Prove that $\operatorname{Gal}(\mathbb{Q}(\sqrt[8]{2}, i)/\mathbb{Q}(\sqrt{-2})) \cong Q_8$

Solution 1:

Hint: The order of $\sigma$ is not 8.

Note that $\sigma(\sqrt{2})=\sigma((\sqrt[8]{2})^4)=(\sigma(\sqrt[8]{2}))^4=\zeta_8^4(\sqrt[8]{2})^4=-\sqrt{2}$.

Note that $\zeta_8=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$.

Now compute $\sigma(\zeta_8)$.

Now compute $\sigma^{2}(\sqrt[8]{2})=\sigma(\zeta_8\sqrt[8]{2})=\sigma(\zeta_8)\sigma(\sqrt[8]{2})$, and then $\sigma^4(\sqrt[8]{2})$.

Solution 2:

Would it be easier to notice that extension $\mathbb{Q}(\sqrt[8]{2},i)$ is equal to $\mathbb{Q}(\sqrt[8]{2},\zeta)$ which is a cyclotomic extension followed by Kummer extension? You can then work out which elements of its Galois group fix $\sqrt{-2}$.