Answer to Question #17359 in Abstract Algebra for sanches

Question

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”?

Answer 1

If the indexing set I is infinite, the answer is “no”. For instance, taking A_p = (p)for primes p in R = Z, we have A =(intersection) A_p= (0). Here, each R/A_p ∼ 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.