Timeline for Does torsion-freeness of class group localize?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Apr 15, 2010 at 7:32 | vote | accept | Hailong Dao | ||
| Apr 15, 2010 at 7:32 | history | bounty ended | Hailong Dao | ||
| Apr 15, 2010 at 7:31 | comment | added | Hailong Dao | This looks good to me, so I will accept. Thanks a lot. | |
| Apr 14, 2010 at 14:54 | comment | added | naf | I have edited the answer. Hope it is clear now. | |
| Apr 14, 2010 at 14:52 | history | edited | naf | CC BY-SA 2.5 |
Added references and some more details
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| Apr 13, 2010 at 17:07 | comment | added | Hailong Dao | Hi again, thanks for the new post. This looks really interesting, but I am a little confused about your $Y$. Do you mean something like $Y=Spec k[x,y,z]/(x^2+y^2+z^2)$? Then I think $Cl(Y)=\mathbb Z/(2)$. | |
| Apr 13, 2010 at 5:08 | history | edited | naf | CC BY-SA 2.5 |
Tried to correct the errors in the earlier example
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| Apr 12, 2010 at 14:58 | comment | added | Hailong Dao | No problem, thanks for being interested! | |
| Apr 12, 2010 at 14:46 | comment | added | naf | You are right. Sorry for being so careless. | |
| Apr 12, 2010 at 13:37 | comment | added | Hailong Dao | The class group of the local ring the origin of the cone over Y is $Cl(Y)/\mathbb Z(H)$, where $H$ is the hyperplane section. So I don't think your example works. | |
| Apr 12, 2010 at 12:57 | history | answered | naf | CC BY-SA 2.5 |