Let $K$ be a number field and $a \in K^\times$ an element that is not an $n$-th power. Is it true that the splitting field of $x^n-a$ is abelian if and only if $\zeta_n \in K$?
I know one direction is true: if $\zeta_n \in K$, then $K(\sqrt[n]{a})$ is the splitting field, hence every automorphism is determined by the image $\sqrt[n]{a} \mapsto \zeta_n^k \sqrt[n]{a} $. Therefore $G$ is a subgroup of $\mathbb{Z}/n\mathbb{Z}$.
The opposite direction is harder. In this case, I would love to say that the extension $K(\sqrt[n]{a})$ is not normal (thus, corresponding to a non-abelian subgroup of $G$). If it were normal, it would contain all roots of the minimal polynomial of $\sqrt[n]{a}$, in particular some $\zeta_n^r\sqrt[n]{a}$ and their quotient $\zeta_n^r$. If $n=p$ is prime, this provides a contradiction, since the dregee of $K(\zeta_p)/K$ is prime to $p$, which is the degree of $K(\sqrt[n]{a})/K$. In the general case two problems arise:
- if $r$ is not prime to $n$, then $\zeta_n^r$ is allowed to be in $K$ and $K(\sqrt[n]{a})$ might be a splitting field!
- if $r$ is prime to $n$ (or, say, we decide we contempt ourselves with the case of $n$ prime) we conclude that $K(\sqrt[n]{a})$ is the splitting field of $x^n-a$, but I am not sure how to exclude the possibility
Any idea?