Equivalent definitions of the Brauer group

Question

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)

Answer 1

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 $A\sim \text{Mat}_n\ D\sim D\sim B$.