# Construction of the Brauer Group

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.

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.