# Why is $\mathbb{Q}(\zeta_7)$ not a radical extension?

My script states that there are $$2$$ non-trivial subfields of $$\mathbb{Q}(\zeta_7)$$, where $$\zeta_7$$ is a $$7$$-th primitive root of unity. These subfields are $$\mathbb{Q}(\zeta_7 +\overline \zeta_7)$$ and $$\mathbb{Q}(\sqrt{-7})$$.

Definition of radical extension I am using:

An extension $$L/K$$ is a radical extension, if $$\exists\gamma_1,...,\gamma_r\in L, n_1,...,n_r\in\mathbb{N}$$, such that $$L=K(\gamma_1,...,\gamma_r)$$ and $$\gamma_i^{n_i}\in K(\gamma_1,...,\gamma_{i-1})\ \forall i$$

Ok, so using the definition on $$\mathbb{Q}(\zeta_7)$$ I obviously have the first condition satisfied and the second with $$\zeta_7^7=1\in\mathbb{Q}$$.

I am not sure what I don't understand at this point. Also I am unsure of why the subfields are even mentioned. In my script the lattice of the subfields is drawn at it says (translated from german): "Not a radical extension because $$\zeta_7$$ is only present at the the top of the lattice". I have no idea what this means.

• @reuns I am sorry, I cannot follow your comment. Do you mean "...hence so is $\mathbb{Q}(\zeta_7,\zeta_3)/\mathbb{Q}(\zeta_3)$? I am unfamiliar with Kummer-Theory, I am hearing a basic algebra class. Why do I need $\zeta_3$? I just want a simple radical extension $\mathbb{Q}(\zeta_7)/\mathbb{Q}$. – EpsilonDelta Feb 25 at 12:37
• $\mathbb{Q}(\zeta_7)/\mathbb{Q}$ is a $7$-radical extension needing a $7$-root of unity, $\mathbb{Q}(\zeta_7,\zeta_3)/\mathbb{Q}$ is a $2,3$-radical extension not needing any root of unity, and none of them is constructible (ie. a $2$-radical extension). – reuns Feb 25 at 12:39
• Every cyclotomic field is both Galois and radical over $\Bbb Q$, see here. – Dietrich Burde Feb 25 at 12:48