1
$\begingroup$

The following question essentially appeared (https://math.stackexchange.com/questions/931801/preservation-of-localness-among-certain-krull-domains) on math.SE a while ago, but nobody has done anything with it, and I'm interested in the answer, so I'll post it here.

Namely, let $R$ be a local Noetherian normal domain, and let $\mathfrak p$ be a height one prime whose class in the divisor class group is non-torsion. Let $$ S := \bigcap_{\mathfrak q \in \operatorname{Spec} R, \operatorname{ht} \mathfrak q = 1, \mathfrak q \neq \mathfrak p} R_{\mathfrak q}. $$ Is $S$ local? And if not, what can we say about $\mathfrak p S$? Is it a proper ideal of $S$?

Improve this question
$\endgroup$
3
  • $\begingroup$ The ring $S$ does not look local to me. Using the characterization of normality, $S$ should contain the functions on $\operatorname{Spec}R$ which are defined away from the divisor corresponding to $\mathfrak{p}$. This looks much like the localization of a local ring, which in general has no reason for being local. Similarly, it seems to me that $\mathfrak{p}S$ would have a tendency of being everything. I have to admit that my intuition comes from the case where $\mathfrak{p}$ is principal, but I also do not see how class group statements would change much here... $\endgroup$
    – Matthias Wendt
    Commented Sep 17, 2014 at 16:54
  • $\begingroup$ I agree with Mathias, but I can't prove anything yet. I suggest you look at the 2-dimensional case. In this case, if you denote by $U$ the complement of the closed point in $\mathrm{Spec}(R)$, then $S$ is the ring of global functions on $V:=U\smallsetminus\{\mathfrak{p}\}$, and I would rephrase Mathias's intuition by saying that $V$ might be affine (in which case, of course, $S$ is not local since $\mathrm{Spec}(S)=V$). $\endgroup$
    – Laurent Moret-Bailly
    Commented Sep 17, 2014 at 19:30
  • $\begingroup$ @Matthias The relevance of torsion in the class group is as follows. It is well known that $\mathfrak p$ is contained in the union of the other height one primes if and only if its class is non-torsion. If $\mathfrak p$ is not contained in said union (i.e. the class is torsion), then $\mathfrak p S$ is not a proper ideal of $S$. I was wondering (and presumably so was the OP at math.SE) whether, in the local case, the converse were true. $\endgroup$
    – Neil Epstein
    Commented Sep 18, 2014 at 14:08

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .