# Answer to Question #17362 in Algebra for Mohammad

Question #17362

Let R = Endk(M) where M is a right module over a ring k. Show that an element f ∈ R is von Neumann regular iff ker(f) and im(f) are both direct summands of M.

Expert's answer

If

*M*issemisimple, then_{k}*R*is von Neumann regular. First, suppose*f**∈**R*is such that ker(*f*)*⊕**P*=*M*= im(*f*)*⊕**Q,*where*P*,*Q*are*k*-submodulesof*M*. Then*f | P*:*P →*im(*f*) is an isomorphism.Defining*g**∈**R*such that*g*(*Q*) = 0 and*g |*im(*f*) is the inverse of*f | P*, we get*f*=*fgf*.Conversely, if*f**∈**R*can bewritten as*fgf*for some*g**∈**R*, then the surjection*M f→*im(*f*) is split by the map*g*(since,for every*m**∈**M*,*fg*(*f*(*m*)) =*f*(*m*)).This shows that*M*= ker(*f*)*⊕**im(**gf*)*.*Dually, the injection im(*f*)*→ M*is split by the map*fg*,so*M*= im(*f*)*⊕**ker(**fg*)*.*Need a fast expert's response?

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