Show that for every $n$-primary root of unity that ${\rm Gal}(K(\zeta),\Bbb Q)$ is solvable.

Question

Let $K=\mathbb{Q}(\sqrt{2},i\sqrt{5})$ and $G={\rm Gal}(K,\mathbb{Q})$. Show that for every $n$-primary root of unity that ${\rm Gal}(K(\zeta),\mathbb{Q})$ is solvable.

My approach untill I got stuck:

I found that $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2$. That means that every subgroup of $G$ is normal under $G$ (and of course every subfield of $K$ is a splitting field over $\mathbb{Q}$, including $K$.

We know that $K$ is a root extension since the series:$\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{2})\subseteq \mathbb{Q}(\sqrt{2},i\sqrt{5})=K$ include every attribute needed.

Also $K(\zeta)$ is for sure a splitting field over $\mathbb{Q}$ since $\mathbb{Q}(\zeta)$ is a splitting field itself.

From the above we can conclude that $G$ is solvable. Yet the question implies that ${\rm Gal}(K(\zeta),Q)$. Well since $K(\zeta)$ is a splitting field over $\mathbb{Q}$ then $G$ is a normal subgroup of ${\rm Gal}(K(\zeta),Q)$. We know for a group to be solvable, other than every subgroup of a subgroup etc being normal under each other, we need ${\rm Gal}(K(\zeta),Q)/G$ be abelian or because we have finite groups, we $|{\rm Gal}(K(\zeta),Q)/G|=p$ (prime number).

From the fundamenta theorem of Galois theory we know that ${\rm Gal}(K(\zeta),Q)/G \simeq{\rm Gal}(K(\zeta),K)$. And thus we need $|{\rm Gal}(K(\zeta),K)|=p$
Since $K(\zeta)$ is a splitting field $\mathbb{Q}$, then naturally $K(\zeta)$ is a splitting field over $K$ and for that reason $|{\rm Gal}(K(\zeta),K)|=[K(\zeta):K]\leq \phi(n).$

This is where I don't know how to follow. I'm aware of a lemma saying:

Let $F$ be a field that contains $n$-primary root of unity and $a \in \mathbb{C}$, for which $a^n \in F$. Then ${\rm Gal}(F(a),F)$ is solvable.

But was unable to use it somehow.

Any help would be appreciated. Thank you in advance

Answer 1

Note that the galois group of $K|\mathbb{Q}$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ because the only operation you can do is the "conjugate", that's it $$\sqrt2\rightarrow -\sqrt2\\ \sqrt5i\rightarrow -\sqrt5i\\$$ So it's an abelian extension. Also the Galois group of $\mathbb{Q}(\zeta)|\mathbb{Q}$ where $\zeta$ is any nth-root of the unity is isomorphic to $\mathbb{Z}_{n}^{*}$, which is also abelian. So the galois group of $K(\zeta)|\mathbb{Q}$ must be abelian (1). It could be that $K(\zeta)=K$, for example if $\zeta=e^{2\pi i/8}$, because $\mathbb{Q}(\zeta)=\mathbb{Q}(\sqrt2,i)$

To prove (1) just see that: Let $ψ:Gal(EL/\mathbb{Q})→Gal(E/\mathbb{Q})×Gal(L/\mathbb{Q})$ be a map such that $ψ(σ)=(σ|E,σ|L)$ for every $\sigma$ automorphism of the composition. $ψ$ is clearly a group homomorphism and it's easy to see that $ψ$ is injective. Hence Gal(KL/k) is abelian.