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Answer on Algebra Question for Mohammad

Question #17360
Let R be the ring of all continuous real-valued functions on a topological space A. Show that R is J-semisimple, but “in most cases” not von Neumann regular.
Expert's answer
The following are clearly maximalideals of R: ma = {f ∈ R : f(a) = 0},where a ∈ A. Therefore, rad R ⊆(intersection) ma = {f ∈ R : f(A) = 0} = 0.
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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