A radical extension with a non-radical subextension

Question

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?

Answer 1

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.

Answer 2

You can prove by contradiction:

It is not too hard to verify that $\mathbb{Q}(\zeta_7+\zeta_7^{-1})$ is fixed by the automorphism $\zeta_7\mapsto \zeta_7^6$, which has order 2 since the Galois group is isomorphic to $(\mathbb{Z}/7\mathbb{Z})^\times$. Therefore $[\mathbb{Q}(\zeta_7):\mathbb{Q}(\zeta_7+\zeta_7^{-1})]=2$ and $\mathbb{Q}(\zeta_7+\zeta_7^{-1})/\mathbb{Q}$ is Galois of degree 3.

Suppose $\mathbb{Q}(\zeta_7+\zeta_7^{-1})=\mathbb{Q}(\alpha)$ is radical. Then there exists positive integer $n$ such that $\alpha^n\in \mathbb{Q}$. Then the minimal polynomial of $\alpha$ over $\mathbb{Q}$ divides the polynomial $x^n-\alpha^n$ in $\mathbb{Q}[x]$. Since the minimal polynomial is of degree 3, it must be of the form $$(x-\alpha)(x-\alpha\zeta_n^a)(x-\alpha\zeta_n^b).$$ The constant coefficient is given by $-\alpha^3\zeta_n^{a+b}\in \mathbb{Q}$. Since $\zeta_7+\zeta_7^{-1}$ is real, $\alpha$ must be real. Hence $\zeta_n^{a+b}$ is real, which implies $\zeta_n^{a+b}=\pm 1$. Hence $\alpha^3\in \mathbb{Q}$. Hence we see that $$\mathbb{Q}(\zeta_7+\zeta_7^{-1})=\mathbb{Q}(\sqrt[3]{\gamma})$$ where $\gamma=\alpha^3$. But since the minimal polynomial of $\sqrt[3]{\gamma}$ over $\mathbb{Q}$ is $x^3-\gamma$ which has a root $\sqrt[3]{\gamma}\zeta_3$ in the complex field, so it is not contained in $\mathbb{Q}(\sqrt[3]{\gamma})$ which is real. Therefore $\mathbb{Q}(\sqrt[3]{\gamma})$ is not normal, hence it cannot be Galois, a contradiction.