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Answer on Abstract Algebra Question for sanches

Question #17359
Let Ai (i ∈ I) be ideals in a ring R, and let A =(intersection on i) Ai. True or False: “If each R/Ai is von Neumann regular, then so is R/A”?
Expert's answer
If the indexing set I is infinite, the answer is “no”. For instance, taking Ap = (p)for primes p in R = Z, we have A =(intersection) Ap= (0). Here, each R/Ap ∼ Z/pZ is a field and hence von Neumann regular, but R/A ∼ Z is not von Neumann regular. To treat the case |I| < ∞,let I = {1, 2, . . . , n}. We claim that here the answer is “yes”. It suffices to prove this for n = 2, and we may assume A1 ∩ A2 = (0). Consider any a ∈ R. Since R/A1 and R/A2 are von Neumann regular, there exist x, y ∈ R such that (1 − ax)a ∈A1 and a(1 − ya) ∈A2. Then
(1 − ax)a(1 − ya)∈A1 ∩ A2 = 0.
This yields a = a(x+ y − xay)a, so R is von Neumann regular.

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