Answer to Question #17271 in Abstract Algebra for sanches

Question #17271
Show that RM is simple iff M ∼ R/m (as left R-modules) for a suitable modular maximal left ideal m ⊂ R.
Expert's answer
Let M be simple. Then Rm is nonzero for some m ∈ M, and so Rm= M. Let ϕ: R → M be the R-epimorphism defined by ϕ(r) = rm, and let m =ker(ϕ). Then m is a maximal left ideal of R. Since m ∈ M = Rm,there exists e ∈ R such that m = em. For any r ∈ R: ϕ(r − re) = (r − re)m = rm− rm = 0. Therefore r ≡ re (mod m), so m is modular, with R/m∼ M. Conversely, let M = R/m, where m is a modular maximal left ideal. Let e ∈ R be such that r ≡ re (mod m) for all r∈ R. Then R e = M; in particular, R·M is nonzero.

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