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Answer to Question #17367 in Algebra for Melvin Henriksen

Question #17367
Let M be a right module over a ring k such that R = Endk(M) is von Neumann regular. Show that R is unit-regular iff, whenever M = K ⊕ N = K’ ⊕ N’ (in the category of k-modules), N ∼ N’ implies K ∼ K’.
Expert's answer
For a ∈ R, write M = ker(a) ⊕ P = Q ⊕ im(a).Since a defines an isomorphism from P to im(a), thehypothesis implies that ker(a) ∼ Q(∼ coker(a)). Defining u ∈ U(R) such that u is an isomorphism from Q to ker(a),and u : im(a) → P is the inverse of a|P : P → im(a),we have a = aua ∈ R. For thenecessity part, assume R is unit-regular. Suppose M = K ⊕ N = K' ⊕ N', where N ∼ N'. Define a ∈ R such thata(K) = 0 and a|N is a fixed isomorphism from N to N'.Write a = aua, where u ∈ U(R).
(∗) M = ker(a) ⊕ im(ua) = K ⊕ u(N').
Since u defines anisomorphism from N' to u(N'), it induces an isomorphismfrom M/N' to M/u(N'). Noting that M/N' ∼ K' and M/u(N') ∼K (from (∗)), weconclude that K ∼ K'.

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