Galois groups via cohomology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T22:52:23Z http://mathoverflow.net/feeds/question/28772 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28772/galois-groups-via-cohomology Galois groups via cohomology Franz Lemmermeyer 2010-06-19T18:40:46Z 2010-06-20T22:21:52Z <p>I would like to know about references for the following result (point 3): </p> <p>Let $K/k$ be a normal extension (I am interested in number fields, but everything should work in fields of characteristic $\ne 2$) with Galois group $G$, and let $L = K(\sqrt{\mu}\,)$ be a quadratic extension.</p> <ol> <li>$L/k$ is normal if and only if for every $\sigma \in G$ there is an $\alpha_\sigma \in K$ such that $\mu^{\sigma-1} = \alpha_\sigma^2$.</li> <li>Let $L/k$ be normal. Then we can define an element $[\beta]$ in the second cohomology group $H^2(G,\mu_2)$ with values in $\mu_2 = {-1,+1}$ by setting $$\beta(\sigma,\tau) = \alpha_\sigma^\tau \alpha_\tau \alpha_{\sigma\tau}^{-1}.$$</li> <li>If $L/k$ is normal, then $[\beta]$ is the element of the second cohomology group attached to the group extension $$1 \rightarrow \mu_2 \rightarrow Gal(L/k) \rightarrow Gal(K/k) \rightarrow 1.$$</li> </ol> http://mathoverflow.net/questions/28772/galois-groups-via-cohomology/28778#28778 Answer by Homology for Galois groups via cohomology Homology 2010-06-19T20:31:38Z 2010-06-20T22:21:52Z <p>I don't have a reference, but it does not seem too hard.</p> <ol> <li><p>Assume $L/k$ normal, and take $\sigma \in G$, which can be extended to an element of $Gal(L/k)$. Then $\sigma(\sqrt{\mu})/\sqrt{\mu} \in L$, and if $\gamma$ is the nontrivial element of $Gal(L/K)$, $\sigma^{-1} \gamma \sigma$ is trivial on $K$, and being nontrivial on $L$ it has to be equal to $\gamma$. So $\gamma\left( \sigma(\sqrt{\mu})/\mu \right) = \sigma(\gamma(\sqrt{\mu}))/\gamma(\sqrt{\mu}) = \sigma(\sqrt{\mu})/\sqrt{\mu}$, so we can take $\alpha_{\sigma} = \sigma(\sqrt{\mu})/\sqrt{\mu}$, it is an element of $K$. Now for the other way, take $\tilde{\sigma} \in Gal(\bar{L}/k)$. Denote the restriction of $\tilde{\sigma}$ to $K$ by $\sigma$. Then $\sigma(\sqrt{\mu})/\sqrt{\mu}= \pm \alpha_{\sigma}$ (this equality is in $\bar{L}$), so $\sigma(\sqrt{\mu}) = \pm \alpha_{\sigma} \sqrt{\mu} \in L$, so $L$ is normal.</p></li> <li><p>Consider a set-theoretic section $\sigma \mapsto \tilde{\sigma}$ for the surjective morphism $Gal(L/k) \rightarrow G$. Then the (up to a coboundary) 2-cocycle $\beta_0$ associated to the group extension is given by the formula $\tilde{\sigma} \tilde{\tau} = \beta_0(\sigma,\tau) \widetilde{\sigma \tau}$. Evaluating at $\sqrt{\mu}$ gives the equality between $\beta$ and $\beta_0$, if for every $\sigma \in G$, $\alpha_{\sigma} = \tilde{\sigma}(\sqrt{\mu})/\sqrt{\mu}$. You can always choose your section so that it is the case (change of section = associating a sign to each $\sigma \in G$).</p></li> </ol> <p>EDIT: There's a left/right action problem, because $\beta_0(\sigma,\tau) = \sigma(\alpha_{\tau}) \alpha_{\sigma} \alpha_{\sigma \tau}^{-1}$ with what I wrote. I think it has to do with the fact that you use exponential notation, so somehow your action is on the right? Maybe you define $x^{\sigma} = \sigma^{-1}(x)$? Otherwise the definition of $\beta$ doesn't make it a 2-cocycle, with the definitions I know. Could you clarify?</p>