Timeline for Filtration over tensor product
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2022 at 13:43 | comment | added | Pierre21 | @FunctionOfX I did gave some extra info thanks ! | |
| Dec 13, 2022 at 0:41 | comment | added | FunctionOfX | I know this is an old question, but this may help: arxiv.org/pdf/1306.1359.pdf | |
| Mar 12, 2021 at 15:23 | comment | added | Pierre21 | I must admit I didn't try such a proof cause I don't have any références on it. Basically I just read an article of Deligne (Hodge Mixte theory 2) and Bourbaki. So if you have anything close even in the finite filtration case I would be happy to read it and learn | |
| Mar 12, 2021 at 14:33 | comment | added | R. van Dobben de Bruyn | Ah, I was thinking of the case of finite length filtrations, where you construct it by induction on the length of one of them. Does that work, or is there an issue with this? I'm not sure about the infinite case. I'm not aware of any references, but that doesn't mean they don't exist (this type of auxiliary result can sometimes be hard to find ― it could be in a paper about something completely different). | |
| Mar 12, 2021 at 10:32 | comment | added | Pierre21 | Hello again, I'm sorry to ask you again but I didn't succeeded to prove it by myself, do you have any references of books or articles that prove such a thing ? Thank you very much | |
| Feb 25, 2021 at 0:24 | comment | added | Pierre21 | Okay ! I can add this conditions to my theorem so thank you very much ! | |
| Feb 24, 2021 at 19:10 | comment | added | R. van Dobben de Bruyn | You can make a similar example where $N_0 \supsetneq N_1 \supsetneq N_2$ is also $R \supset I \supset 0$. I think a better condition would be that $M_i/M_{i+1}$ and $N_i/N_{i+1}$ are flat, and then the construction you suggest should probably work. | |
| Feb 24, 2021 at 10:20 | comment | added | Pierre21 | Thanks for the answer. I didn't thought about such examples in the first place. Is there any chance it would be true if we add the condition that M and N are flat ? | |
| Feb 23, 2021 at 23:23 | comment | added | R. van Dobben de Bruyn | Then $M \otimes N \cong R/\mathfrak p$ and $M_0/M_1 \otimes N \cong R/\mathfrak p \cong M_1/M_2 \otimes N$. Since $K_0(R) \cong \operatorname{Cl}(R) \oplus \mathbf Z$ (taking $R/\mathfrak p$ to $(\mathfrak p,0)$ and $R$ to $(0,1)$), we see that the two do not coincide. (There might be a positive statement in general if you use the derived tensor product $M \otimes_R^{\mathbf L} N$.) | |
| Feb 23, 2021 at 23:22 | comment | added | R. van Dobben de Bruyn | If this were possible, it would in particular imply that $[M \otimes_R N] = \sum_{i,j} [\operatorname{Gr}_i M \otimes \operatorname{Gr}_j N]$ in the Grothendieck group of $R$. But this is not always true; for example let $R$ be a Dedekind domain admitting a prime ideal $\mathfrak p \subseteq R$ of order $>1$ in the class group, and let $M_0 \supsetneq M_1 \supsetneq M_2$ be $R \supseteq I \supseteq 0$ and $N_0 \supsetneq N_1$ be $R/\mathfrak p \supseteq 0$. | |
| Feb 23, 2021 at 23:00 | history | asked | Pierre21 | CC BY-SA 4.0 |