# Faithful flatness for rings

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?

• 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 Sep 14, 2021 at 17:17

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.
• 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 19:19