Timeline for How to compute the integer corresponding to a class in $G_0(B_{\mathrm{red}})$ for a commutative noetherian ring $B$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 7, 2023 at 23:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 9, 2023 at 22:08 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 7, 2023 at 20:35 | answer | added | Mohan | timeline score: 1 | |
| Feb 7, 2023 at 0:31 | comment | added | Boris | I have tried to compute the quotients $I/I^2$, $I^2/I^3$ and so on as $B/I$-modules to show that they are all isomorphic to $B/I$ in the general case. But I am stuck in this process. | |
| Feb 6, 2023 at 22:17 | comment | added | Mohan | From the comment above, I hope you have worked out that the integer is just $m$. | |
| Feb 4, 2023 at 14:19 | comment | added | Boris | Thank you so much for your sample reasoning and calculations. I will try to see how the general case works. | |
| Feb 4, 2023 at 6:28 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Feb 4, 2023 at 4:30 | comment | added | Eoin | ... in this case, $[S'/I] + [I/I^2] +[I^2] = 3 [S'']$, if I didn't mess up any of the computations. | |
| Feb 4, 2023 at 4:28 | comment | added | Eoin | Here's an example when $m=3$: let $S=k[t_0,t_1,t_2,t_3]$ and take $\phi:S\rightarrow R$ with $\phi(t_i)=xy^i$. Then the kernel of $\phi$ is (I think; I didn't try to prove these were all the relations) the ideal $(t_1^2=t_0t_2, t_1t_2=t_0t_3, t_2^2=t_1t_3, t_1^3=t_3t_0, t_2^3=t_3^2t_0)$. So $R/xR \cong S/(t_1^2,t_1t_2,t_2^2=t_1t_3, t_2^3):=S'$. And $S'':=S'_{red}\cong k[t_3]$ as you said. The generator of $G_0(S'_{red})$ is $[S'']$. Here if $I=(t_1,t_2)$ in $S'$ is the nilradical, then $S'/I \cong S''$; $I/I^2\cong S''/t_3S'' \oplus S''$ so $[I/I^2]=[S'']$; $I^3=0$ and $I^2\cong S''$ so ... | |
| Feb 4, 2023 at 2:25 | history | asked | Boris | CC BY-SA 4.0 |