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

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...

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.

• Nice, +1. A minor quibble: when I first read this, I took the parenthetical to mean that the characteristic assumption was necessary for the Galois group of $t^n-a$ to be solvable, as opposed to the separability of the polynomial and hence the Galoisity of its splitting field. May 13 '13 at 2:19
• Is the fact that no abelian Galois extensions imply no solvable Galois extensions standard knowledge? May 13 '13 at 2:25
• @Clayton: maybe not. It came up for me when I was thinking about the maximal solvable extension of $\mathbb{Q}$. May 13 '13 at 2:36
• Dear, Pete! Could you explain a bit more how to get a countere-example in part b)? I understood that separable closure of $\mathbb{F}_p(t)$ has no non-trivial separable extension. But why it is not algebraically closed? And how to deduce counterexample? Would be very grateful for help!
– ZFR
May 3 '19 at 21:21

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.

• In other words, if $K$ has no abelian extension then it must contain a primitive $n$th root of unity. May 13 '13 at 2:01