Let $\omega_{13}$ be a primitive $13^{th}$ root of unity. Suppose $f: \mathbb{Q}(\omega_{13}) \rightarrow \mathbb{C}$ is a field homomorphism. It's really intuitive that $f(\mathbb{Q}(\omega_{13}+ \omega_{13}^{-1})$ should be inside $\mathbb{R},$ but how can I prove it rigorously?
I can clearly see that $\mathbb{Q}(\omega_{13})/ \mathbb{Q}$ is a Galois extension with Galois group $\mathbb{Z_{13}}^{*}.$ Moreover, $ \mathbb{Q}(\omega_{13})/\mathbb{Q}(\omega_{13}+ \omega_{13}^{-1})$ is also Galois extension by the fundamental theorem of Galois theory, and it's the fixed field of the subgroup generated by the complex conjugation restricted to $\mathbb{Q}(\omega_{13}).$ Since the subgroup generated by the complex conjugation is normal in $Gal(\mathbb{Q}(\omega_{13})/ \mathbb{Q},$ $\mathbb{Q}(\omega_{13}+ \omega_{13}^{-1})/ \mathbb{Q}$ is also Galois, and the corresponding Galois group is isomophic to $\mathbb{Z_{\frac{p-1}{2}}}$. I cannot procede further. By the way, it's clear that $\mathbb{Q}(\omega_{13}+ \omega_{13}^{-1})$ is inside $\mathbb{R},$ but why $f(\mathbb{Q}(\omega_{13}+ \omega_{13}^{-1})$ has to be in $\mathbb{R}?$ My second question is that is this true for for any prime $p$?
Thank you in advance.
$f(\Bbb Q(\omega+\omega^{-1}))=\Bbb Q(f(\omega+\omega^{-1}))$. But $$f(\omega+\omega^{-1})=f(\omega)+f(\omega)^{-1}=\zeta+\zeta^{-1}$$ where $\zeta=f(\omega)$. But $\zeta$ is also a non-trivial $13$-th root of unity, and so $\zeta+\zeta^{-1}$ is real.