Timeline for Duality for rank one modules over a number ring
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 5, 2013 at 14:33 | comment | added | Tommaso Centeleghe | @Marguaux: I decided to ask on MO the original question I had, which motivates the one asked here. Here is the link: mathoverflow.net/questions/141340/… | |
| Sep 4, 2013 at 18:55 | comment | added | Marguax | @Filippo: My main point was that it is probably "wrong" to ask for a projective $X$ if $R$ isn't Gorenstein, and that the correct context for the question is that of the appropriate derived category. (I am dubious that a projective $X$ can work when $R$ isn't Gorenstein, but haven't tried to make a proof.) Since we don't know the actual motivating application, it is impossible to judge what is really appropriate to do. | |
| Sep 4, 2013 at 16:31 | comment | added | Filippo Alberto Edoardo | @ Marguaux: By the way, as you say $X$ would be (concentrated in degree $0$ and) projective iff $R$ is Gorenstein. But Tommaso insisted that $X\in\mathcal{M}$ so I think he wants it to be projective. Doesn't it somehow forces $R$ to be Gorenstein? Of course, it depends if satisfying duality for all modules in $\mathcal{M}$ is equivalent to being the dualizing module, which I ignore. | |
| Sep 4, 2013 at 16:21 | comment | added | Tommaso Centeleghe | Thanks for the nice answer and for giving my question some more appropriate context (I got the question from looking at subgroups of ordinary abelian varieties over a finite field. In the situation I had in mind R is Z[\pi], where \pi is an ordinary Weil-number). | |
| Sep 4, 2013 at 15:14 | comment | added | Filippo Alberto Edoardo | Marguaux: Ok, then I agree. I thought you were claiming something special about the quadratic case. | |
| Sep 4, 2013 at 13:05 | comment | added | Marguax | @Filippo: Completion commutes in an evident manner with module-finite algebras, and beyond the quadratic case there will be tons of non-monogenic orders (so one cannot make a uniform claim of the Gorenstein property based just on degree beyond the quadratic case). | |
| Sep 4, 2013 at 8:55 | comment | added | Filippo Alberto Edoardo | Nice answer, but I don't follow completely the argument for the quadratic case being Gorenstein. I guess you want to use that if $R$ becomes Gorenstein after completion, it was already so: but you are completing at any prime of $\mathcal{O}_{K_0}$, right? Then you use that $\mathcal{O}_{K_0}$ is a DVR and hence any monogenic extension of it is still such; but in the local case not only quadratic ones are monogenic, and also I do not understand how to go back to $R$. | |
| Sep 4, 2013 at 3:34 | review | First posts | |||
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| Sep 4, 2013 at 3:26 | history | edited | Marguax | CC BY-SA 3.0 |
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| Sep 4, 2013 at 3:19 | history | answered | Marguax | CC BY-SA 3.0 |