Let $L/K$ be a finite field extension with degree $m$. And let $n\in \mathbb N$ such that $m$ and $n$ are coprime. Show the following:
If there is a $a\in \mathbb K$ such that an $n$-th root of $a$ lies in $L$ then we have already $a\in \mathbb K$.
My attempt:
The field extension $K(\sqrt[n]{a})/K$ has degree smaller $n$. The minimal polynomial of $\sqrt[n]{a}$ namely $m_{\sqrt[n]{a}}(X)\in K[X]$ divides $X^n-a$. I.e. let $k$ be the degree of the minimal polynomial, then $k|n$.
But because of the formula $[L:K]=[L:K(\sqrt[n]{a}]\cdot [K(\sqrt[n]{a}):K]$ $k|m$, hence $k=1$ and hence our conclusion follows because $[K(\sqrt[n]{a}):K]=1$ yields $\sqrt[n]{a}\in K$ .
Can someone go through it? Thanks.
The non-separable case is actually not the problem: Let $p$ be the characteristic.
If $X^n-a$ is not separable, we have $p|n$ and we can write $X^n-a=(X^d-\sqrt[p^r]{a})^{p^r}$ with the inner polynomial being separable. By assumption $\sqrt[n]{a} \in L$, hence $\sqrt[p^r]{a} \in L$. But $\sqrt[p^r]{a}$ is purely inseparable over $K$ and seprable over $K$, because it lies in the separable extension $L/K$. This extension is separable because $m$ and $p$ must be co-prime. Thus $\sqrt[p^r]{a} \in K$. So we are reduced to the separable case with the polynomial $X^d-\sqrt[p^r]{a} \in K[X]$, since $d$ and $m$ are co-prime.