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I am trying to figure out a good way to see the equivalence in definitions of a brauer group of a field. The two are usually offered: either the brauer group has as elements, central simple algebras modulo morita equivalence, or it's offered as central simple algebras modulo the relation that we can tensor A, B with some matrix ring over the field and have the resulting algebras be isomorphic.

How are these two the same? Thanks! (The hard direction is Morita equivalent implies second definition)

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For any ring $R$ and any $n\geq 1$, $R$ is Morita equivalent to $\text{Mat}_n\ R$ through the $R-\text{Mat}_n\ R$-bimodule $R^{\oplus n}$. In particular, the relation $\sim$ on central simple algebras induced by (i) $A\sim B$ for $A\cong B$, and (ii) $A\sim\text{Mat}_n\ A$ for $n\geq 1$, is coarser than the relation $\sim_{\text{Mor}}$ of Morita equivalence.

Conversely, suppose $A$ and $B=\text{Mat}_m\ D$ are CSA, with $D$ a division algebra, and assume $A$ and $B$ are Morita equivalent. Then, since $B\sim_{\text{Mor}} D$, also $A$ and $D$ are Morita equivalent. An equivalence ${\mathscr F}: A\text{-Mod}\cong D\text{-Mod}$ sends the (Noetherian) regular left $A$-module $_A A$ to a $D$-module of the form $_DD^{\oplus n}$ for some $n$ (all $D$-modules have a basis), hence $$A^{\text{op}}\cong\text{Hom}_{A\text{-Mod}}(_AA,\ _AA)\xrightarrow{\mathscr F}\text{Hom}_{D\text{-Mod}}(_DD^{\oplus n},\ _DD^{\oplus n})\\\quad\cong\text{Mat}_n(\text{Hom}_{D\text{-Mod}}(_DD,_DD))\cong\text{Mat}_n(D^{\text{op}})\cong\text{Mat}_n(D)^{\text{op}},$$ the last isomorphism by virtue of the transpose map. Hence $A\cong\text{Mat}_n\ D$, so $A\sim \text{Mat}_n\ D\sim D\sim B$.

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  • $\begingroup$ Thank you, I worked it out with a friend last night and arrived at the same proof, using the characterization that if R and R' are morita equivalent, then there is a f.g. projective P that is a (R, R')-bimodule such that R is End_R'( P ) $\endgroup$
    – Elliot
    Dec 4, 2014 at 12:46

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