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Answer to Question #17271 in Abstract Algebra for sanches

Question #17271
Show that RM is simple iff M &sim; R/m (as left R-modules) for a suitable modular maximal left ideal m &sub; R.
1
2012-10-30T10:34:12-0400
Let M be simple. Then Rm is nonzero for some m &isin; M, and so Rm= M. Let ϕ: R &rarr; 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 &isin; M = Rm,there exists e &isin; R such that m = em. For any r &isin; R: ϕ(r &minus; re) = (r &minus; re)m = rm&minus; rm = 0. Therefore r &equiv; re (mod m), so m is modular, with R/m&sim; M. Conversely, let M = R/m, where m is a modular maximal left ideal. Let e &isin; R be such that r &equiv; re (mod m) for all r&isin; R. Then R e = M; in particular, R&middot;M is nonzero.

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