It is known that $M_n(R/J(R))\simeq M_n(R)/M_n(J(R))=M_n(R)/J(M_n(R))$. I tried to prove the same "isomorphism" replacing $J(R)$ by $Soc(R_R)$, where $J(R)$ and $Soc(R_R)$ stand for the Jacobson radical and the right socle of the ring $R$ (as a right $R$-module), but to no avail. Any help or suggestion would be welcome!
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You can use the Morita equivalence between the categories $Mod(R)$ and $Mod(M_n(R))$. The equiavlence is given by tensoring a right $R$ module with the bimodule $R^n$ over $R$ (in the other direction we also tensor with $R^n$, but over $M_n(R)$). Under this equivalence, the $M_n(R)$-module $M_n(R)$ corresponds to the $R$-module $R^n$. Since being simple is a categorical property, and subobjects can also be defined categorically, we get that $Soc(R^n)=Soc(R)^n$, and this corresponds to the $M_n(R)$-submodule $Soc(R)^n\otimes_R R^n=Soc(R)^{n^2}$. In other words, we get that the socle of $M_n(R)$ as an $M_n(R)$-module is exactly the matrices whose entries are in $Soc(R)$.
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$\begingroup$ Thanks for the answer! Now, how do we conclude the isomorphism $M_n(R/Soc(R_R))\simeq M_n(R)/M_n(Soc(R_R))$? $\endgroup$ Commented Jan 22, 2017 at 15:09
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1$\begingroup$ Well, for any $R$-submodule $A\subseteq R$ you have an isomorphism of $M_n(R)$ modules: $M_n(R/A)\cong M_n(R)/M_n(A)$ given in the obvious way. $\endgroup$ Commented Jan 22, 2017 at 15:30
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$\begingroup$ Meier: By Morita equivalence, how is it deduced that the ring $M_n(R/Soc(R_R))$ is Boolean if the ring $R/Soc(R_R)$ is Boolean, or I am wrong...? $\endgroup$ Commented Jan 22, 2017 at 16:48
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$\begingroup$ by Boolean you mean that all elements are idempotents? because in this case the answer is no, take for example the ring $\mathbb{Z}/2$. $\endgroup$ Commented Jan 22, 2017 at 16:49
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1$\begingroup$ Take then $R=\mathbb{Z}/4$. The quotient by the socle is $\mathbb{Z}/2$. $\endgroup$ Commented Jan 22, 2017 at 16:57