Timeline for If Serre's intersection multiplicity $\chi(R/I, R/J)$ equals $\operatorname{length}_R (R/(I+J))$, then are $R/I, R/J$ Cohen-Macaulay?
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5 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2022 at 5:12 | comment | added | Snake Eyes | Instead of Auslander and Buchsbaum, it should be Auslander and Lichtenbaum I think ... respectively projecteuclid.org/journals/illinois-journal-of-mathematics/… and projecteuclid.org/journals/illinois-journal-of-mathematics/… ... and later in slightly more generality by Hochster as I have mentioned in my answer | |
| Oct 6, 2022 at 15:22 | comment | added | Nawaj | My bad. But again, this is true in equal characteristic according to Serre. Given your hypothesis, $\chi_1(R/I, R/J) = 0$ and this is if and only if $\text{Tor}_{i+1}(R/I, R/J) =0$. Serre attributes this result, which holds more generally for $\chi_r$, to Auslander & Buchsbaum. | |
| Oct 6, 2022 at 2:18 | comment | added | Alex | This is not at all what I asked ... I asked if $\chi(R/I,R/J)=\mathcal l_R(R/(I+J))$ implies $\text{Tor}^R_{>0}(R/I, R/J)=0$ or not ... | |
| S Oct 5, 2022 at 19:36 | review | First answers | |||
| Oct 5, 2022 at 19:44 | |||||
| S Oct 5, 2022 at 19:36 | history | answered | Nawaj | CC BY-SA 4.0 |