Skip to main content
20 events
when toggle format what by license comment
Apr 18, 2019 at 11:04 answer added David Corfield timeline score: 4
Mar 5, 2019 at 3:53 vote accept CommunityBot moved from User.Id=30211 by developer User.Id=481663
Feb 27, 2019 at 18:08 comment added user30211 You're right. I changed example 3 now.
Feb 27, 2019 at 18:08 history edited user30211 CC BY-SA 4.0
added 30 characters in body
Feb 27, 2019 at 18:03 comment added Friedrich Knop The statement 3 is not correct. Just take $A=M=\mathbb Z$ and the two filtrations $0\subset M$ and $0\subset 2M\subset M$. There is the primary decomposition but that's different.
Feb 27, 2019 at 17:56 answer added Friedrich Knop timeline score: 14
Feb 27, 2019 at 1:30 comment added user30211 $A$ is a noetherian ring.
Feb 26, 2019 at 22:11 comment added Richard Lyons @DeanYoung: In example 3, what is $A$?
Feb 26, 2019 at 3:00 comment added Benjamin Steinberg I am not so familiar with example 3 to know if the modular lattice result applies
Feb 26, 2019 at 2:19 comment added user30211 That's interesting, Benjamin. Do you think by chance that it could be made to include example 3 above?
Feb 26, 2019 at 1:09 history edited user30211 CC BY-SA 4.0
deleted 67 characters in body
Feb 25, 2019 at 21:32 comment added Benjamin Steinberg I believe many Jordan-Hölder type theorems can be deduced from the version for modular lattices.
Feb 25, 2019 at 21:00 answer added spin timeline score: 17
Feb 25, 2019 at 19:07 history edited user30211 CC BY-SA 4.0
added 316 characters in body
Feb 25, 2019 at 18:56 history edited user30211 CC BY-SA 4.0
added 52 characters in body
Feb 25, 2019 at 18:49 history edited user30211 CC BY-SA 4.0
added 64 characters in body
Feb 25, 2019 at 18:43 history edited user30211 CC BY-SA 4.0
deleted 210 characters in body
Feb 25, 2019 at 18:13 history edited user30211 CC BY-SA 4.0
added 507 characters in body
Feb 25, 2019 at 18:02 history edited user30211 CC BY-SA 4.0
added 1277 characters in body
Feb 25, 2019 at 17:31 history asked user30211 CC BY-SA 4.0