Let $\zeta_{32}$ be the 32th primitive root of unity. I want to calculate the Galois group of the extension $\mathbb Q(\zeta_{32} + \zeta_{32}^{-1})/\mathbb Q$.
First of all, the Galois group of $\zeta_{32}$ is the units of $\mathbb Z_{32}$, so an abelian group of order
$16$ consisting of $1, 3, 5, 7, 9, 11, 13, 25, 17, 19, 21, 23, 25, 27, 29, 31$. It can be checked brutally by hand that $\zeta_{32} + \zeta_{32}^{-1}$ is fixed by $\langle 31 \rangle$. Note $\mathbb Q(\zeta_{32} + \zeta_{32}^{-1})/\mathbb Q$ is galois because its fixed field is a subgroup of an abelian group thus normal. Then its degree of extension should be $8$.
How do I calculate its Galois group?
Best Answer
Hint: Recall from elementary number theory $(\mathbb{Z}/2^n\mathbb{Z})^\times\cong C_{2^{n-2}}\times C_2$ for $n\geq 3$, with the $C_{2^{n-2}}$ generated by $5$ and the $C_2$ generated by $-1$.