Timeline for Picard group of a cusp [duplicate]
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 12, 2023 at 6:39 | history | closed |
Steven Landsburg Max Horn Friedrich Knop Bugs Bunny CommunityBot |
Duplicate of When $R $ is a cusp then $K_0(R) \ncong K_0(R[s])$ | |
| Sep 6, 2023 at 4:34 | review | Close votes | |||
| Sep 12, 2023 at 6:41 | |||||
| Mar 25, 2023 at 1:48 | comment | added | user443060 | @Stahl yes, there was a notational clash, I have fixed it | |
| Mar 25, 2023 at 1:43 | history | edited | user443060 | CC BY-SA 4.0 |
edited body
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| Mar 25, 2023 at 0:34 | comment | added | Stahl | @Wojowu Ah yes, that would make more sense. | |
| Mar 25, 2023 at 0:33 | comment | added | Wojowu | @Stahl I suspect there's a notational clash, and OP means to take the polynomial ring over $R$, which probably should be denoted by $R[T]$ or some other letter. | |
| Mar 25, 2023 at 0:32 | comment | added | Stahl | Am I missing something here? If $R = k[t^2,t^3],$ then $R[t] = k[t],$ so you're just looking at the Picard group of $k[t].$ | |
| Mar 24, 2023 at 23:57 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| Mar 24, 2023 at 18:42 | history | asked | user443060 | CC BY-SA 4.0 |