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A commutative ring is said to be r-Noetherian if every regular ideal is finitely generated, where an ideal is said to be regular if 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 Dedekind if 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.

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I believe I have answered my question in the positive. Let $R = \mathbb{Z}+\varepsilon\mathbb{Q}[\varepsilon]$, where $\mathbb{Q}[\varepsilon]$ denotes the ring of dual numbers over $\mathbb{Q}$. Then $R$ is the integral closure of $\mathbb{Z}$ in the total quotient ring $\mathbb{Q}[\varepsilon]$ 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 $\mathbb{Q}[\varepsilon]$ of $R$ is Noetherian.

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