Does there exist a non-Noetherian integral domain of Krull dimension one that has a finitely generated maximal ideal?

There is an example of a non-Noetherian valuation ring of Krull dimension one here but its maximal ideal appears to be infinitely generated.


1 Answer 1


[Edited to remove false information]

In his paper Overrings of Prüfer Domains,$^1$ Gilmer constructs a ring which is a locally a DVR and has all but one maximal ideal finitely generated. It is the last example in that paper. In general, "most" almost Dedekind domains (aka locally DVR rings) which are non-Noetherian will have some sharp primes. The literature on almost Dedekind domains is a good place to look for examples of the sort in question.

However, there cannot be any non-noetherian 1-dimensional local domain with a finitely generated maximal ideal. This follows, as user 26857 noted in the comments, from Cohen's theorem which states that a ring is Noetherian iff its prime ideals are finitely generated.

I am not sure whether there can be f.g. maximal ideals in non-Noetherian $1$-dimensional semi-local domains.

$^1$ Overrings of Prufer domains, I, J. Algebra 4 (1966), 331-340

  • $\begingroup$ For proving Part (2) one can also invoke Cohen characterization of noetherian rings using prime ideals. $\endgroup$
    – user26857
    Aug 9, 2019 at 20:13
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    $\begingroup$ @user26857 hmm yea that's a good point, and for this purpose a much better explanation than the one I gave. I tend to think of this as a special case of the fact that the localization of a completely integrally closed domain at a divisorial prime is a DVR, which I guess got me sidetracked. $\endgroup$ Aug 9, 2019 at 20:27
  • $\begingroup$ @BadamBaplan: I think that what you say in part (1) is not true. If I am not missing something, you make a mistake arguing that $xB[x]$ is a principal ideal of $R$. For, assume that $xB[x] = (f_1,...,f_n)_R$ and let $b_1,...,b_n \in B$ be the coefficient standing in front of $x$ in $f_1,...,f_n$. Let $b \in B$. Then $bx \in xB[x]$, so there exist $g_1,...,g_n \in R$ s.t. $bx = g_1f_1+...+g_nf_n$. If $c_1,...,c_n \in A$ are the constant terms of $g_1,...,g_n$, we get $b = c_1b_1 + ... + c_nb_n$. So $B$ would be f.g. as an $A$-module. Contradiction to our assumption. $\endgroup$
    – Daniel W.
    Feb 5, 2020 at 16:12
  • $\begingroup$ $xB[x]$ is of course principal in $B[x]$ but not in general in $R$. I forgot to say that $A \subseteq B$ should be an infinite dimensional field extension, as in your example. $\endgroup$
    – Daniel W.
    Feb 5, 2020 at 16:14
  • $\begingroup$ @DanielW you are of course right. What I wrote is bogus and also the polynomial construction is unnecessarily complicated. Hopefully this answer hasn't caused too much damage, I'll edit it with correct examples once I get to a computer. $\endgroup$ Feb 6, 2020 at 21:18

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