I want to prove that if $X^{n}-a\in K[X]$ is irreducible and $\beta$ is a root of $X^{n}-a$ in one extension of $K$ and $m$ divides $n$ then $\left[K\left(\beta^{m}\right):K\right]=n/m$. I want to calculate the irreducible polynomial of degree $n/m$ if possible, because maybe with that it would be easier to conclude that $\left[K\left(\beta^{m}\right):K\right]=n/m$
I've managed to do the following...
If $\beta$ is a root of $X^{n}-a\Rightarrow$ $\beta^{n}=a$ in its extension. Considering the fact that $m|n$ then $n=m\cdot t$. Thus $\left(\beta^{m}\right)^{t}=\beta^{n}=a$ and $\beta^{m}$ is a root of $X^{t}-a$. In conclusion, $\left[K\left(\beta^{m}\right):K\right]\leq t=n/m$ but I don't know how to prove the other inequality. Any ideas? Thanks in advance!
$f(x):=x^n-a$ and $g(x):=x^{\frac{n}{m}}-a$ are $\it{both}$ irreducible over $K$. Otherwise, $$g(x)=\prod_{j=1}^rh_j(x)$$ for some $r>1$ and irreducibles $h_1(x),\dots,h_r(x)\in K[x]$, which would then imply $$f(x)=g(x^m)=\prod_{j=1}^rh_j(x^m)$$ violating the fact that $f(x)$ is actually irreducible over $K$. Therefore, $g(x)$ is irreducible and $g(\beta^m)=f(\beta)=\mathbf 0$. $$\therefore [K(\beta^m):K]=\deg(g(x))=\frac{n}{m}$$.