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Prepended: OK, here is an elementary argument based only on the simple submodules. We begin by assuming $R$ is a direct sum of $n$ minimal right ideals, and we know that $n$ is unique. If $ab=1$, then the homomorphism $x\mapsto bx$ from $R\to R$ is injective. Thus $bR\cong R$ as $R$ modules. Find a complement $C$ to $bR$ so that $R=bR\oplus C$. If \$ba\neq ...

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