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

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

• For any root of unity $\zeta$, the extension $K(\zeta)/\Bbb Q$ is the compositum of $\Bbb Q(\zeta)/\Bbb Q$ and $K/\Bbb Q$. Since both are abelian extensions, their compositum is again an abelian extension and hence solvable. Is this what you want? Jun 19, 2022 at 18:19
• Yeah that does the trick . How can I prove that $K(\zeta)$ is abelian since $\mathbb{Q}(\zeta)$ and $K$ are abelian ? Does it have something to do with the Chinese Remainder Theorem? I understand it by sense,but mathematically is there a straightforward way to prove it ? Thank you for the reply!
– GGG
Jun 19, 2022 at 18:24

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.
• Are you sure that $Gal(K,\mathbb{Q})$ is not isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$ ?Also is the fact that the Galois group of $K(\zeta)$ abelian trivial using that $Gal(K,\mathbb{Q})$ and $Gal(\mathbb{Q}(\zeta),Q)$ are abelian ?Thank you for your time !
• Just note that $[\mathbb{Q}(\sqrt p, \sqrt q):\mathbb{Q}]=4$ if $gcd(p,q)=1$. Jun 19, 2022 at 19:03
• Ups! Sorry i did not read it well. My apologies! Of course the group is $\mathbb{Z}_2\times\mathbb{Z}_2$ Jun 19, 2022 at 19:06
• In general it can be extended to a numerable family of primes $\mathbb{Q}(\sqrt p_1,....,\sqrt p_n)=2^n$ Jun 19, 2022 at 19:08