Timeline for constructive Serre classes
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Feb 11, 2016 at 17:03 | comment | added | Mike Shulman | And I agree with Ingo's interpretation of "finitely generated". Another way to say it would be that it has a finitely indexed subset in terms of which every other element can be written. | |
| Feb 11, 2016 at 17:02 | comment | added | Mike Shulman | @IngoBlechschmidt okay, thanks! I updated the question. | |
| Feb 11, 2016 at 17:02 | history | edited | Mike Shulman | CC BY-SA 3.0 |
updated to ask the real question
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| Feb 11, 2016 at 11:27 | comment | added | Ingo Blechschmidt | @David I would say a quotient of $\mathbb{Z}^n$, where $n$ is a natural number. Similarly, a finitely presented abelian group is a cokernel of a linear map $\mathbb{Z}^m \to \mathbb{Z}^n$. For those cokernels we constructively have the structure theorem (using the Smith normal form). Also this notion of finitely presented coincides with the general categorical notion of a compact object in a category. | |
| Feb 11, 2016 at 2:11 | comment | added | David Roberts♦ | What do you mean by finitely generated anyway? Is a quotient of a free abelian group on a subfinitely indexed set? | |
| Feb 10, 2016 at 23:51 | comment | added | Ingo Blechschmidt | @Mike Did so. But be encouraged to not accept the answer, as the true question is still open. I'm quite interested in a non-cheating answer myself. | |
| Feb 10, 2016 at 23:49 | answer | added | Ingo Blechschmidt | timeline score: 4 | |
| Feb 7, 2016 at 15:36 | comment | added | Mike Shulman | @IngoBlechschmidt I suppose that technically answers the question, so if you posted it as an answer I'd accept it. But I'd prefer something more explicit. | |
| Feb 7, 2016 at 14:13 | comment | added | Ingo Blechschmidt | Any subclass $\mathcal{C}$ of an abelian category determines a smallest Serre class containing it, by iteratively adding (the zero object and) the object $Y$ for any exact sequence $X \to Y \to Z$ where $X$ and $Z$ are objects of the previous stage. Note the missing zeros; alternatively, one can iteratively add the zero object, subobjects, quotients, and extensions. Anyway, this construction can in particular be performed with the class $\mathcal{C}$ of finitely generated abelian groups. Classically, its closure will coincide with $\mathcal{C}$, as $\mathcal{C}$ is already a Serre class. | |
| Jun 5, 2013 at 23:27 | comment | added | Mike Shulman | Hmm, well, it should still be true constructively that a quotient of a subgroup of a group is also the subgroup of a quotient, and vice versa, right? So the class of subquotients of finitely generated groups will be closed under quotients and subgroups. I'm not so sure about extensions, though. | |
| Jun 1, 2013 at 19:12 | comment | added | Andrej Bauer | So if you take quotients of subgroups of finitely generated groups, how far are we from a Serre class? | |
| Jun 1, 2013 at 19:07 | comment | added | Andrej Bauer | Just one word of warning. When working with constructive groups it makes sense to consider antisubgroups as well as subgroups. | |
| May 30, 2013 at 5:21 | comment | added | Mike Shulman | Yeah, I specifically don't want to change the definition of Serre class. I don't think the application I have in mind (spectral sequences) will yield any decidability conditions. | |
| May 30, 2013 at 0:15 | comment | added | François G. Dorais | I was about to ask something along the lines of Andrej's comment but his formulation is better than mine. I have a hard time imagining what quotients by non-decidable subgroups. What does $\mathbb{R}/\mathbb{Q}$ look like constructively? | |
| May 29, 2013 at 23:49 | comment | added | Andrej Bauer | Silly answer: define a Serre class to be one closed under decidable subgroups and quotients by decidable normal subgroups. But you probably need fairly arbitrary quotients, yes? | |
| May 29, 2013 at 21:37 | history | asked | Mike Shulman | CC BY-SA 3.0 |