Galois Extension of Even Order

Question

Suppose $E=F(\alpha)$ is a proper Galois extension. Let $\sigma \in Gal(E/F)$ such that $\sigma(\alpha)=\alpha^{-1}$. Show that $[E:F]$ is even and $[F(\alpha + \alpha^{-1}):F]=\frac{1}{2} [E:F]$.

I was wondering that if $f(x)$ is the minimal polynomial of $\alpha$ then is the reciprocal polynomial the minimal polynomial of $\alpha^{-1}$?

Answer 1

Extended hints/steps:

Let $K$ be the fixed field of $\sigma$, and let $L=F(\alpha+\alpha^{-1})$.

1. Show that $\sigma$ is of order two (assuming "proper extension" means that $F$ is a proper subfield of $E$).
2. Show that $[E:K]=2$.
3. Show that as $E=L(\alpha)$, we also have $[E:L]\le2$.
4. Show that $L\subseteq K$.
5. Show that $L=K$.