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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.

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This Google search may be of interest: at.yorku.ca/cgi-bin/… –  Fredrik Meyer Oct 17 '11 at 19:31
    
Thank you very much! –  Daniele A Oct 17 '11 at 20:18

2 Answers 2

up vote 15 down vote accepted

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):

  1. A Boolean ring is Noetherian iff it is finite.

  2. 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.

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Sorry that it took me so much to accept your very interesting answer! –  Daniele A May 19 '13 at 19:59

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

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