Prove that $E=F[\alpha^2]$

Question

Let $E=F[\alpha]$, $\alpha$ is algebraic over $F$ and $[E:F]$ is odd. Prove that $E=F[\alpha^2]$.

Now clearly $[F[\alpha^2]:F]\mid[E:F]$, so $[F[\alpha^2]:F]$ is also odd. But how can we show that $E=F[\alpha^2]$?

Answer 1

Since $\alpha^2\in F[\alpha]$, $F[\alpha^2]\subseteq F[\alpha]$. Thus $$[F[\alpha]:F]=[F[\alpha]:F[\alpha^2]][F[\alpha^2]:F].$$ Since $\alpha$ is a root of $x^2-\alpha^2$ in$F[\alpha^2][x]$, the extension $[F[\alpha]:F[\alpha^2]]\leq 2$. Thus it must be 1 since the total extension is of odd degree showing that $F[\alpha]=F[\alpha^2]$.