$R_R$ is semisimple $\implies$ $M_R$ is semisimple

For any ring $R$ if the right regular $R$-module $R_R$ is semisimple then all right $R$-modules are semisimple.

Let $M$ be an artirary right $R$-module. In order to show that $M$ is semisimple I have considered a submodule $N \subseteq M$. I have to show that $N$ is a direct summand of $M$. On the other hand, I have a useful characterization of semisimple modules:

$M$ is semisimple iff it is the sum of a family of simple submodules.

Using these instruments how can obtain desired result

You're overthinking this. If $R = \sum_{i \in I} R_i$ then $$M = \sum_{m \in M} mR = \sum_{m \in M}\sum_{i \in I}mR_i$$ and each $mR_i$ is either zero or simple.
• Do you not mean $mR$? – K.Power Jun 21 '17 at 21:50
Since $R_R$ is semisimple it is the sum of simple submodules. As a free $R$ module is the direct sum of $R_R$, it follows that each free $R$ module is a sum of simples, so semisimple. Now any $R$ module $V$ is the homomorphic image of a free $R$ module (easy exercise to prove), say $F$. However, we showed that $F$ must semisimple, and the homomorphic image of a semisimple module must be semisimple (another simple exercise to try), so $V$ must be semisimple.
• This feels like the essentially same approach. The fact that a module is a quotient of a free module is the map $$\bigoplus_{m \in M} e_mR \longrightarrow \sum_{m \in M} mR = M$$ where the $e_m$'s are given no relations and the map sends $e_m \mapsto m$. – Trevor Gunn Jun 21 '17 at 22:46