Galois Extensions and $n^{\text{th}}$ Roots

Question

I've been studying for my prelims lately, and this problem has me stuck:

(a) Let $K$ be a field with no abelian Galois extensions. Suppose that $n$ is a positive integer and either $char(K)=0$ or the characteristic is relatively prime to $n$. Prove that every element of $K$ has an $n^\text{th}$ root in $K$.

(b) Is the statement still true if $n$ is not relatively prime to the characteristic of $K$?

For part $(a)$, I was thinking about the polynomial $f(x)=x^n-a$ for a given $a\in K$. If it has a root in $K$, then $a$ has an $n^\text{th}$ root. I wasn't able to make it much further than this, I only know some conditions when the polynomial is irreducible, which doesn't necessarily say anything about the roots. As for part $(b)$, I'm pretty unsure...

Answer 1

If a field has no abelian Galois extensions it also has no solvable Galois extensions. This solves part a) since (with the characteristic assumption) the Galois group of $t^n - a$ is solvable.

As for part b), start with an imperfect field like $\mathbb{F}_p(t)$ and then pass to its separable closure. This field is separably closed -- so has no nontrivial Galois extensions period -- but not algebraically closed. Every element of the algebraic closure is obtained by taking $p^n$th roots of elements of your ground field, so you'll find a counterexample here.

Answer 2

This is just a special case of Kummer extensions. If $a\not\in K^n$, where the hypothesis apply, then $K(\sqrt[n]{a})/K$ is a Kummer extension. These are all abelian.