First assume that some isotypiccomponent, say M1, is not finitely generated. Then M1 is an infinitedirect sum of a simple k-module, so it is easy to find anepimorphism f1 : M1 → M1 which is not an isomorphism.Extending f1 by the identity map on the other Mi’s, we get an f: M → M which is an epimorphism but not an isomorphism. Now for anysplitting g for f, we have fg = 1 not equal gf.So, R cannot be unit-regular.
Next, assume all Mi’s arefinitely generated. M = K ⊕ N = K' ⊕ N' and N ∼ N' ⇒ K ∼ K'.
We conclude that R isunit-regular. Alternatively, we can give a more direct argument. Since Mi isfinitely generated, Ri : = Endk(Mi) is a simple artinianring, so it is unit-regular. It is easy to see that R = Endk(⊕iMi) ∼
(product)Ri, so it followsthat R is also unit-regular.
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