I've proven that the $K$ tensor product of two central simple $K$ algebras is itself central simple, and I've proven Wedderburn's theorem, but I now need to construct the Brauer group. I've been told that two algebras $A\cong M_n(D)$ and $B\cong M_m(D')$ are Brauer equivalent if $D\cong D'$. The operation on equivalence classes is defined as $[A][B]=[A\otimes B]$. Having shown closure, I need to show:

That the operation is independent of representative

That $[K]$ is the identity in the Brauer group. If the composition is independent of representative then I can use $k$ and all I have to show is that $A\otimes K\cong A$. I think an argument on dimensions does this.

That every equivalence class has an inverse. I have been told that the inverse of $A$ is $A^{op}$, so I'd have to show that $A\otimes A^{op}=M_n(K)$ for some $n$.

Most of the resources I've found online state these as fact and don't bother proving them. If anyone could refer me to a resource that covers the construction of the Brauer group in detail, I'd be very appreciative.


2 Answers 2


IMHO the authors of the Stacks project do a good job in proving certain theorems on CSA's and Brauer groups. The proofs are very short and to the point. Here's a link to the chapter on Brauer groups.

In your particular case though I think you would benefit from the relation $$A\otimes_K\operatorname{Mat}_n(K)\cong_K\operatorname{Mat}_n(A),$$ for a $K$-algebra $A$ (so CSA's over $K$ in particular). Together with Wedderburn's theorem, the proofs of your first two statements come rolling out prety easily.


You can use the book of Rotman "Advanced modern algebra" it covers the Brauer group of a field.

There is an article of Auslander and Goldman called "The Brauer group of a commutative ring," which generalizes the result of the book of Rotman to the context of commutative rings.


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