Prove that $\sqrt[5]{5} \notin \mathbb{Q}(e^\frac{2 \pi i}{25})$ I need to prove that $\sqrt[5]{5} \notin \mathbb{Q}(e^\frac{2 \pi i}{25})$. Let's call $\xi = e^\frac{2 \pi i}{25}$. I think I have found a solution using the field trace operator but I am not quite sure it is correct.
Let's call $T$ the field trace operator over $\mathbb{Q}$ and suppose $\sqrt[5]{5} \in \mathbb{Q}(\xi)$. Then there would exist $\alpha_i$ ($i \neq 5, 10, 15, 20$) such that:
$$ \sqrt[5]{5} = \alpha_0 + \alpha_1 \xi + \alpha_2 \xi^2 + \dots \alpha_{24} \xi^{24}$$
The trace of $\sqrt[5]{5}$ is $\sqrt{5} (1 + \xi^5 + \xi^{10} + \xi^{15} + \xi^{20}) = 0$ since it is the sum of the 5th roots of unity. On the other hand all the $\xi^j$ with $j \neq 5,10,15,20$ are conjugates between themselves so the trace of any of them is:
$$ T(\xi^j) = (\xi+\xi^2 + \dots \xi^{24}) = -(1+\xi^5+\xi^{10}+\xi^{15}+\xi^{20})=0$$
so taking traces at the original expression we find that $\alpha_0 = 0$.
Usually I multiply by some suitable expression at both sides and repeat this process so I can remove enough coefficients. In this case, however once i reach:
$$ \sqrt[5]{5} = \xi (\alpha_1 + \alpha_2 \xi + \dots + \alpha_{24} \xi^{23})$$
My only idea is to take norms and deduce that:
$$ \sqrt[5]{5} = |\alpha_1 + \alpha_2 \xi + \dots + \alpha_{24} \xi^{23}|$$
and iterate the process above to remove all coeficients (which would lead to an obvious contradiction). Would that be ok?
 A: On one side, ${X^5} - 5$ is an Eisenstein polynomial, hence is irreducible over Q, so K = Q($\sqrt [5]5$) has degree 5 over Q; but K /Q is not galois because K does not contain any 5th root of unity. On the other side, because the given cyclotomic field L/Q is an abelian extension, all its subextensions are galois over Q, hence K is not contained in L .
A: What you mean by "taking the norm" looks like taking the complex absolute value. This however doesn't allow you to take trace anymore.
Usually this can be proved with Galois theory. A sketch of the procedure:

*

*Let $K$ be the field $\Bbb Q(\sqrt[5]5, \zeta_5)$ where $\zeta_5$ is a primitive fifth root of unity. Show that $K/\Bbb Q$ is Galois.


*Show that the two subextensions $\Bbb Q(\sqrt[5]5)/\Bbb Q$ and $\Bbb Q(\zeta_5)/\Bbb Q$ of $K/\Bbb Q$ have degrees $5$ and $4$ respectively.


*Show that $K/\Bbb Q$ has degree $20$ and deduce that its Galois group is isomorphic to a semidirect product $\Bbb Z/5 \Bbb Z \rtimes \Bbb Z/4\Bbb Z$ which is non-commutative.


*Conclude that $\sqrt[5]5$ cannot be contained in $\Bbb Q(\zeta_{25})$ because $\operatorname{Gal}(\Bbb Q(\zeta_{25})/\Bbb Q)$ is commutative.
