Radical field extension is soluble, **counter-example?**

Question

There is a theorem that says that if $E/F$ is a normal and radical extension with $\text{char}(K)=0$ then then $\text{Aut}(E/F)$ is soluble. But why do we need normal and $\text{char}(K)=0?$ What would happen if $E/F$ is not normal? For example $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not normal but its Galois group ${Id}$ is solvable? What would be a counter-example if $\text{char}(K) \neq 0$?

Answer 1

See e.g. here for a proof of the more general theorem that requires neither normality nor zero characteristic.