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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.

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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.

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    $\begingroup$ You will have each left (or right) ideal generated by an idempotent. But in general $Re$ and $eR$ are completely different. @user45765 $\endgroup$ – Lord Shark the Unknown Mar 29 at 17:29
  • $\begingroup$ I see. Thanks a lot. $\endgroup$ – user45765 Mar 29 at 17:32

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