Let $A$ be a central, simple $K$-algebra and let $P$ be a finitely generated projective $A$-module.
I want to show that the endomorphism ring $\operatorname{End}_A(P)$ is also a central, simple $K$-algebra.
I know that $P$ is a semisimple $A$-module since $A$ is a simple, Artinian ring and I tried using Wedderburn's theorem on $\operatorname{End}_A(A^n)$ where $P\oplus Q\cong A^n$ but I couldn't get very far.
Would someone be able to provide a proof of this?
Many thanks.
Note that since $A$ is not only semi-simple but also simple, there is a unique simple $A$-module. Call it $I$. Then any finitely generated $A$-module is isomorphic to $I^n$ for some $n$ (in particular they are all projective).
Then $\operatorname{End}_A(P) \simeq \operatorname{End}_A(I^n) = M_n(\operatorname{End}_A(I))$ (the last equality is valid for any module and any ring).
Now for any $K$-algebra $A$ and any simple module $I$, $\operatorname{End}_A(I)$ is a division algebra ; call it $D$.So $\operatorname{End}_A(P)\simeq M_n(D)$.
The fact that $D$ is central can be seen by taking $P=A$ : if $A\simeq I^r$, $A^{op} = \operatorname{End}_A(A) = M_r(D)$ is central and $Z\left( M_r(D)\right) = Z(D)$, so $D$ is central.
Then $\operatorname{End}_A(P)\simeq M_n(D)$ with $D$ a central division algebra over $K$ : it's easy to check that this implies $\operatorname{End}_A(P)$ is central simple.