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Apr 13, 2017 at 12:19 history edited CommunityBot
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Sep 18, 2014 at 14:08 comment added Neil Epstein @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.
Sep 17, 2014 at 19:30 comment added Laurent Moret-Bailly 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 $\mathrm{Spec}(S)=V$).
Sep 17, 2014 at 16:54 comment added Matthias Wendt 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...
S Sep 17, 2014 at 16:43 history suggested user26857 CC BY-SA 3.0
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Sep 17, 2014 at 16:22 review Suggested edits
S Sep 17, 2014 at 16:43
Sep 16, 2014 at 17:23 history asked Neil Epstein CC BY-SA 3.0