# 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.Need a fast expert's response?

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