Timeline for When $R $ is a cusp then $K_0(R) \ncong K_0(R[s])$
Current License: CC BY-SA 4.0
10 events
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| Sep 11, 2023 at 6:19 | comment | added | user443060 | Ok so since $R$ is commutative $K_0(R) \cong H_0(R) \oplus \tilde{K_0(R))}$, and what you have shown is for $R = k[t^2,t^3,s]$. | |
| Sep 10, 2023 at 12:17 | comment | added | user443060 | I just wanted to clarify one thing, how did we show that $K_0(k[t^2,t^3]) \ncong K_0(k[t^2,t^3,s])$? | |
| Sep 8, 2023 at 2:19 | history | edited | Steven Landsburg | CC BY-SA 4.0 |
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| Sep 6, 2023 at 6:58 | comment | added | user443060 | Thank you very much. | |
| Sep 6, 2023 at 5:04 | vote | accept | user443060 | ||
| Sep 6, 2023 at 4:59 | comment | added | Steven Landsburg | And yes, what you are saying is right. | |
| Sep 6, 2023 at 4:51 | comment | added | Steven Landsburg | Sorry, I meant to mod out the units coming from $k[t]$. That leaves you with units of the form $1+\alpha\epsilon$, which form a group isomorphic to the additive group of $k$. I've edited to fix this. | |
| Sep 6, 2023 at 4:47 | history | edited | Steven Landsburg | CC BY-SA 4.0 |
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| Sep 6, 2023 at 4:43 | comment | added | user443060 | If $k$ is a field then $Pic(k[t^2,t^3])$ should be $k$, right? Am I making some mistakes here? So now by the same conductor ideal formula, we are getting $k[s]$ for $Pic(k[t^2,t^3,s])$ and we have an isomorphism class of a projective module that is exclusive to $K_0(k[t^2,t^3,s])$. Right? | |
| Sep 6, 2023 at 4:08 | history | answered | Steven Landsburg | CC BY-SA 4.0 |