If $M$ is a simple $R$-module, and an $F$-space, why does $End_F(M)\cong M^{\oplus\dim_F(M)}$? Suppose a ring $R$ is an $F$-algebra for $F$ a field, and $M$ is a simple $R$-module and a finite dimensional $F$-vector space. 
We can endow $\operatorname{End}_F(M)$ with an $R$-module structure by $r\cdot\varphi(m)=r\varphi(m)$. Then why does $\operatorname{End}_F(M)$ decompose as $M^{\oplus\dim_F(M)}$ as an $R$-module? Is it some sort of Wedderburn type application?
This is a line in Proposition 2 in Daniel Bump's notes on Hecke Algebras.
 A: Let $R$ be an $k$-algebra and $M$ a simple $R$-module of dimension $n$, such that $M$ is also a finite vector space over $k$. Then the space 
$$
Hom_{k}(M,M)\cong Hom(k^{n},k^{n})
$$
has dimension $n^2$ over $k$ as a $k$-vector space, thus the dimension of the result above. 
To see that $Hom_{k}(M,M)\cong M^{n}$ as an $R$-module, it suffice to construction explicit isomorphism maps by choosing a basis $e_{i}$ of $M$. We know that the homomorphisms generated by
$$
f_i:e_{i}\rightarrow \sum a_{ij} e_{j}
$$
is an $R$-module by $rf_i(e_{i})=\sum a_{ij}re_{j}$. And the image is an $R$-submodule of $M$, which must be equal to $M$ since $M$ is simple. So $\langle Rf_{i}\rangle\cong M$. We also know that all maps from $Hom_{k}(M,M)$ can be viewed as a direct sum of type $\bigoplus \langle Rf_{i}\rangle$. Thus you have the isomorphism to be $M^{n}$. 
I do not know if this is related to Wedderburn's theorem about division algebras, but I think if you think about $\mathbb{C}G$ decomposition into irreducible pieces, then the statement is quite similar. Not sure if helps. 
