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’.
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Expert's answer
2012-10-31T08:56:09-0400
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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