TITLE: How to prove that a ring $R$ is Morita-equivalent to $M_n(R)$?
QUESTION [5 upvotes]: Let $R$ be a ring and $M_n(R)$ the ring of $n \times n$ matrices with elements in $R$. Let $\textrm{mod} R$ denote the category of left modules over $R$.
How to define a equivalent functor $F: \textrm{mod} R \to \textrm{mod} M_n(R)$?  
In Example, one defines $F(X) = X^n \in \textrm{mod} M_n(R)$ for a left $R$-module $X$. 
I do not know how to define the inverse functor $F^{-1}$. In case that $n=2$, $(M_1, M_2)^{T} \in \textrm{mod} M_2(R)$, then what is $F^{-1}((M_1, M_2)^{T})$?

REPLY [3 votes]: Given any $M_n(R)$-module $Y$, we can treat it like an $R$-module (since $R\hookrightarrow M_n(R)$ diagonally), in which case $Y=\bigoplus_i e_{ii}Y$ as $R$-modules. To see why $Y=\sum_i e_{ii}Y$, simply note $I_n=\sum_i e_{ii}$ and to see the sum is direct, suppose $y$ is in both $e_{ii}Y$ and $\bigoplus_{j\ne i}e_{jj}y$, then apply $e_{ii}$ to $y$ to get $y=0$ (since $e_{ii}$ acts as the identity on $e_{ii}Y$ and as $0$ on all other $e_{jj}Y$s). Moreover, observe $e_{ii}Y\cong e_{jj}Y$ as $R$-modules, with multiplications by $e_{ij}$ and $e_{ji}$ as mutually inverse homomorphisms (hence isomorphisms), from which we conclude $Y\cong X^n$ as $M_n(R)$-modules with $X=e_{ii}Y$.
So the inverse of $X\mapsto X^n$ is $Y\mapsto e_{11}Y$.