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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?

edited Jul 6 2010 at 9:57

Charles Matthews
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asked Jul 6 2010 at 3:39

TmobiusX
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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 2010 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 2010 at 4:19

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Maybe someone should write up that ring in the Counterexamples In Algebra thread. – Gerry Myerson Jul 6 2010 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 2010 at 7:44

Bill Dubuque · Accepted Answer · 2010-07-07 19:26:40Z UTC

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

A remark on Cohen’s theorem

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