The proofs I know of the fact $\operatorname{Br}(K) = H^2(G,K^*)$ ($G= \operatorname{Gal}(K^s/K))$ involve non-abelian group cohomology of $H^1(G,PGL_n(K))$. Are there any nice conceptual proofs which don't non-abelian cohomology in anyway?
Proof of $\operatorname{Br}(K) = H^2(G,K^*)$
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abstract-algebra
number-theory
group-cohomology
brauer-group
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0**The** conceptual proof is through nonabelian cohomology. CSAs are forms of matrix algebras, and they are classified by Galois cohomology, nonabelian because the automorphism group of matriz algebras is nonabelian. – 2017-01-17
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Yes. You can use crossed product algebras to do this. See the book by Pierce on Associative Algebras, for example.
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0(Most people would not call this conceptual, I guess. Emmy Noether would tell them to deal with it ;-) It is conceptual in that it is so concrete that you can understand what you are talking about, which is the sinequanon of concepts) – 2017-01-17