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I would like to know about references for the following result (point 3):

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.

  1. $L/k$ is normal if and only if for every $\sigma \in G$ there is an $\alpha_\sigma \in K$ such that $\mu^{\sigma-1} = \alpha_\sigma^2$.
  2. 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 $$ \beta(\sigma,\tau) = \alpha_\sigma^\tau \alpha_\tau \alpha_{\sigma\tau}^{-1}. $$
  3. If $L/k$ is normal, then $[\beta]$ is the element of the second cohomology group attached to the group extension $$ 1 \rightarrow \mu_2 \rightarrow Gal(L/k) \rightarrow Gal(K/k) \rightarrow 1. $$
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1 Answer

I don't have a reference, but it does not seem too hard.

  1. 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.

  2. 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 $\sigma \in G$).

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?

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Just to expand on the answer above (with "references"): 1. This is Kummer Theory. 2/3. I would look particularly at Proposition 1.6.6 and Theorems 2.4.3 and 2.4.4 in Cohomology of Number Fields. –  Adam Topaz Jun 19 '10 at 20:48
    
Homology, when I apply $\sqrt{u}$ to your formula, I get $\beta_0=\alpha_\tau^\sigma\alpha_\sigma\alpha_{\sigma\tau}^{-1}$. I don't see how this differs from what Franz wanted by a coboundary. Also, it makes no difference here, but the formula for the class associated to an extension is usually given by $\beta_0\widetilde{\sigma\tau}=\tilde{\sigma} \tilde{\tau}$. –  Kevin Ventullo Jun 20 '10 at 20:36
    
You're right about the formula, I corrected it. The section defining $\alpha_{\sigma}$ doesn't have to be the same as the one used to define the 2-cocycle associated to the extension, that's why there is a coboundary involved. Anyway, your comment makes me realize there's some problem with left/right action, so thank you. –  Homology Jun 20 '10 at 22:19
    
@Adam: google tells me that these results are about the transgression map; how is this map related to the problem except that it lands in the group I'm interested in? –  Franz Lemmermeyer Jun 22 '10 at 15:21
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