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It is well known as Cohen's theorem that a commutative ring is Noetherian if all its prime ideals are finitely generated. Is this statement true or false when prime ideals are replaced by maximal ideals?

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Presumably it is not true or not known, since if it were known to be true that fact would surely appear in the books which prove Cohen's theorem. Puttering around the internet I find a specific counterexample (f.g. maximal ideal and non-f.g. prime ideals) described in math.purdue.edu/~heinzer/preprints/nonfg17.pdf although I have not actually looked closely at this. –  KConrad Jul 6 '10 at 4:18
    
The example in that link is a local ring, so there's just one maximal ideal to worry about. –  KConrad Jul 6 '10 at 4:19
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Maybe someone should write up that ring in the Counterexamples In Algebra thread. –  Gerry Myerson Jul 6 '10 at 5:42
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I believe that any valuation ring with value group $\mathbb{Z} \oplus \mathbb{Z}$ (lexicographically ordered) gives a counterexample. The unique maximal ideal is principal -- generated by any element of valuation $(0,1)$ -- but the ring is not Noetherian, since if so it would be a DVR. –  Pete L. Clark Jul 6 '10 at 7:44
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2 Answers 2

up vote 1 down vote accepted

See the following paper (and search for its citations for related work)

Gilmer, R; Heinzer W.
A non-Noetherian two-dimensional Hilbert domain with principal maximal ideals,
Michigan J. Math. 23 (1976), 353-362
http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.mmj/1029001770

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Here is another nice ring which shows that the answer is: "false". It is due to B. Osofsky and can serve as (counter)example in other situations.

Start with the $p$-adics $\mathbb{Z}_p$ and $\mathbb{Q}_p$. (Any other complete DVR with its fraction field would do.) Our ring, call it $R$, is the additive group $\mathbb{Z}_p \oplus \mathbb{Q}_p/\mathbb{Z}_p$ with multiplication $(a,b) \cdot (c,d) = (ac, ad + cb)$. (This is called the "trivial extension" of the ring $\mathbb{Z}_p$ by its module $\mathbb{Q}_p/\mathbb{Z}_p$.) The ideals in $R$ form the chain

$R \supsetneq pR \supsetneq p^2R \supsetneq \; ... \; (0, \mathbb{Q}_p/\mathbb{Z}_p) \; ... \; \supsetneq (0, p^{-2}\mathbb{Z}_p/\mathbb{Z}_p) \supsetneq (0, p^{-1}\mathbb{Z}_p/\mathbb{Z}_p) \supsetneq 0$

Among other interesting properties that $R$ has (like being a cogenerator ring, in particular self-injective), it is local with principal maximal ideal $pR$; but it has exactly one non-maximal prime ideal -- the lonely one there in the middle of the chain, having no neighbours --, and this sad thing is not finitely generated.

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Another source of examples: There are lots of valuation rings where the unique maximal ideal is principal. Any valuation ring that is discrete but not of rank one will do. –  Neil Epstein Jan 21 at 16:16
    
@Neil Epstein: Yes, that generalises Pete L. Clark's comment to the original post. Contrary to Osofsky's ring, those valuation rings are even domains which give counterexamples. –  Torsten Schoeneberg Jan 21 at 23:14
    
Good point; I hadn't noticed his comment. Actually come to think of it, even many non-discrete valuation rings will work. Let $G_1, \dotsc, G_n$ be subgroups of the reals, with $G_n = Z$ (the integers). Then any valuation ring with value group $G_1 \oplus \cdots \oplus G_n$ has principal maximal ideal, but no such ring is Noetherian unless $n=1$. –  Neil Epstein Jan 22 at 0:47
    
Ah, and now that I come to think of it: Osofsky's ring has Krull dimension 1, whereas all the other counterexamples here have dimension $\ge 2$ -- since a domain of dimension 1 (or any ring of dimension 0) of course can never be a counterexample! –  Torsten Schoeneberg Jan 22 at 1:07
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