Brauer group

Remarks

The operation is well defined, and $\mathrm{Br}(K)$ is an abelian group with identity $[K]$ and inverse $[A]^{-1} = [A^{\mathrm{op}}]$, the opposite algebra; this rests on $A \otimes_K A^{\mathrm{op}} \cong M_d(K)$ with $d = \dim_K A$. By Wedderburn's theorem each central simple algebra is $M_n(D)$ for a central division algebra $D$ unique up to isomorphism, so each class contains exactly one such $D$. Example: $\mathrm{Br}(\R)$ has two elements, $[\R]$ and $[\mathbb{H}]$, by the Frobenius theorem. Example: $\mathrm{Br}(K) = 0$ for $K$ algebraically closed and for $K$ finite. Non-example: the $K$-algebra $K \times K$ has no class in $\mathrm{Br}(K)$, since its center is $K \times K$, not $K$, and $K \times \set{0}$ is a two-sided ideal. There is an isomorphism $\mathrm{Br}(K) \cong H^2(K, (K^s)^\times)$ with Galois cohomology, where $K^s$ is a separable closure of $K$.