Timeline for Does torsion-freeness of class group localize?
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 15, 2010 at 7:32 | vote | accept | Hailong Dao | ||
| Apr 15, 2010 at 7:32 | history | bounty ended | Hailong Dao | ||
| Apr 13, 2010 at 17:57 | comment | added | Hailong Dao | @hibertthm90:I don't think normality imply what you want | |
| Apr 13, 2010 at 3:29 | comment | added | Matt | Ah. I guess the set doesn't have to be open. It is just the intersection of all open sets containing $P$. I think if we had a condition on $R$ that guaranteed all spectra of localizations be isomorphic to an open subset (the image of that natural map), this would extend to a proof. Does normality imply this for some reason? | |
| Apr 12, 2010 at 22:23 | comment | added | Matt | Sorry. That was sloppy wording due to lack of space. It is a isomorphism onto an open subset. This group is an isomorphism invariant, so is it possible to then just work on this open set? I realized the subtlety from the last comment. If $pD \sim (f)$ on U (the open set), then we'd hope $pD\sim (f)$ considering $f\in K(SpecR)=K(U)$. It seems possible that on the whole space it could have more vanishing and poles. Is there an example of that? It might lead to your example. If there isn't it might lead to a proof (I suspect the latter) | |
| Apr 12, 2010 at 12:57 | answer | added | naf | timeline score: 3 | |
| Apr 12, 2010 at 11:55 | comment | added | Gerald Edgar | A little off-topic grammar remark. Why say "torson-freeness" and not "torsion-freedom" or something? | |
| Apr 12, 2010 at 7:11 | comment | added | Hailong Dao | hilbertthm90: The Spec map is not inclusion of open set. | |
| Apr 12, 2010 at 6:48 | comment | added | Matt | I'm not incredibly comfortable with this stuff, so forgive me if this is way off. $R\hookrightarrow R_P$ induces $Spec(R_P)\to Spec(R)$ and then $Cl(R)\to Cl(R_P)$ is the pullback under the previous map. But geometrically isn't the Spec map an inclusion of an open set, so the pullback is just taking a divisor on the open set and considering it as a divisor on the whole space. There is probably some subtlety I'm missing, but it seems if the above is correct that since it is really the "same" divisor, that $pD=0$ on the open should imply $pD=0$ on $Cl(R)$. | |
| Apr 12, 2010 at 4:20 | history | bounty started | Hailong Dao | ||
| Jan 12, 2010 at 9:28 | answer | added | Maharana | timeline score: 5 | |
| Jan 11, 2010 at 23:38 | comment | added | Maharana | For the longest of time I couldn't even find an example of $R$ other than the affine quadric threefold such that it has torsion-free Class group. But now that I have some of them, they all turn out with local class groups $0$. Your question is so tough! | |
| Dec 26, 2009 at 7:57 | history | asked | Hailong Dao | CC BY-SA 2.5 |