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Timeline for Faithful flatness for rings

Current License: CC BY-SA 4.0

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Sep 15, 2021 at 7:05 vote accept Tim Montegue
Sep 14, 2021 at 19:45 comment added Benjamin Steinberg @MaximeRamzi, that's no problem. Arguing directly about flatness is fine. For me the argument that there is no idempotent projecting to the trivial module is more immediate than trying to prove flatness directly but for a single example an abelian group is easiest since H_1 is obviously non-trivial.
Sep 14, 2021 at 19:32 comment added Maxime Ramzi Yes, definitely there are such groups - but since you were just giving an example, finite is enough :) but of course, what's easier for me need not be for everyone else (and need not be the most general example), I just wanted to point out this alternative argument (which was the one I came up with when reading your example, before reading your argument)
Sep 14, 2021 at 19:30 comment added Benjamin Steinberg This question mathoverflow.net/questions/291786/acyclic-finite-groups gives examples of non-trivial groups for which the trivial module is flat.
Sep 14, 2021 at 19:23 comment added Benjamin Steinberg @MaximeRamzi, that's true but my impression was people are usually more familiar with group cohomology than homology. I should add that I don't really need the group to be finite. Finitely generated is good enough to have the trivial module finitely presented. I'm not sure if the trivial module is ever flat for a non trivial group but I didn't want to think too hard
Sep 14, 2021 at 19:19 comment added Maxime Ramzi As a left $R$-module, $M=\bigoplus_{g\in G}\mathbb Z$, so if it were flat, so would be $\mathbb Z$, in other words the orbits functor would be exact - it suffices to take a finite group with nontrivial homology (that seems easier to me than a finitely presented => projective argument, but that might be personal - allow me to mention it anyways :) )
Sep 14, 2021 at 17:25 history answered Benjamin Steinberg CC BY-SA 4.0