Answer to Question #17360 in Algebra for Mohammad
To see that in most cases R isnot von Neumann regular, consider any nonsingleton connected compact Hausdorff
space A. Then the only idempotents in R are 0 and 1. Assume R isvon Neumann regular. For any nonzero f ∈ R, fR = eR for some idempotent e ∈ R, so we must have fR = R, i.e. f∈ U(R). Therefore, R is a field. The known classificationtheorem for maximal ideals of R then implies that |A| = 1, acontradiction.
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