# How do you prove that the Galois Group of a radical field extension is always soluble/solvable?

The question is asking how to prove (not necessarily in high detail but concisely) that the Galois Group (the group of Q-automorphisms of F where Q is the base field and F is a field extension of the base field) of a radical field extension (an extension of a field K that is obtained by adjoining a sequence of nth roots - radicals - of elements) is always soluble/solvable (a group which has a normal series such that each normal factor is abelian).

Thank you

Here is a rough sketch: You use induction on the degree of the field extension, and it is useful to allow the base field to vary. So we need to consider the case that $F = L[a^{\frac{1}{n}}]$ for some $a$, and some intermediate field $L$, itself a radical extension. Set $b = a^{\frac{1}{n}}$, a particular fixed choice of $n$-th root of $a \in L$. Any field automorphism $\alpha$ of $F$ which fixes $L$ ( ie $\alpha \in {\rm Gal}(F/L) )$ must send $b$ to another $n$-th root of $a$ and this must have the form $\omega b$ for some $n$-th root of unity $\omega$ and we have $\omega \in F$. It follows from this that ${\rm Gal}(F/L)$ is Abelian ( it is isomorphic to a group roots of unity under multiplication, which is certainly Abelian). Now the basic idea is to show that ${\rm Gal}(F/Q)/{\rm Gal}(L/Q)$ is isomorphic to ${\rm Gal}(F/L)$, and to use inductive assumption that ${\rm Gal}(L/Q)$ is solvable to conclude that ${\rm Gal}(F/Q)$ is solvable. However, there are quite a lot of technicalities needed to make the induction work, and it is usual to work with separable normal intermediate extensions, to ensure that they are left invariant under enough automorphisms ( for example, if you don't make some assumptions, it might be that an automorphism of $F$ which fixes $Q$ does not send elements of $L$ back into $L$: eg, think about the case that $F = Q[\omega,2^{\frac{1}{3}}]$ where $\omega$ is a complex cube root of unity and the cube root of $2$ is the real one. Take $L$ to be $F \cap \mathbb{R}$. Then any automorphism of $F$ which does not fix the real cube root of $2$ must send that cube root somewhere outside $L$).