# Answer to Question #25255 in Abstract Algebra for Tsit Lam

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.Need a fast expert's response?

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