Morita equivalence and Brauer equivalence

Question

Let $k$ be a field and $A,B$ be two (finite-dimensional) central simple $k$-algebras. We usually say that $A$ and $B$ are Brauer equivalent (or similar) if their underlying division algebras (given by Wedderburn's theorem) are isomorphic or, equivalently, if $M_n(A)\cong M_m(B)$ for some integers $n,m$.

The Wikipedia article about the Brauer group says that this is equivalent to demanding that the categories $A$-Mod and $B$-Mod of left modules are equivalent. However, I couldn't prove this fact neither did I found it in any books.

(PS: I know absolutely nothing about Morita equivalence, so I would prefer a direct answer to an answer that uses a basic fact from the theory of Morita equivalence.)

Why is it true?

Answer 1

If $A$ and $B$ are Morita equivalent, then there exists an $n$ and a full idempotent $e\in M_n(A)$ such that $B\cong eM_n(A)e$.

Now if $A\cong M_{n'}(D)$ using the Artin-Wedderburn theorem, rewrite the right hand side via an isomorphism to be $fM_{n'}(D)f$.

There must exist some unit $u$ diagonalizing $f$ to a matrix $\hat f$ which is some number of $1$'s on the diagonal followed by zeros on the diagonal (this just amounts to selecting an eigenbasis for the transformation, putting the nonzero eigenvectors up front.) Conjugation by $u$ makes the right hand side isomorphic to $ufu^{-1}M_{n'}(D)u^{-1}ufu^{-1}=\hat fM_{n'}(D)\hat f$, but as you can see the diagonalized idempotent $\hat f$ simply selects out some upper left corner of $M_{n'}(D)$, which is apparently of the form $M_m(D)$ for some $m < n'$. Going back through the isomorphisms, you have $B\cong M_m(D)$. From there it is easy that $M_{m}(A)\cong M_{n'}(B)$.

The other direction has already been discussed above: if $M_n(A)\cong M_m(B)$, then they are a priori Morita equivalent, and it is well known that $R$ is equivalent to $M_n(R)$ for any ring, so by transitivity of equivalence you get that $A$ is Morita equivalant to $B$ if $A$ is Brauer equivalent to $B$.

Answer 2

All you need to know is that $M_n(R)$ and $R$ are Morita equivalent. What this says explicitly is that the category of modules over $M_n(R)$ and over $R$ are equivalent, and this equivalence can be written down explicitly: it sends

$$\text{Mod}(R) \ni M \mapsto M \otimes_R R^n \cong M^n \in \text{Mod}(M_n(R)).$$

Actually we only need that this is an equivalence for $R$ a central simple algebra, but it's true in general. Now:

$\Rightarrow$: if two central simple algebras $A, B$ are Brauer equivalent then $A \cong M_n(D)$ and $B \cong M_m(D)$ for some central division algebra $D$, and $\text{Mod}(A) \cong \text{Mod}(D) \cong \text{Mod}(B)$.

$\Leftarrow$: if $A \cong M_n(D)$ and $B \cong M_n(E)$ where $D, E$ are central division algebras, then $\text{Mod}(A) \cong \text{Mod}(D)$ and $\text{Mod}(B) \cong \text{Mod}(E)$. A division algebra can be recovered from its category of modules: it's the algebra of endomorphisms of the unique simple module. So $\text{Mod}(D) \cong \text{Mod}(E)$ implies $D \cong E$.

(We need the isomorphism to be $k$-linear to get that $D \cong E$ over $k$.)

The last observation can be extended to give a module-theoretic characterization of module categories over central simple algebras, or equivalently of module categories over division rings: they are precisely the semisimple module categories with a unique simple object.