Timeline for Example of a ring with non-finitely generated unit group?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 14, 2018 at 22:12 | comment | added | Aurel | @YCor Thanks, I was away for a few days so I could not write up the details. | |
| Jul 12, 2018 at 19:16 | comment | added | YCor | @LeeMosher I expanded the argument in a separate answer. | |
| Jul 12, 2018 at 18:36 | comment | added | Lee Mosher | May I ask someone to be more specific about which bit of the work of Borel and Harish-Chandra is being applied in the comment of @YCor and in this answer? | |
| Jul 6, 2018 at 17:41 | comment | added | Aurel | Not sure what the weakest assumption possible is. I don't have time to think about it now, but I will ask Alex if he has an idea. | |
| Jul 6, 2018 at 13:27 | vote | accept | Denis T | ||
| Jul 6, 2018 at 12:54 | comment | added | YCor | I'd be curious how the assumption that $I$ is finite can be relaxed (for the surjectivity statement). At least, $I$ of finite length as left $A$-module is enough? | |
| Jul 6, 2018 at 11:27 | comment | added | Aurel | @YCor Yes, you are right. I think the general case is worth knowing about, so I will leave the reference like this if you don't mind! | |
| Jul 6, 2018 at 11:23 | comment | added | YCor | Great. Rather it's Lemma 3.6 in this paper that we need (the general case makes use of it). It says that if $A\to B$ is a surjective ring homomorphism with finite kernel, then the group homomorphism $A^\times\to B^\times$ is also surjective (clearly it has finite kernel). | |
| Jul 6, 2018 at 8:51 | history | answered | Aurel | CC BY-SA 4.0 |