Finding the Galois group over $\Bbb{Q}$.

Solution 1:

$\zeta:=\zeta_8$ is of degree $2$ over $\mathbb{Q}(\sqrt[8]{2})$, hence $K:=\mathbb{Q}(\sqrt[8]{2},\zeta)$ is of degree $16$ over $\mathbb{Q}$ and of degree $4$ over $\mathbb{Q}(\zeta)$. The minimal polynomial of $\sqrt[8]{2}$ over $\mathbb{Q}(\zeta)$ is $X^4-\sqrt{2} = X^4 - (\zeta + \zeta^{-1})$.

Hence $Gal(K/\mathbb{Q})$ is an extension of $(\mathbb Z_8)^\times$ by $\mathbb Z_4$.

If $\sigma \in Gal(K/\mathbb{Q})$, then $\sigma$ satisfies: $\sigma(\zeta) = \zeta^a$ and $\sigma(\sqrt[8]{2})=\zeta^b \sqrt[8]{2}$ for some $a \in \mathbb Z_8^\times$, $b \in \mathbb Z_8$ such that $\zeta^{4b} \sqrt{2}= \zeta^a + \zeta^{-a}$ which means $b = \tfrac{a-1}{2} \pmod 2$.

EDIT: Note that $\zeta_8 = \exp(2i\pi/8) = \exp(2i\pi/8)$. We know that take any value in $\{ 1,3,5,7 \} =\mathbb Z_8^\times$, and we have :

• $\exp(1 \times i \pi/4) + \exp(-1.i \pi/4) =\sqrt{2}$.
• $\exp(7 \times i \pi/4) + \exp(-7.i \pi/4) =\sqrt{2}$.
• $\exp(3 \times i \pi/4) + \exp(-3.i \pi/4) =-\sqrt{2}$.
• $\exp(5 \times i \pi/4) + \exp(-5.i \pi/4) =-\sqrt{2}$.

Depending on $a$, the value $(\zeta^b)^4$ must be $+1$ or $-1$. In any case they are exactly $4$ values of $b$ allowed.