Question #25255

Let R be a left primitive ring. Show that for any nonzero idempotent e ∈ R, the ring A = eRe is also left primitive

Expert's answer

Let *V *be a faithful simpleleft *R*-module. It suffices to show that *U *= *eV *is afaithful simple left *A*-module. First*A · U *= *eRe · eV *= *eReV **⊆** eV *= *U, *so *U *is indeed an *A*-module.Let *a *= *ere **∈** A*, where *r**∈** R*. Then *ae *= *ere*^{2} = *a*.If *aU *= 0, then 0 = *aeV *= *aV *implies that *a *= 0, so*AU *is faithful. To check that *AU *is simple, let us show that for0 *<>* *u **∈** U *and *u'**∈** U*, we have *u' **∈** Au*. Note that *u, u' **∈** eV *implies *u *= *eu, u' *= *eu'*.We have *u' *= *ru *for some *r **∈** R*, so *u' *= *eu' *= *eru *= (*ere*)*u**∈** Au, *as desired.

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