Is there an example of a field extension that is solvable, but not radical? I also would like an example of an extension that is radical but not solvable. Been trying to come up with these examples and have been making little to no progress.

An extension $K/k$ is said to be radical if there exists a tower of sub extensions $$k_0 \subseteq k_1 \subseteq ... \subseteq k_n \subseteq K,$$ where $k_{j+1} = k_j(a_j)$ for some $a_j \in \mathbb{C}$ with $a_j^{n} \in k_i$ for some $n \geq 1$.

An extension $L/k$ is said to be solvable if there is a radical extension $K/k$ with $L \subseteq K$.

• So from the definition it is clear that any radical extension is solvable since it is contained in itself. – Tobias Kildetoft Jun 8 '16 at 8:09
• – user319128 Jun 8 '16 at 8:30
• @elliot I don't see how the other question is relevant here. Those extensions are solvable and radical. – mercio Jun 8 '16 at 8:34

The answer is yes : there do exist solvable extensions which are not radical. For example $\mathbf{Q}(\cos \frac{2\pi}{7})/\mathbf{Q}$. See there.

• By the way, if you do not trust the link, you can also refer to David Cox, Galois Theory, example 8.2.3 p197. – BrL Jun 8 '16 at 8:29