Timeline for $M \oplus N$ is of finite type if $M,N$ are of finite type?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 16, 2010 at 12:31 | vote | accept | Martin Brandenburg | ||
| Nov 15, 2010 at 20:28 | answer | added | Sándor Kovács | timeline score: 1 | |
| Nov 15, 2010 at 20:15 | answer | added | Torsten Ekedahl | timeline score: 3 | |
| Nov 15, 2010 at 20:07 | comment | added | Martin Brandenburg | @Kevin: Oh yes! We may reduce to the case of a small abelian category (consider the smallest exact abelian subcategory containing $M,N$ and the $P_i$) and then apply this embedding theorem ... but of course this is cheating! | |
| Nov 15, 2010 at 20:00 | comment | added | Kevin Buzzard | Can one cheat and use some sort of Freyd-Mitchell embedding theorem? If not then I give up :-) | |
| Nov 15, 2010 at 19:57 | comment | added | Kevin Buzzard | Let me think out loud. Given your $P_i$ I want to consider some sort of subobjects of $M$ (the kernel) of the form ((finite sum of $P_i$) intersect $M$). Do we even know that the sum of these things is $M$? | |
| Nov 15, 2010 at 19:55 | comment | added | Kevin Buzzard | @Martin: :-(. Then perhaps it is false in general? Ouch. I thought about it a bit and I can't see an element-free proof either. | |
| Nov 15, 2010 at 19:53 | comment | added | Martin Brandenburg | @Kevin: I know the approach does not work. But also for modules I don't know an element-free proof. | |
| Nov 15, 2010 at 19:52 | comment | added | Kevin Buzzard | @Martin: You seem to have outlined quite a bad way of attacking the problem---your approach seems to not even work in the category of modules over a ring. What happens if you try to generalise the standard proof for the category of modules over a ring, rather than an idea that doesn't work in this setting? | |
| Nov 15, 2010 at 19:45 | comment | added | Harry Gindi | I think that the preferred notion in general is finite presentation, which reduces to finite generation in the Noetherian case. It has a very nice categorical definition, namely that the contravariant hom functor defined by $A$, that is, $Hom(A,-)$ commutes with filtered colimits. | |
| Nov 15, 2010 at 19:09 | history | asked | Martin Brandenburg | CC BY-SA 2.5 |