# Is every finite extension radical?

To me it is seemingly true, e.g. $\mathbb{Q}(\sqrt2))$,$\mathbb{Q}(\sqrt[3]2))$ are all examples. Also, $$\mathbb{Q}(\sqrt{2}+\sqrt{3})= \mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2})(\sqrt{3})$$ Then we can see that it is a radical extension $$\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{2})\subseteq(\mathbb{Q}(\sqrt{2}))(\sqrt{3})$$ Even though the numbers are coupled, we can decouple them.

Actually, it is not very easy (for me) to prove some extension is not radical. Because we have to run over all possible towers of fields. Is there any examples?

• The splitting field of a suitable (namely such that the Galois group is $A_5$) 5th degree polynomial is not solvabel by radicals. – Hagen von Eitzen Jun 20 '18 at 15:24
• – lhf Jun 20 '18 at 15:47