Prove that $K(\alpha)=K(\alpha^6)$ when $[K(\alpha):K]=2011$

Question

Let $L/K$ be a finite extension and let $\alpha \in L$ so that $[K(\alpha):K]=2011$. Prove that $K(\alpha)=K(\alpha^6)$.

My idea is as follows:

$K \subset K(\alpha^6) \subset K(\alpha)$, therefore $[K(\alpha):K] = [K(\alpha):K(\alpha^6)][K(\alpha^6):K]$. We have two options:

First, $[K(\alpha):K(\alpha^6)] =2011$ and $[K(\alpha^6):K] =1$. In that case ,$K(\alpha^6)=K $ and $\alpha^6 \in K$. The polynomial $f(x)=x^6-\alpha^6 \in K[X]$ has the the root $\alpha$ which is not possible since $deg(irr(\alpha,K)=2011$. (*)

We are with $[K(\alpha):K(\alpha^6)] =1$ so $K(\alpha)=K(\alpha^6)$.

I'm not really sure about the part (*), is it correct?

Thanks.

Answer 1

Yes, $\alpha^6 \in K$ implies $[K(\alpha):K] \le6$, which contradicts $[K(\alpha):K]=2011$.