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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3$\begingroup$ 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. $\endgroup$– KConradCommented Jul 6, 2010 at 4:18
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$\begingroup$ The example in that link is a local ring, so there's just one maximal ideal to worry about. $\endgroup$– KConradCommented Jul 6, 2010 at 4:19
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1$\begingroup$ Maybe someone should write up that ring in the Counterexamples In Algebra thread. $\endgroup$– Gerry MyersonCommented Jul 6, 2010 at 5:42
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13$\begingroup$ 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. $\endgroup$– Pete L. ClarkCommented Jul 6, 2010 at 7:44
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3 Answers
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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
Link
answered Jul 7, 2010 at 19:26
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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.
answered Jan 21, 2014 at 10:59
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$\begingroup$ 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. $\endgroup$ Commented Jan 21, 2014 at 16:16
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$\begingroup$ @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. $\endgroup$ Commented Jan 21, 2014 at 23:14
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$\begingroup$ 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 $n=1$. $\endgroup$ Commented Jan 22, 2014 at 0:47
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$\begingroup$ 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! $\endgroup$ Commented Jan 22, 2014 at 1:07
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This is an old question, but Julian Rosen and I observed an example that is a bit simpler than those written above. I am posting it in case it's helpful to others.
It is a bit closer to algebraic geometry, in particular a one-dimensional subquotient of $k[x,y]$ where $k$ is an algebraically closed field, basically a thickening of $k[x]$ at the origin.
Let $\displaystyle{R = \frac{k[x,y,\tfrac{y}{x}, \tfrac{y}{x^2}, \ldots]}{(y)}.}$
Claim 1. The maximal ideals of $R$ are $(x-c)$ for $c \in k$.
Claim 2. The only other prime ideal is the nilradical of $R$, which is $\displaystyle{(\tfrac{y}{x}, \tfrac{y}{x^2}, \ldots)}$.
Proof: By the description of the nilradical, $R / \mathrm{rad}(R) \cong k[x]$, which makes it easy to list all the prime ideals. It suffices to check their generators in $R$, i.e. each $(x-c) \subset R$ actually is maximal and that the nilradical is as described.
For the nilradical claim, $(\tfrac{y}{x^n})^2 = y \cdot \tfrac{y}{x^{2n}} = 0$ for all $n$. Quotienting by them leaves $k[x]$ which is reduced.
For the maximal ideals claim, set $x=c$. If $c \ne 0$, then $\tfrac{y}{x^n} \cdot c^n = \tfrac{y}{x^n} \cdot x^n = y = 0$, so $\tfrac{y}{x^n} = 0$ for all $n$, which leaves a field. On the other hand if $c = 0$, then $\tfrac{y}{x^n} = \tfrac{y}{x^{n+1}} \cdot x = 0$ for all $n$. Either way, we get a field.
