Prove that : $K(\alpha)=K(\alpha^{2})$ , $K$ commutative field

Question

Problem :

Let $K$ be a commutative field, and $\alpha$ an

aldebraic element of $K$ of odd degree.

Proof that : $K(\alpha)=K(\alpha^{2})$

My simple try

Since $\alpha$ algebraic in $K$ so

$\exists F(x)=x-\alpha$ such that $d°F≥1$ and

$F(\alpha)=0$

But I don't know how I complete this work

Please give me ideas or hints to approach it

Answer 1

$[K(\alpha):K(\alpha^2)]\leq 2$ since $X^2-\alpha^2$ has $\alpha$ as a root. Is $[K(\alpha):K(\alpha^2)]=2$ possible? (Hint: tower-law)