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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