Simple radical extension Let $F/K$ be a Galois (finite) extension with solvable group. Must $F$ be a simple radical extension of $K$? or at least have an intermediate field which is a simple radical extension?
If $F/K$ is finite and Galois then $F$ is simple. Can we ensure it is radical if the Galois group is solvable?
Simple radical extension A field extension $T/S$ is called a simple radical extension if $T=S(a)$ where $a^{n} \in T$ for some positive integer $n$.
 A: No. The following example answers all the questions.
(See Page 270 in Lang's algebra )
Let $f(x)=x^3-3x+1$. Then the Galois group of $f$ is cyclic of order
$3$. (so it not a simple radical extension. )
I.e., Let $\alpha$ be a root of $f$, $F=\mathbb{Q}[\alpha]$, then $F/\mathbb{Q}$ is Galois which Galois group is cyclic and of order $3$. And if $F/\mathbb{Q}$ is simple radical, say $F=\mathbb{\beta}$ with $\beta^3\in \mathbb{Q}$, then $\beta\omega\in F$ which implies $\omega\in F$(where $\omega$ is a primitive 3th root of 1), a contradiction. 
However, the Kummer theory will tell when an extension is cyclic. See following proposition.
PROPOSITION 5.25.
Let $F$ be a field containing a primitive $n$th root
of 1. Let $E=F[\alpha]$ where $\alpha^n\in F$ and no smaller power of
$\alpha$ is in $F$. Then $E/F$ is a Galois extension with cyclic
Galois group of order $n$. Conversely, if $E/F$ is a cyclic extension
of degree $n$, then $E=F[\alpha]$ for some $\alpha$ with $\alpha^n\in
F$.
(See Proposition 5.25 in J.S.Milne's note ``Fields and Galois Theory'' in
Version 4.30, April 15, 2012)
