Let $a$ be a square free integer and $P(X)=X^n-a$.
By Einstein's criterion P is irreducible. Now let $k$ be an extension pf $\Bbb{Q}$ of degree $d$, and $\beta$ the positive root of $P.$
I would like to prove that if $[k(\beta):\Bbb{Q}(\beta)]=[k:\Bbb{Q}]$ then $X^n-a$ is irreducible over $k[X]$. Already, I first proved the converse, but I am stuck visualizing how the fact that $[k(\beta):\Bbb{Q}(\beta)]=d$ implies the result.
$P$ is a degree $n$ so if we have $P=QR$ the degree of $Q,R$ is less than $n\le 2.$
I have another question (it's a exercise). I proved that if $(n,d)=1$ then $X^n-a$ is irreducible, now I would to prove that $P$ is irreducible over $\Bbb{Q}(i)[X].$ So I need to prove that $(n,2)$ are coprime but I don't see why is true.
Edit: Perhaps I misunderstand the definition of square-free, I would translate it literally: $a$ is not divisible by a square. So it implies that for $a=p_1^{\alpha_1}\dots p_n^{\alpha_n}$ that $\alpha_j=1$ for at least one $j.$
Hint: Note that $[k(\beta):{\mathbb Q}]$ is equal to both $[k(\beta):k][k:{\mathbb Q}]$ and $[k(\beta):{\mathbb Q}(\beta)][{\mathbb Q}(\beta):{\mathbb Q}]$.