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Suppose $L|\mathbb{Q}$ is an abelian extension of number fields. Then, all the roots of unity are certainly contained in the maximal abelian extension $L^{ab}$ of $L$. Why is it obvious that if $L \ne \mathbb{Q}$ then $L^{ab} \ne \mathbb{Q}^{ab}$.

flag	asked Nov 19 at 4:48 LMN 1,665●10

What prevents $L$ from having an abelian extension which is not even galoisian over $\mathbf{Q}$, let alone abelian over $\mathbf{Q}$ ? Consider for example quadratic extensions of quadratic extensions. – Chandan Singh Dalawat Nov 19 at 5:08

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Thanks Chandan! I guess the argument you had in mind is the same as the one below. – LMN Nov 19 at 5:27

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Related post: mathoverflow.net/questions/85775/… – KConrad Nov 19 at 7:08

Thanks for the link Keith. I think the comments of Lavender Honey and SGP there give a reasonably complete answer. Let P a prime of $\mathbb{Q}$ split in $L$, and consider a quadratic extension $K$ of $L$ ramified at one prime $\mathcal{P}$ over $P$ but not at another $\mathcal{P}'|P$. Suppose $K \subset L*\mathbb{Q}(\mu_\infty)$, then $K$ is obtained by adjoining some element $\alpha \in \mathbb{Q}(\mu_\infty)$ to $L$. However, looking locally at primes of $L(\alpha)$ over $P$ we see that all the local extensions are the same (infact, they are all $Q_p(\alpha)$)$. But, this is diff. from K|L – LMN Nov 19 at 17:13

unknown (google) · Accepted Answer · 2012-11-19 17:34:42Z UTC

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~~Pick some $\gamma_1\in L\setminus\mathbb Q$ which is not a square.~~ Pick some $\gamma\in L^\times/(L^\times)^2$ which is not fixed by $\operatorname{Gal}(L/\mathbb Q)$ and fix a lift $\gamma_1\in L$. Let $\gamma_1,\ldots,\gamma_n$ be the orbit of $\gamma_1$ under $\operatorname{Gal}(L/\mathbb Q)$. Then it is an easy exercise in Galois theory to show that:

$L(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_n})/L$ is Galois with abelian Galois group $\subseteq(\mathbb Z/2\mathbb Z)^n$.
$L(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_n})/\mathbb Q$ is Galois with nonabelian Galois group $\subseteq\operatorname{Gal}(L/\mathbb Q)\ltimes(\mathbb Z/2\mathbb Z)^n$.

edited Nov 19 at 17:34

answered Nov 19 at 5:18

unknown (google)
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Thanks unknown(google)! – LMN Nov 19 at 5:26

In #2 is it really obvious that the extension is non-abelian? – LMN Nov 19 at 5:32

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Yes. It is easy to compute the action by conjugation of $\operatorname{Gal}(L/\mathbb Q)$ on $(\mathbb Z/2 \mathbb Z)^n$, and it is nontrivial. – Will Sawin Nov 19 at 5:43

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I think this argument has a little hole. Consider an A4-extension K/Q of the rationals, let L be its cyclic cubic subfield, and F/L a quadratic extensions of F/L. If the square root of c in L generates F, then your argument predicts that c has an orbit of length 2 under the action of the cyclic Galois group of order 3. For getting rid of this problem you probably have to consider orbits in L^* modulo squares. – Franz Lemmermeyer Nov 19 at 5:53

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@unknown: you're right, I did not see the inclusion signs. There is, by the way, a beautiful article by Waterhouse (The normal closures of certain Kummer extensions, Can. Math. Bull. 37, No.1, 133-139 (1994)) dealing with problems like this one. – Franz Lemmermeyer Nov 19 at 18:12

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