$\zeta = \zeta_{17}$. As stated, the set up is looking at $Gal(\mathbb{Q}(\zeta_{17})/\mathbb{Q}) \simeq \mathbb{Z}/16\mathbb{Z},$ generated by $\sigma: \zeta \to \zeta^2$. I'm looking for the corresponding subfield of $<\sigma^8>$. Shouldn't it be
$\mathbb{Q}(\zeta + \sigma^8(\zeta))?$ But $\sigma^8(\zeta) = \zeta^{2^8} = \zeta.$
Surely $\sigma^8$ does not fix the $\mathbb{Q}(\zeta)$, right?
What am I doing wrong?