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. 



How to compute that Socle of the full matrix rings
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.


