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 A_{p} = (*p*)for primes *p *in *R *= Z, we have A =(intersection) A_{p}= (0). Here, each *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.

(1

This yields

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