Version of Wedderburn's theorem on central simple algebras

Question

Suppose that $A$ be a central simple algebra over a field $k$. Then by Wedderburn's theorem $A\cong M_n(D)$ for some division $k$-algbera $D$. But to define the 'Brauer equivalence' I need that $D$ is finite dimensional and center of $D$ is $k$. How do I prove the previous statement? Thanks in advance.

Answer 1

Note that $\dim A = n^2 \dim D$, so $A$ is finite dimensional if and only if $D$ is finite dimensional. It shouldn't be hard for you to show that if $A\simeq M_n(D)$ then the center of $A$ (which is $k$) is isomorphic to the center of $M_n(D)$. But it is a rather easy exercise to show that the center or $M_n(R)$ for any ring $R$ is isomorphic to the center of $R$ itself. Hence $$k=Z(A)\simeq Z(M_n(D))\simeq Z(D)$$