Timeline for When is $\Omega^1$ an equivalence?
Current License: CC BY-SA 4.0
17 events
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| Oct 22, 2018 at 20:23 | comment | added | David White | @Mare: No worries. I'd say now that everything has been answered, it's best to just leave it as is. Cheers. David. | |
| Oct 22, 2018 at 20:22 | comment | added | Mare | @DavidWhite Well, I deleted it after you told me that its not wise to ask a new question. And also this other question was somehow included in the first question. If you want I can write the second question again. | |
| Oct 22, 2018 at 20:19 | comment | added | David White | @tj_ Yes, that's right. Which, at the moment I wrote that comment, the OP had edited to ask. Now he's edited again to go back to the original question. Anyway, Jeremy's direct argument below is much nicer than the sketchy citation I gave to an indirect argument of Hovey. | |
| Oct 22, 2018 at 18:55 | comment | added | tj_ | @David: By your 2nd comment: 'not qF implies not triangulated' (i.e. 'triangulated implies qF') and '$\Omega^1$ is an equivalence implies triangulated'. In combination: '$\Omega^1$ is an equivalence implies qF'. Together with your answer below, you have shown that $\Omega^1$ is an equivalence iff R is qF. | |
| Oct 21, 2018 at 21:24 | vote | accept | Mare | ||
| Oct 20, 2018 at 21:26 | history | edited | Mare | CC BY-SA 4.0 |
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| Oct 20, 2018 at 18:30 | answer | added | Jeremy Rickard | timeline score: 5 | |
| Oct 20, 2018 at 13:32 | comment | added | David White | But, if you mean "factor out all modules of finite projective dimension" then the answer is yes. You can work with Gorenstein rings, or indeed any ring, and using either Hovey or the Bravo, Gillespie, Hovey machinery, $\Omega^1$ will be an equivalence. | |
| Oct 20, 2018 at 13:31 | comment | added | David White | Generally speaking, I don't think it's wise to edit a question to ask a new question, after an answer has been given. Usually people start a new question for that sort of thing. But, I'll try to answer. It depends on what you mean by "stable module category" - if you mean "factor out projectives" then this is not triangulated for rings that are not quasi-Frobenius, and indeed $\Omega^1$ is not an equivalence (if it were, that would cause the stable module category to be triangulated, by sec 7.1 of Hovey's book, see also 9.3 in Hovey's cotorsion paper referenced below). | |
| Oct 20, 2018 at 13:00 | history | edited | Mare | CC BY-SA 4.0 |
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| Oct 20, 2018 at 12:41 | comment | added | Mare | Not sure whether the tag "triangulated-categories" fits too well since the stable category is triangulated iff the algebra is quasi-Frobenius. But maybe it is quasi-Frobenius iff $\Omega^1$ is an equivalence iff the stable category is triangulated? | |
| Oct 20, 2018 at 12:16 | history | edited | David White | CC BY-SA 4.0 |
triangulated-categories; edited tags
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| Oct 20, 2018 at 12:15 | answer | added | David White | timeline score: 4 | |
| Oct 20, 2018 at 6:12 | history | edited | Mare | CC BY-SA 4.0 |
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| Oct 19, 2018 at 23:32 | history | edited | Mare | CC BY-SA 4.0 |
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| Oct 19, 2018 at 23:18 | history | edited | Mare | CC BY-SA 4.0 |
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| Oct 19, 2018 at 23:12 | history | asked | Mare | CC BY-SA 4.0 |