theory course I followed this year :)