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*_{k} issemisimple, then *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*)*.*

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