most recent 30 from http://mathoverflow.net 2013-05-25T19:08:16Z http://mathoverflow.net/feeds/question/113794 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/113794/non-cyclotomic-abelian-extensions LMN 2012-11-19T04:48:27Z 2012-11-19T17:34:42Z

<p>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 $L^{ab} \ne \mathbb{Q}^{ab}$.</p>

http://mathoverflow.net/questions/113794/non-cyclotomic-abelian-extensions/113796#113796 unknown (google) 2012-11-19T05:18:00Z 2012-11-19T17:34:42Z

<p><strike>Pick some $\gamma_1\in L\setminus\mathbb Q$ which is not a square.</strike> 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:</p> <ol> <li>$L(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_n})/L$ is Galois with abelian Galois group $\subseteq(\mathbb Z/2\mathbb Z)^n$.</li> <li>$L(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_n})/\mathbb Q$ is Galois with nonabelian Galois group $\subseteq\operatorname{Gal}(L/\mathbb Q)\ltimes(\mathbb Z/2\mathbb Z)^n$.</li> </ol>