Theorem 59.61.6. Let $K$ be a field with separable algebraic closure $K^{sep}$. The map $\delta : \text{Br}(K) \to H^2(\text{Gal}(K^{sep}/K), (K^{sep})^*)$ defined above is a group isomorphism.

**Sketch of proof.**
To prove that $\delta $ defines a group homomorphism, i.e., that $\delta (A \otimes _ K B) = \delta (A) + \delta (B)$, one computes directly with cocycles.

Injectivity of $\delta $. In the abelian case ($d = 1$), one has the identification

the latter of which is trivial by fpqc descent. If this were true in the non-abelian case, this would readily imply injectivity of $\delta $. (See [SGA4.5].) Rather, to prove this, one can reinterpret $\delta ([A])$ as the obstruction to the existence of a $K$-vector space $V$ with a left $A$-module structure and such that $\dim _ K V = \deg A$. In the case where $V$ exists, one has $A \cong \text{End}_ K(V)$.

For surjectivity, pick a cohomology class $\xi \in H^2(\text{Gal}(K^{sep}/K), (K^{sep})^*)$, then there exists a finite Galois extension $K^{sep}/K'/K$ such that $\xi $ is the image of some $\xi ' \in H^2(\text{Gal}(K'|K), (K')^*)$. Then write down an explicit central simple algebra over $K$ using the data $K', \xi '$. $\square$

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