Galois Extension of Even Order 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 $\alpha^{-1}$?

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


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*Show that $\sigma$ is of order two  (assuming "proper extension" means that $F$ is a proper subfield of $E$).

*Show that $[E:K]=2$.

*Show that as $E=L(\alpha)$, we also have $[E:L]\le2$.

*Show that $L\subseteq K$.

*Show that $L=K$.

