TITLE: Prove that if $[F(\alpha):F]$ is odd then $F(\alpha)=F(\alpha^2)$
QUESTION [12 upvotes]: Prove that if $[F(\alpha):F]$ is odd then $F(\alpha)=F(\alpha^2)$.
My justification for this question is as follows;
Suppose $F(\alpha^2)\subsetneq F(\alpha)$, we have $F \subsetneq F(\alpha^2) \subsetneq F(\alpha)$.
As $[F(\alpha):F]=[F(\alpha):F(\alpha^2)][F(\alpha^2):F]$ and $[F(\alpha):F(\alpha^2)]=2$ we would be end up with the case that $[F(\alpha):F]$ is a multiple of 2 i.e., even which contradicts given condition of $[F(\alpha):F]$ being odd. 
So, $F(\alpha^2) =F(\alpha)$.
After doing this, i have seen for a solution on some webpage where he/she used criterion of minimal polynomial.
Please let me know Is my justification clear or there are any gaps in between?

REPLY [18 votes]: Your proof is correct but you don't need contradiction. If $F(\alpha) \neq F(\alpha^2)$ then $\alpha \notin F(\alpha^2)$. Thus consider the tower $[F(\alpha) : F(\alpha^2)][F(\alpha^2) : F]$. The first factor is at most two (because $\alpha$ is a root of $x^2 - \alpha^2$ and it is at least two because $\alpha \notin F(\alpha^2)$. By the tower law it follows $[F(\alpha) : F]$ is even.