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 $n$.

• Can you explain what does a simple radical extension mean ? So that I can try to make some efforts. Thanks. – user119882 Mar 15 '14 at 6:34
• œuser119882: sure, done! – user10 Mar 15 '14 at 6:37
• Thanks! Now it is easy to construct an example that answer your first question negetively. Say $F=\mathbb{Q}[\sqrt{2}+\sqrt{3}]$. If $F=\mathbb{Q}[\alpha]$ such that $\alpha^4\in \mathbb{Q}$ then $\alpha\mathbf{i}\in F$, so $\mathbf{i}\in F$ but it is not. – user119882 Mar 15 '14 at 6:47

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.
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)