Timeline for derived tensor product and finite projective dimension
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 6, 2021 at 23:53 | vote | accept | strat | ||
| May 24, 2021 at 11:17 | comment | added | მამუკა ჯიბლაძე | Oh I see now, you just split out the redundant stuff, thanks! | |
| May 24, 2021 at 10:55 | comment | added | Maxime Ramzi | @მამუკაჯიბლაძე Sure, but I don't think this changes the fact that the Stacks lemma is applicable, since it's an existence statement, and a statement about representatives. Namely, if $X\otimes k$ is perfect, then it will have a representative of the form $E$ as in the Stacks lemma, and then you're rolling | |
| May 24, 2021 at 10:50 | comment | added | მამუკა ჯიბლაძე | Well you do have lots of acyclic complexes over a field too, don't you? Sure they split into sums of shifts of $...\to0\to V\to V\to0\to...$, but you still can add these to $X$ the same | |
| May 24, 2021 at 10:39 | comment | added | Maxime Ramzi | Well, I should say that in general "a bounded complex of finite stably free modules" is stronger than perfect but not over a field | |
| May 24, 2021 at 10:27 | comment | added | Maxime Ramzi | In fact you don't even need $X$ to be bounded below (which is obvious from my proof, I don't know why I added it), just $X\otimes^L_R k$ | |
| May 24, 2021 at 10:26 | comment | added | Maxime Ramzi | @მამუკაჯიბლაძე I think so, together with the Nakayama lemma. You need to replace Stacks' "bounded above" with bounded below (cohomological vs homological), and then if E is bounded in both directions (corresponding to perfectness, as E is a complex of finite stably free modules), then so must P be (because $P\otimes R/I\cong E$ - an honest isomorphism, if I'm reading Stacks correctly - with Nakayama's lemma). And therefore $P$ is perfect as well (it's a bounded complex of finite stably free modules). And Stacks says that $P$ is quasi-isomorphic to $K$ | |
| May 24, 2021 at 9:22 | comment | added | მამუკა ჯიბლაძე | Sorry I became confused. So does after all the Proposition 1 here follow from Stacks' Lemma 15.74.4? | |
| May 24, 2021 at 8:48 | comment | added | Z. M | I think that the Proposition 1 is much more important than the original question. What led to my confusion is the general question when every finite projective module of a quotient ring lifts to a finite projective module of the ring itself. I know the result for Henselian pairs, and the exact stupid thing that hovered in my mind was that this might not be true for $(R,\mathfrak m)$, but ignored that modules over fields are free... | |
| May 24, 2021 at 7:37 | comment | added | Maxime Ramzi | @Z.M. : No problem ! the conclusion also uses that we are over a field for Künneth and a local ring (for $x\otimes y = 0$ to imply $x=0$ or $y=0$). So the more general intermediary results might hold, but not the result in the end anyways | |
| May 24, 2021 at 7:34 | comment | added | Z. M | Thanks for clarification. I did nkt take into account that the residue is a field (what I wrote by $R$ is in fact $k$), but just thinking about the situation of modding out an ideal in the Jacobson radical... | |
| May 24, 2021 at 7:19 | comment | added | Maxime Ramzi | @Z.M. : I'm not inducting on the homology groups of $X$, but those of $X\otimes_R^L k$, and there I can assume I have a bijection | |
| May 23, 2021 at 23:55 | comment | added | Jeremy Rickard | @Z.M $R$ is local, so every perfect object is isomorphic to a bounded complex of finite free modules, isn’t it? | |
| May 23, 2021 at 23:03 | comment | added | Z. M | @JeremyRickard In case that you are interested: I was mistaken. See above. | |
| May 23, 2021 at 22:55 | comment | added | Z. M | This argument seems flawed: $Y$ does not necessarily have less nonzero homology groups. Consider the case that $X$ is finite projective with a surjection $f\colon R^n\to X$. The cofiber will create a nonzero homology if $f$ is not injective. | |
| May 23, 2021 at 22:51 | comment | added | Z. M | @მამუკაჯიბლაძე I was incorrect. I mistakenly thought that every perfect object has a bounded chain complex of finite free modules as a representative. | |
| May 23, 2021 at 21:37 | comment | added | Jeremy Rickard | @მამუკაჯიბლაძე The Stacks Project lemma is using cohomological notation (i.e., cochain complexes), but the OP and Z. M. are using homological notation (i.e., chain complexes), so they both involve the same sort of boundedness. | |
| May 23, 2021 at 19:31 | comment | added | მამუკა ჯიბლაძე | @Z.M Is it clear that boundedness from below can be maintained in that lemma? | |
| May 23, 2021 at 17:17 | comment | added | Z. M | A reference for your Proposition: stacks.math.columbia.edu/tag/0BCB | |
| May 23, 2021 at 12:46 | history | answered | Maxime Ramzi | CC BY-SA 4.0 |