# Let $R$ be a semi-simple ring. All left modules are semi-simple. Then all right modules are semi-simple. So all modules are semi-simple on both sides?

Let $$R$$ be a semi-simple ring not necessarily commutative. Let $$R-Mod$$ be the category of left $$R-$$modules and $$Mod-R$$ be the category of right $$R$$ modules.

Then I have both $$R-Mod$$ and $$Mod-R$$ composed of semi-simple modules on left and right correspondingly. Suppose $$M$$ is $$(R,R)$$ bimodule on both-sides. Say $$M_{1R}$$ is one of the left simple module factor of $$M$$(i.e. $$M_{1R}$$ is one of the direct summand of $$M$$ as right $$R-$$module.) Now consider simple $$_{R}M_j$$ as some direct summand of $$M$$ as left $$R-$$module.

$$\textbf{Q:}$$ Is $$M_{1R}$$ necessarily agrees with some $$_R M_j$$ for some $$j$$? In other words, if a module is left-semi simple and it is a bi-module over the same ring, then it is also right-semi simple with the same direct summand? Consider ring $$R$$ itself. Since $$R$$ is semi-simple, say it decomposes into $$\oplus_jI_j$$ with $$I_j$$ simple. Pick out orthogonal idempotents $$e_j$$ as left module.(Is it even obvious to pick out orthogonal idempotents in general?) If it is possible to pick out orthogonal idempotents, then it is clear that $$e_je_k=\delta_{jk}e_j$$. Since left decomposition of $$R$$ by those orthogonal idempotents is also a right decomposition of $$R$$ by those orthogonal idempotents, decomposition in left and right for $$R$$ is the same.

Let $$R=M_2(k)$$, the $$2\times 2$$ matrices over a field $$k$$. This is a classic example of a semi-simple ring. A classic example of an $$R$$-bimodule is $$R$$ itself. Consider the set $$N$$ of matrices of the form $$\pmatrix{*&*\\0&0}.$$ Then $$N$$ is a simple right $$R$$-submodule of $$R$$, but is certainly not a left $$R$$-submodule.
• You will have each left (or right) ideal generated by an idempotent. But in general $Re$ and $eR$ are completely different. @user45765 – Lord Shark the Unknown Mar 29 at 17:29