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

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 $$\text{char}(K) \neq 0$$?

• Perhaps you mean solvable in the title? – Yanko Dec 28 '18 at 15:37
• @Yanko solvable = soluble (British vs American English, don't know which one is which) – Kenny Lau Dec 28 '18 at 15:47