$\mathbb{Q}<\mathbb{F}<\mathbb{C}$ - field extensions, such that $[\mathbb{F}:\mathbb{Q}]=m \in \mathbb{N}$
p is a prime number, $\alpha=p^{\frac{1}{n}}$
gcd(m,n)=1
prove n divides $[\mathbb{F}[\alpha]:\mathbb{Q}]$
My thoughts:
$\mathbb{F}<\mathbb{R}$ since otherwise $m=[\mathbb{F}:\mathbb{Q}]\ge [\mathbb{R}:\mathbb{Q}] = \infty$, which is a contradiction.
moreover if $[\mathbb{F}[\alpha]:\mathbb{Q}]$ is finite then $[\mathbb{F}[\alpha]:\mathbb{Q}]=[\mathbb{F}[\alpha]:\mathbb{F}]*[\mathbb{F}:\mathbb{Q}]=[\mathbb{F}[\alpha]:\mathbb{F}]*m$
$\{ 1,\alpha,\alpha^2,... \}$ is clearly linearly dependent over $\mathbb{F}$ since $\alpha^k$ (for $k\ge n$) is a multiplication of $\alpha^{k-n}$ by a rational constant.
if $B=\{ 1,\alpha,\alpha^2,..., \alpha^{n-1} \}$ is independent its span (with coefficients from $\mathbb{F}$) is $\mathbb{F}[\alpha]$ and therefore $[\mathbb{F}[\alpha]:\mathbb{F}] = n$
Is it about right? and how can I tell the set B is linearly independent over $\mathbb{F}$?
thanks