Idempotent elements in matrix ring Is there a relation between the idempotent elements of a ring $R$ and those of $M_{n}(R)$ - the ring of $n \times n$ matrices over $R$?
 A: Any idempotent $e$ of $R$ induces an idempotent $E=\mathop{diag}(e,\ldots,e)$ of $M_n(R)$. In fact, if $e_i$ are idempotents in $R$, then $E=\mathop{diag}(e_1,\ldots,e_n)$ is an idempotent of $M_n(R)$.
Conversely, if $R$ is nice enough, an idempotent $E$ in $M_n(R)$ can be diagonalized to $E=U^{-1} \cdot \mathop{diag}(e_1,\ldots,e_n) \cdot U$ for some $U \in M_n(R)$ and idempotents $e_i$ in $R$.
Of course, this relies crucially on $R$ being nice enough. One sufficient "nicety" condition is that $R$ is an AW*-algebra; see for example this paper. 
A: Yes. You can see the relationship easily in the following way. Suppose the ring $R=R_1\times R_2$ so that there are two primitive idempotents $e_1$ and $e_2$ with $1=e_1 +e_2$. Next note that $M_n(R)\approxeq Hom_R(R^n,R^n)$. Then $M_n(R)\approxeq Hom_{R_1\times R_2}(\(R_1\times R_2\)^n,\(R_1\times R_2\)^n)\approxeq Hom_{R_1}(R_1^n,R_1^n)\times Hom_{R_2}(R_2^n,R_2^n) $ since $e_1 \centerdot  Hom_{R_2}(R_2^n,R_2^n)=0$ and similarly for $e_2$. Thus idempotents are calculated relative to each factor in the ring decomposition.
