I am trying to prove the following:
For fields $\mathbb{Q}\subset F \subset \mathbb{Q}(\xi_n)$, with $\xi_n$ a primitive $n$-th root of unity, prove that $\sigma(\overline{z})=\overline{\sigma(z)}$ for every $\mathbb{Q}$-automorphism $\sigma$ of $F$ and $z\in F$. (The bar indicates complex conjugation).
My issue with this problem is the fact that I am being asked to prove this for a subfield of $\mathbb{Q}(\xi_n)$ and not for $\mathbb{Q}(\xi_n)$ itself.
This is what I would do if $F$ were $\mathbb{Q}(\xi_n)$:
Since there is a $\mathbb{Q}$-basis for $\mathbb{Q}(\xi_n)$ made of powers of $\xi_n$, I think it is enough to prove that $\sigma(\overline{\xi_n})=\overline{\sigma(\xi_n)}$. Since $\xi_n^n=1$, we have $1=|\xi_n|=\xi_n\overline{\xi_n}$ and therefore $\xi_n^{-1}=\overline{\xi_n}$. It is also true that $\overline{\sigma(\xi_n)} = \sigma(\xi_n)^{-1}$ because $\sigma$ takes roots of $x^n-1$ to roots of $x^n-1$. Then $\sigma(\overline{\xi_n})=\sigma(\xi_n^{-1})=\sigma(\xi_n)^{-1}=\overline{\sigma(\xi_n)}$.
However, if $F$ is not equal to $\mathbb{Q}(\xi_n)$, I am concerned that this proof would not work because it may happen that $\xi_n$ is not even in $F$, so its image by $\sigma:F\rightarrow F$ would not be defined.
I would find it very helpful if someone could explain how I could fix my proof so that I can use it in this second case or maybe point out a different approach that would work in that situation. Thanks in advance.