A non-noetherian ring with all localizations noetherian If for a ring $A$ every localization $A_\mathfrak{p}$ by a prime $\mathfrak{p}\subseteq A$ is noetherian, is it true that $A$ is noetherian? 
I believe not but I can't find a good counterexample.
 A: If you don't mind integral domains this link might help. This is what I said there.
There do exist integral domains $D$ such that $D_M$ is a Dedekind domain
for every maximal ideal $M.$ These were called almost Dedekind domains
by Robert Gilmer in Integral domains which are almost Dedekind, Proc. Amer. Math. Soc. 15(1964) 813-818. That some of these domains are non-Noetherian is also indicated in the above reference. Note that (a) a Dedekind domain is Noetherian and (b) an almost Dedekind domain is a one dimensional Prüfer domain. (c) a domain $D$ is Prüfer if every finitely generated nonzero ideal of $D$ is invertible.
I hope this helps.
Muhammad
A: Recall that a ring is Boolean if every element is idempotent: for all $x \in R$, $x^2 = x$.  And in fact a Boolean ring is necessarily commutative.  Here are two further rather easy facts about Boolean rings (for proofs see e.g. Section 9 of these notes):

*

*A Boolean ring is Noetherian iff it is finite.


*A Boolean ring is local iff it is a domain iff it is a field iff it is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.
Combining these facts, one sees that any infinite Boolean ring -- e.g. the product of infinitely many copies of $\mathbb{Z}/2\mathbb{Z}$ -- will be non-Noetherian but everywhere locally Noetherian.
