Skip to main content

Timeline for Exotic principal ideal domains

Current License: CC BY-SA 4.0

12 events
when toggle format what by license comment
Jun 22, 2022 at 8:13 history edited CommunityBot
replaced http://math.uga.edu/~pete with http://alpha.math.uga.edu/~pete
Jan 10, 2019 at 8:58 comment added Matthieu Romagny Dear Pete, do you have a reason to link to commutative algebra notes (integral.pdf) different from those one finds on your webpage (integral2015.pdf) ?
Jan 10, 2019 at 6:35 history edited Martin Sleziak CC BY-SA 4.0
corrected link - it was hidden under proxy
May 16, 2011 at 13:40 vote accept Qiaochu Yuan
Feb 24, 2011 at 17:48 comment added Pete L. Clark @JSE: Really? Do I use the word "finicky" a lot? Or is it just the kind of word I would use? (If so, you must know me pretty well...)
Feb 24, 2011 at 16:42 comment added JSE As soon as I saw the word "finicky" I knew this was one of yours, Pete.
Feb 24, 2011 at 16:11 comment added Qiaochu Yuan Agh, yes, what I meant to say was "all the Bezout domains I care about are Noetherian anyway."
Feb 24, 2011 at 15:44 comment added Chris Eagle All PIDs are Noetherian. A Bezout domain is Noetherian iff it is a PID.
Feb 24, 2011 at 15:30 comment added Qiaochu Yuan I see. I suppose all the PIDs I care about are Noetherian anyway, so I am totally okay with taking the perspective that Bezout domains are the more fundamental concept.
Feb 24, 2011 at 15:18 comment added Emil Jeřábek With regards to Bézout domains that are not (elementary submodels of) unltraproducts of PIDs, an example which I think is easier to verify is $R=\bigcup_nF[x^{1/n}]$, where $F$ is a field. $R$ is Bézout because any its finitely generated subring is contained in some $F[x^{1/n}]$, which is a PID. OTOH, $x$ is not a unit in $R$, but it has no prime divisor.
Feb 24, 2011 at 15:10 history edited Pete L. Clark CC BY-SA 2.5
added 260 characters in body
Feb 24, 2011 at 15:05 history answered Pete L. Clark CC BY-SA 2.5