# To prove any real Galois extension of $\mathbb{Q}$ of odd degree must not be radical

I'm trying to prove this using the statement:

Let $$K$$ be a field containing n distinct roots of unity. An extension of $$K$$ of degree $$n$$ is a radical extension generated by an $$n$$th root of an element of $$K$$ if and only if it is a Galois extension whose Galois group is a cyclic group of order $$n$$.

but I can't find the link between radical extension and odd degree of the Galois extension

• It seems like $K = \mathbb{Q}(\cos(2\pi /7))$ (which is Galois of degree 3 over $\mathbb{Q}$) is a counter example, but maybe I misunderstood the question. – user420261 Oct 21 '18 at 20:13
• @user420261 Is it radical extension ? – Xin Hu Oct 21 '18 at 22:13

Suppose $$L$$ is a non-trivial real radical Galois extension of $$\mathbb Q$$ of odd degree. (I added the condition "non-trivial" to rule out the case where $$L = \mathbb Q$$.)
Since $$L : \mathbb Q$$ is radical of degree $$> 1$$, there exists a tower of fields $$E_0 = \mathbb Q \subsetneq E_1 \subset \dots \subsetneq E_{n-1}\subsetneq E_n = L$$
with $$n \geq 1$$, where, for each $$i \in \{1, \dots, n \}$$, there exist $$\alpha_i \in E_{i}$$, $$\beta_i \in E_{i-1}$$, $$k_i \in \mathbb N$$ such that $$\alpha_i^{k_i} = \beta_i , \ \ \ E_i = E_{i-1}(\alpha_i).$$
Since $$[L : \mathbb Q]$$ is odd, $$[E_1 : \mathbb Q]$$ is also odd, by the tower law. As $$[E_1 : \mathbb Q] > 1$$ by assumption, this implies that $$[E_1 : \mathbb Q] \geq 3.$$ (The same applies to $$[E_i : E_{i-1}]$$ for each $$i \in \{1, \dots, n \}$$, but we won't need this.)
Let $$m_1(X) \in \mathbb Q[X]$$ be the minimal polynomial for $$\alpha_1$$ over $$\mathbb Q$$. Then $$m_1(X)$$ is a polynomial of degree $$\geq 3$$, irreducible over $$\mathbb Q$$, which divides the polynomial $$X^{k_1} - \beta_1$$.
Since $$L$$ is Galois over $$\mathbb Q$$, and since $$L$$ contains the root $$\alpha_1$$ of $$m_1(X)$$, the polynomial $$m_1(X)$$ splits completely in $$L$$. The roots of $$m_1(X)$$ in $$L$$ must also be roots of $$X^{k_1} - \beta_1$$, i.e. each root of $$m_1(X)$$ is equal to $$\alpha_1$$ multiplied by some $$k_1$$th root of unity. As $$m_1(X)$$ has at least three distinct roots in $$L$$, at least one of these roots must be non-real. This contradicts the assumption that $$L$$ is a real field.