Galois group of $x^{8}+16$

I was trying to do that problem. I first found that the roots of the polynomial are $e^{\frac{k\pi i}{8}}\sqrt{2}$ with $k=1,3,5,7,9,11,13,15$. Then, I say that if $\sigma$ is an element of the Galois group then $\sigma(e^{\frac{\pi i}{8}}\sqrt{2})=e^{\frac{k\pi i}{8}}\sqrt{2}$ for another root. Then I claim that this will uniquely determine $\sigma$ since $\sigma(e^{\frac{(2m+1)\pi i}{8}}\sqrt{2})=\frac{1}{2^m}\sigma((e^{\frac{\pi i}{8}}\sqrt{2})^{2m+1})$, and since we can send it to any other root then we get that the Galois group has $8$ elements. So I think since we are mapping $1$ into an odd number defined modulo $16$ then would the galois group be isomorphic to $(\mathbb{Z}/16\mathbb{Z})^\times$?

Solution 1:

I am slightly expanding on your own work. Consider, as you did, $$ \alpha = \sqrt{2} \cdot \left(\frac{1}{2} \sqrt{2 + \sqrt{2}} + i \frac{1}{2} \sqrt{2 - \sqrt{2}}\right) = \sqrt{2} \cdot (\cos(\pi/8) + i \sin(\pi/8)). $$ We have $\alpha^{8} = -16$, and thus $\alpha$ is a root of $f = x^8 + 16$. Note that $$\eta = \cos(\pi/8) + i \sin(\pi/8)$$ is a primitive $16$-th root of unity, and that $$\sqrt{2} = (\eta + \eta^{-1})^2 - 2 = \eta^2 + \eta^{-2} \in \mathbf{Q}[\eta^2].$$

Moreover, $\dfrac{\alpha^2}{2} = \eta^{2}$, so $\eta^2 \in \mathbf{Q}[ \alpha ]$. It follows $$ \mathbf{Q}[ \sqrt{2} ] \subseteq \mathbf{Q}[ \eta^2 ] \subseteq \mathbf{Q}[\alpha] \subseteq \mathbf{Q}[ \eta ]. $$ Since $\alpha/\sqrt{2} = \eta$, it follows $\mathbf{Q}[\alpha] = \mathbf{Q}[ \eta ]$. In particular, $\mathbf{Q}[\alpha]$ is the splitting field of $f$ over $\mathbf{Q}$.

Now it is well-known that $\lvert \mathbf{Q}[ \eta ] : \mathbf{Q} \rvert = \varphi(16) = 8$, so that $f$ is irreducible over $\mathbf{Q}$, and the Galois group is isomorphic to the group of units of $\mathbf{Z}_{16}$, and it is thus isomorphic to $\mathbf{Z}_{4} \times \mathbf{Z}_{2}$.