Let $R$ be a ring and let $M$ be a right module over $R$. We say that $M$ is faithfully flat as a right module if the functor $M \otimes_R -$ from left $R$-modules to abelian groups that preserves and reflects exact sequences. Faithful flatness for a left $R$-module is defined analogously.

What is an example of an $R$-bimodule that is faithfully flat as a right module, but not faithfully flat as a left module? Or what is an example of an $R$-bimodule that is faithfully flat as a left module, but not faithfully flat as a right module?

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    $\begingroup$ What about $\mathbb ZG$ with G a non-trivial finite group viewed as a trivial left ZG-module and a free right ZG -module? This should be faithfully flat on the right and not even flat on the left. Of course you can dualize $\endgroup$ Sep 14, 2021 at 17:17

1 Answer 1


Let $G$ be a non-trivial finite group and let $R=\mathbb ZG$. We can view $M=\mathbb ZG$ as an $R$-bimodule via the right regular module structure on the right and the trivial module structure on the left (so left multiplication by $g\in G$ fixes $M$). Then $M$ is faithfully flat as a right module because $M\otimes_R ()$ is the underlying abelian group functor. But $M$ is not flat as a left $R$-module. Indeed $M$ is finitely presented (cf. the bar resolution) and so if it were flat it would be projective. But then the trivial module $\mathbb Z$ would be projective which is never the case for a non-trivial finite group as no idempotent $e$ satisfies $ge=e$ for all $g\in G$ in a group algebra unless the order of $G$ is a unit in the coefficient ring.

Of course you can switch the roles of right and left.

  • $\begingroup$ 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 :) ) $\endgroup$ Sep 14, 2021 at 19:19
  • $\begingroup$ @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 $\endgroup$ Sep 14, 2021 at 19:23
  • $\begingroup$ This question mathoverflow.net/questions/291786/acyclic-finite-groups gives examples of non-trivial groups for which the trivial module is flat. $\endgroup$ Sep 14, 2021 at 19:30
  • $\begingroup$ 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) $\endgroup$ Sep 14, 2021 at 19:32
  • $\begingroup$ @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. $\endgroup$ Sep 14, 2021 at 19:45

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