TITLE: Is a domain all of whose localizations are noetherian itself noetherian ? 
QUESTION [25 upvotes]: 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: 

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. 

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.

REPLY [13 votes]: All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for 

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian. 

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=\mathbb{Z} \setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.  
Remark. The foregoing shows that $\dim D=2$.