non-Dedekind Domain in which every ideal is generated by at most two elements - MathOverflow

non-Dedekind Domain in which every ideal is generated by at most two elements

2010-01-25T21:40:37Z

Does anyone know of such a domain?

Answer by Ben Weiss for non-Dedekind Domain in which every ideal is generated by at most two elements

2010-01-25T21:46:33Z

[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.

Answer by Ben Linowitz for non-Dedekind Domain in which every ideal is generated by at most two elements

2010-01-25T22:10:32Z

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 R_M 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.

Answer by Pete L. Clark for non-Dedekind Domain in which every ideal is generated by at most two elements

2010-01-26T10:15:40Z

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.

Answer by Carl Weisman for non-Dedekind Domain in which every ideal is generated by at most two elements

2010-01-26T20:28:23Z

H. Sah was probably Chih Han Sah. Obituary: http://www.nytimes.com/1997/08/18/nyregion/chih-han-sah-62-mathematics-professor.html?pagewanted=1