Question

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.

Answer 1

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 andg | 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).