12
$\begingroup$

Which integral domains have the property that $R = \bigcap R_P$, the intersection being taken over all height one prime ideals of $R$?

It is a standard fact that Krull domains, and thus noetherian normal rings, have this property. But Krull domains satisfy two additional properties, namely:

  1. $R_P$ is a discrete valuation ring for all height one primes $P$.
  2. Every non-zero element is contained in only a finite number of height one prime ideals.

What happens if I drop these two assumptions? Do I get anything new? Is this true for non-noetherian normal rings (I fear not)?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ A one-dimensional, local, integral domain has this property. So there are many examples, e.g., the local ring of $k[x,y]/\langle y^2 - x^3 \rangle$ at the maximal ideal $\langle x,y\rangle$. $\endgroup$
    – Jason Starr
    Commented Sep 1, 2015 at 16:19

2 Answers 2

Reset to default
4
$\begingroup$

This property has been studied before: see http://www.kkms.org/kkms/vol08_1/08110.pdf.

In particular, the author G. W. Chang says that an integral domain $R$ satisfies APIT (i.e., the associated prime ideal theorem) if every (weak Bourbaki) associated prime of $R$ has height one.

Let $R$ be an integral domain, and let $X^1(R)$ be the set of all height one primes of $R$, and let $\text{Ass}(R)$ denote the set of all associated primes of $R$ (which by definition are the prime ideals that are minimal over some nonzero conductor ideal $(aR:_R bR)$ with $a,b \in R$). It is easy to show that $$X^1(R) \subseteq \text{Ass}(R)$$ and $$R = \bigcap_{P \in \text{Ass}(R)} R_P.$$ Moreover, one has $R = \bigcap_{P \in X^1(R)} R_P$ if and only if $X^1(R) = \text{Ass}(R)$, if and only if every associated prime of $R$ has height one, i.e., iff $R$ satisfies APIT. Example 1 of the referenced Chang paper is an integrally closed two-dimensional non-Noetherian local domain that does not satisfy APIT, but nevertheless every prime ideal of $R$ that is minimal over a nonzero principal ideal of $R$ has height one.

Here is a closely related MO post whose answer gives a Noetherian example of a domain that does not satisfy APIT: Primes associated to a principal ideal

For more on weak Bourbaki associated primes, see https://stacks.math.columbia.edu/tag/0546.

Improve this answer
$\endgroup$
1
$\begingroup$

I cannot fully answer the question but some keywords:

  • If you drop 1. what you get is called a weakly Krull domain.

  • If you just weaken 1. to valuation ring what you get is called a generalized Krull domain.

There are a bunch of related notions. Thus, yes, you get something new and these types of rings got studied.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ I didn't know about "weakly Krull domains". Thank you, I'll take a look at this. $\endgroup$
    – user78294
    Commented Sep 1, 2015 at 15:57
  • $\begingroup$ After browsing through the literature I have to admit that I haven't seen any interesting non-obvious examples of weakly Krull domains. That's a bit annoying. $\endgroup$
    – user78294
    Commented Sep 3, 2015 at 12:06
  • $\begingroup$ I am not sure which examples you saw but one-dimensional noetherian domains are weakly Krull (but perhaps this is in the 'obvious' category). IIIRC Cohen-Macaulay domains are also weakly Krull. Some constructions can be found in projecteuclid.org/euclid.rmjm/1181069485 $\endgroup$
    – user9072
    Commented Sep 3, 2015 at 12:21
  • $\begingroup$ Alright, quite a lot in there indeed, thanks. Do you happen to know whether there exist non-normal non-noetherian non-onedimensional rings with $R = \bigcap R_P$? Perhaps not all non- have to be satisfied at the same time. $\endgroup$
    – user78294
    Commented Sep 3, 2015 at 12:24
  • 1
    $\begingroup$ I would try to look at the subring of $C[x,y]$ generated by monomials $x^my^n$ with $m\leq n\sqrt 2$ or something like that. $\endgroup$
    – Lev Borisov
    Commented Feb 9, 2016 at 22:35

You must log in to answer this question.