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Does anyone know of such a domain?

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3 Answers 3

up vote 8 down vote accepted

You may find Matlis' paper The Two-Generator Problem for Ideals to be interesting, as its main theorem concerns the class of integral domains in which every ideal is generated by two elements.

It was proven by Cohen (in Commutative Rings with Restricted Minimum Condition ) that an integral domain with the property that there exists an integer $n$ such that every ideal can be generated with fewer than $n$ elements must be Noetherian and of Krull dimension 1.

Say that an integral domain R has property FD if every finitely generated torsion free R-module is direct sum of modules of rank 1. Moreover, say that R has property FD locally if RM has property FD for every maximal ideal M of R.

Theorem (simplified form) - Let R be an arbitrary integral domain. Then every ideal of R can be generated by two elements if and only if R is a noetherian ring that has property FD locally.

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[Edited to restrict to the case of quadratic orders. --PLC]

Take any non-maximal order of a quadratic number field. This is not Dedekind because it fails to be integrally closed in its field of fractions. Every ideal is a free abelian subgroup of rank at most $2$.

For example: $\mathbb{Z}[\sqrt{-3}].$

I hope this answers your question. For further reading on Dedekind domains, and non-maximal Orders, I highly recommend the chapter on it in Neukirch's Algebraic Number Theory.

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This certainly works for any order in a quadratic number field, but I believe it fails in general. (This was a problem on a problem set I did some time ago: but I don't have the example I came up with on hand.) – Alison Miller Jan 25 '10 at 22:35
You are probably correct, I have only been thinking about quadratic extensions lately, and didn't think to check the general case. I'll try to think of a counter-example myself. – Ben Weiss Jan 25 '10 at 23:10
Alison is right. Consider the function field analogue: the number of elements needed to generate the maximal ideal of a point on an affine curve is at least the dimension of the Zariski tangent space, which can be arbitrarily large at a singularity. Here is an explicit number ring example: Let A=Z[2^{1/3}], let R=Z+2A, and let m=2A, viewed as an R-ideal. Then m/m^2 is 3-dimensional over R/m=F_2, so m is not generated by fewer than 3 elements. – Bjorn Poonen Jan 26 '10 at 3:15
Thank you all for the corrections and example. – Ben Weiss Jan 26 '10 at 3:43

By way of comparison, Dedekind domains are characterized by an even stronger property, sometimes referred to colloquially as "$1+\epsilon$''-generation of ideals. Namely:

Theorem: For an integral domain $R$, the following are equivalent:
(i) $R$ is a Dedekind domain.
(ii) For every nonzero ideal $I$ of $R$ and $0 \neq a \in I$, there exists $b \in I$ such that $I = \langle a,b \rangle$.

The proof of (i) $\implies$ (ii) is such a standard exercise that maybe I shouldn't ruin it by giving the proof here. That (ii) $\implies$ (i) is not nearly as well known, although sufficiently faithful readers of Jacobon's Basic Algebra will know it: he gives the result as Exercise 3 in Volume II, Section 10.2 -- "Characterizations of Dedekind domains" -- and attributes it to H. Sah. (A MathSciNet search for such a person turned up nothing.) The argument is as follows: certainly the condition implies that $R$ is Noetherian, and a Noetherian domain is a Dedekind domain iff its localization at every maximal ideal is a DVR. The condition (ii) passes to ideals in the localization, and the killing blow is dealt by Nakayama's Lemma.

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H. Sah was probably Chih Han Sah. Obituary:… – Carl Weisman Jan 26 '10 at 20:28
Thanks, Carl. That seems very reasonable, although I have no way of knowing for sure. I did a MathSciNet search for C.-H. Sah and found plenty of papers, although nothing which jumped out as containing this result. – Pete L. Clark Jan 26 '10 at 20:46

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