Is a domain $D$, all of whose localizations $D\_P$ for $P \in Spec(D)$ are noetherian, itself noetherian ? 


The question is motivated by proposition 11.5 of Neukirch's Algebraic Number Theory: 



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> Let $\mathfrak{o}$ be a noetherian integral domain. $\mathfrak{o}$ is a Dedekind domain if and only if, for all prime ideals $\mathfrak{p}\neq 0$, the localizations $\mathfrak{o}\_\mathfrak{p}$ are discrete valuation rings. 
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If the question above has a positive answer, this proposition would give an unconditioned (i.e. without precondition "noetherian") characterization of Dedekind domains by a local property. 


By googling I found a counterexample for a ring with zero-divisors: 


<https://math.stackexchange.com/questions/73421/a-non-noetherian-ring-with-all-localizations-noetherian>


But I couldn't find a counterexample for a domain. 



Is a domain all of whose localizations are noetherian itself noetherian ? 
I had the exact same question not too long ago. Apparently if you drop the noetherian precondition in Neukirch's definition of "Dedekind domain" then you get what some people call an "almost Dedekind domain". There are indeed examples of almost Dedekind domains that aren't Dedekind (i.e. aren't noetherian). The first of these was given by Nakano (*J. Sci. Hiroshima Univ. Ser. A. **16***, 425–439 (1953)): take the integral closure of $\mathbb Z$ in the field obtained by adjoining to $\mathbb Q$ the $p$th roots of unity for all primes $p$.


