# Answer to Question #17359 in Abstract Algebra 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

(1

This yields

*I*is infinite, the answer is “no”. For instance, taking A*= (*_{p}*p*)for primes*p*in*R*= Z, we have A =(intersection) A*= (0). Here, each*_{p}*R/*A_{p}*∼*Z*/p*Z 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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