For a Galois Theory class I've been asked to find a radical extension with a non-radical subextension (all over $\mathbb{Q}$). So, I'm looking at the splitting field of $x^7 - 1$, namely $\mathbb{Q}(\zeta_7)$. This is obviously a radical extension, and it has a normal subfield $L = \mathbb{Q}(\zeta_7 + \zeta_7^{-1}) = \mathbb{Q}(\alpha)$. I'd like to show that $L$ is not radical.
Suppose it were. Then $L = \mathbb{Q}(l)$ with $l^3 = q \in \mathbb{Q}$ hence because $\{1, \alpha, \alpha^2\}$ form a basis of $L$ we have $a + b\alpha + c\alpha^2 = l$, and cubing we obtain that $\sum_{i = 1}^6 a_i\zeta_7^i \in \mathbb{Q}$ with $a_i \in \mathbb{Q}$. The exact value of that expansion is not fun. However I'd like to say the following: $a_1 = a_6, a_2 = a_5, a_3 = a_4$. This statement would imply that $\zeta_7^3 + \zeta_7^4 \in \mathbb{Q}$ which I can show to be false. The reasoning behind my statement is as follows: we should have that the imaginary part of $\sum_{i = 1}^6 a_i\zeta_7^i$ is zero and the only powers of $\zeta_7$ that could cancel would be those opposite over the $x$-axis. Does this many any sense?
Julien, the easy way to show $L$ in your notation is not radical is to notice that $L = \mathbb{Q}(\zeta_7) \cap \mathbb{R}$ since $\zeta_7 + \zeta_7^{-1} \in \mathbb{R}$. So if $L$ were radical, $\cos(2\pi/7)$ would be expressible in terms of real radicals. $\cos(2\pi/7)$ is a root of $8 y^3+4 y^2-4 y-1 = 0$. But by casus irreducibilis, this cubic is unsolvable by radicals over $\mathbb{R}$; you need radicals over $\mathbb{C}$ even though the expressions are ultimately real-valued.