A commutative ring is said to be

r-Noetherianif every regular ideal is finitely generated, where an ideal is said to beregularif it contains a non-zerodivisor. Does there exist a non-Noetherian r-Noetherian commutative ring whose total quotient ring is Noetherian?EDIT: A commutative ring is

Dedekindif every regular ideal is invertible. (A domain is Dedekind if and only if it is a Dedekind domain.) Does there exist a non-Noetherian Dedekind ring whose total quotient ring is Noetherian?The motivation for this question is that conditions like r-Noetherian and Dedekind control only the regular ideals of a ring and one can try to use the total quotient ring to control the non-regular ideals. By this philosophy one would hope that an r-Noetherian ring is Noetherian if its total quotient ring is Noetherian. But there is no obvious proof of this, leading to the desire for a counterexample.

**Answer**

I believe I have answered my question in the positive. Let R=Z+εQ[ε], where Q[ε] denotes the ring of dual numbers over Q. Then R is the integral closure of Z in the total quotient ring Q[ε] of R, so in particular R is integrally closed. Every regular ideal of R is principal, and every finitely generated ideal of R is principal, R is non-Noetherian, and yet the total quotient ring Q[ε] of R is Noetherian.

**Attribution***Source : Link , Question Author : Jesse Elliott , Answer Author : Jesse Elliott*