Is it true that a von Neumann regular (VNR) ring having no infinite set of orthogonal idempotents is semisimple?
I know for the proof of the claim it is sufficient to show that each rght ideal of the ring is a direct summand thereof. Hence if one proves that each right ideal is finitely generated the answer to the question is "yes". Thanks for cooperation!
Yes it is true, because it is the same as having the ACC/DCC on summands, which in a VNR ring means finite uniform dimension, hence semisimple.
You can go along the lines you were thinking by considering a right ideal that isn't finitely generated, and show how the ascending chain of summands produced inside can be used to manufacture orthogonal idempotents.