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I know that for any finite Galois extension K of $\mathbb{Q}$ the group $H^2(Gal(K/\mathbb{Q}),K^*)$ is isomorphic to the Brauer group $Br(K/\mathbb{Q})$. The isomorphism goes as follows: to a 2-cocyle $a_{s,t}$ you associate the crossed product $(K,\mathbb{Q},a_{s,t})$ which is a central simple algebra over $\mathbb{Q}$.

I read that $H^2(Gal(\overline{\mathbb{Q}}/\mathbb{Q}),\overline{\mathbb{Q}}^*)$ is isomorphic to $Br(\mathbb{Q})$ but I can't manage to understand how the isomorphism works. My question is: for a 2-cocyle $a_{s,t}\in H^2(Gal(\overline{\mathbb{Q}}/\mathbb{Q}),\overline{\mathbb{Q}}^*)$ what is the division algebra corresponding in $Br(\mathbb{Q})$?

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    $\begingroup$ By definition, the cohomology group is a direct limit over all cohomology with respect to finite field extensions. So you can reduce tot the case you understand. $\endgroup$ Aug 26, 2021 at 17:14
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    $\begingroup$ The same what @DanielLoughran wrote, but in other words: Your Galois group is profinite, in particular, compact, and your cocycle is locally constant. It follows that it takes finitely many values. Thus it comes from a function in two variables $a^K_{s,t}$ on ${\rm Gal}(K/{\Bbb Q})$ for some finite Galois extension $K/{\Bbb Q}$, and it takes values in some finite Galois extension $K'/{\Bbb Q}$ containing $K$. Now you can inflate $a^K_{s,t}$ to a function $a^{K'}_{s,t}$ on ${\rm Gal}(K'/{\Bbb Q})$ with values in $(K')^\times$ and use it in order to construct a central simple algebra. $\endgroup$ Aug 26, 2021 at 18:19
  • $\begingroup$ That's way more clear now. Thanks! $\endgroup$
    – Jacques
    Aug 27, 2021 at 10:05

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