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

score 3 · Answer 1 · answered May 20 '19 at 08:49

3

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

answered May 20 '19 at 08:49

user10354138

Sir can u explain why $degree ≤2$ – Ellen Ellen May 20 '19 at 09:09
I don't understand – Ellen Ellen May 20 '19 at 09:09
@EllenEllen This is because $f = x^2-\alpha^2$ has root $\alpha$ in $K(\alpha^2)$. If $\alpha \in K(\alpha^2)$, the minimal polynomial would simply be $x-\alpha$ with degree 1. Otherwise f would be the minimal polynomial with degree 2. – Nancium Apr 18 '22 at 10:39

1 Answers1