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9 events

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Sep 1, 2021 at 15:30	comment	added	Pavel Čoupek		Torsion-free are flat for Bézout domains, i.e. when all the finitely generated ideals are principal and generated by a non-zero divisor.
Aug 30, 2021 at 15:21	history	edited	Stefan Kohl♦		edited tags
Aug 30, 2021 at 15:20	history	made wiki			Post Made Community Wiki by Stefan Kohl♦
May 18, 2021 at 16:48	comment	added	abx		The rings (not necessarily commutative) for which every flat module is projective are called perfect rings. For commutative rings, it means semi-local + some nilpotence condition on the radical.
May 8, 2021 at 19:30	comment	added	Badam Baplan		A commutative ring has every flat module projective iff $A = \prod A_i$ where each $A_i$ has exactly one prime ideal $P_i$ and $P_i$ is $T$-nilpotent.
May 6, 2021 at 21:38	comment	added	Mohan		Let me assume that the rings are commutative and Noetherian. Since torsion-freeness is nicer for domains, let me also assume domain. Further, if the ring is a field, all of the above are true, let me also assume non-field. Then, the first question means the ring must be a Dedekind domain. No such rings (only fields) for the second. Third, there are probably many (like polynomial rings over a field?).
May 6, 2021 at 15:10	comment	added	Pace Nielsen		The last class of rings goes by the name, happily enough, "projective-free". Cohn's books mention a few classes of (commutative and noncommutative) rings with this property. For the second question you might start with the paper "When is a flat module projective" and search the relevant literature.
May 6, 2021 at 13:03	comment	added	Gro-Tsen		This should probably be community wiki, but I didn't remember how this works: was I supposed to mark the question as such? (I didn't see any checkbox.) Or do I need to ask a moderator to do it?
May 6, 2021 at 13:01	history	asked	Gro-Tsen	CC BY-SA 4.0