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

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    $\begingroup$ 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? $\endgroup$
    – WhatsUp
    Jun 19, 2022 at 18:19
  • $\begingroup$ 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! $\endgroup$
    – GGG
    Jun 19, 2022 at 18:24

1 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.

  • $\begingroup$ 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 ! $\endgroup$
    – GGG
    Jun 19, 2022 at 18:42
  • $\begingroup$ Yeah im sure, note that the field extension is generated by 3 elements, every K-automorphism is defined by the images of the generators, and their images must be roots of their minimal polynomial, so we have 8 different automorphisms. By the second part i think that yes, its trivial $\endgroup$ Jun 19, 2022 at 18:45
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    $\begingroup$ Just note that $[\mathbb{Q}(\sqrt p, \sqrt q):\mathbb{Q}]=4$ if $gcd(p,q)=1$. $\endgroup$ Jun 19, 2022 at 19:03
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    $\begingroup$ Ups! Sorry i did not read it well. My apologies! Of course the group is $\mathbb{Z}_2\times\mathbb{Z}_2$ $\endgroup$ Jun 19, 2022 at 19:06
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    $\begingroup$ In general it can be extended to a numerable family of primes $\mathbb{Q}(\sqrt p_1,....,\sqrt p_n)=2^n$ $\endgroup$ Jun 19, 2022 at 19:08

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