How to compute that Socle of the full matrix rings

Question

Ler $R$ be a ring with identity. I believed that $M_n(\mbox{Soc}(R_R)) = \mbox{Soc}(M_n(R)_{M_n(R)})$ but I can not prove it. Any help would be helpful.

Answer 1

Let $N$ be a simple right $R$-module. We endow $N^{\oplus n}$ (written as the set of rows) with an obvious structure of a right $M_n(R)$-module. Using the matrix units $e_{ij}\in M_n(R)$ and the simplicity of $N$, it is easy to see that $N^{\oplus n}$ is a simple $M_n(R)$-module. Therefore, if $N$ is a simple submodule in $R_R$, then $M_n(N)$ is the direct sum of $n$ simple submodules of $M_n(R)_{M_n(R)}$ (every matrix is the sum of its rows). We have shown that $M_n\big(\text{Soc}(R_R)\big)\subset\text{Soc}\big(M_n(R)_{M_n(R)}\big)$.

Let $M$ be a simple submodule in $M_n(R)_{M_n(R)}$. For the opposite inclusion, it suffices to show that $e_{1i}Me_{j1}\subset\text{Soc}(R_R)$ for any $i,j$. We have an obvious epimorphism $M\to e_{1i}M$ of right $M_n(R)$-modules. If $e_{1i}Me_{j1}=0$, we are done. Otherwise, $N:=e_{1i}Me_{j1}\ne0$ and $e_{1j}M$ is a simple $M_n(R)$-module. Clearly, $N$ is a right $R$-module. Let $L\subset N$ be an $R$-submodule. It is easy to see that $e_{1j}M\supset Le_{11}+Le_{12}+\dots+Le_{1n}$ is a right $M_n(R)$-submodule. From the simplicity of $e_{1j}M$, we conclude that $N$ is a simple $R$-module.