Irreducible polynomial with coefficients in a real subfield of $\mathbb{C}$ with roots on the unit circle The problem is the next
Let $F \leq \mathbb{R}$ be a field, and $f \in F[x]$ an irreducible polynomial with a non real root $\alpha$ of absolute value 1. Then $\frac{1}{\beta}$ is a root of $f$ for every non real root $\beta$ of $f$.
I don't know how to begin, if I could show that the fact that one of the roots is in the unit circle forces all the other roots to be there it would be done, since the fraction would just be conjugation, but I couldn't prove that (may be false anyway).
 A: Hint: There is an isomorphism $F(\alpha)\to F(\beta)$ sending $\alpha$ to $\beta$.
More details are hidden below.

 Since $f$ is the minimal polynomial over $F$ of both $\alpha$ and $\beta$, $F(\alpha)\cong F[x]/(f)\cong F(\beta)$ with the isomorphisms fixing $F$ and sending $\alpha$ and to $x$ and then $x$ to $\beta$.  Now $1/\alpha=\overline{\alpha}$ is also a root of $f$ since $F\subseteq\mathbb{R}$.  Applying the isomorphism $F(\alpha)\cong F(\beta)$ to the equation $f(1/\alpha)=0$ then gives $f(1/\beta)=0$.  Note that the assumption that $\alpha$ and $\beta$ are non-real is unnecessary.

A: We can write $f=(X-c_1)...(X-c_n)$, Let $L=F(c_1,...,c_n)$ is Galois extension and $G$ is Galois group. $G$ acts transitively on the roots. Remark that the complex conjugation is a Galois automorphism, this implies that if $|c_1|=1$, $\bar c_1$ is a root and remark that $\bar c_1c_1=1$. Let $c_i$ be a root there exists $g\in G$ with $g(c_1)=c_i$, $g(c_1\bar c_1)=g(c_1)g(\bar c_1)=1=c_ig(\bar c_1))$ implies that ${1\over {c_i}}$ is a root.
