# Composition of faithfully flat ring extensions

Let $$R$$ be a not necessarily commutative, unital, ring, and for simplicity let module always mean right module. We say that a unital ring extension $$R \hookrightarrow S$$ is flat, or faithfully flat, if $$S$$ is flat, or respectively faithfully flat, as an $$R$$-module.

Is the composition of two flat, faithfully flat, ring extensions again flat, respectively faithfully flat?

Edit: It seems that in the commutative case this is true. See

https://stacks.math.columbia.edu/tag/00H9

for a proof. Does the argument extend to the noncommutative setting?

• Well if the proof extends to the non-commutative case (and I agree that it does) it must be true... What is your question again? Sep 20, 2021 at 8:47
• I guess my question is if the commutative proof extends to the noncommutative setting. Sep 20, 2021 at 8:49
• You didn't write that ;-) Anyway: It clearly does. Of course, you have to make sure to carefully distinguish between "$S$ is (faithfully) flat as a left-$R$-module" and "$S$ is faithfully flat as a right-$R$-module", but the proof goes through if $M'$ and $S$ are both considered as right-modules. Sep 20, 2021 at 8:52