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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?

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  • $\begingroup$ 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? $\endgroup$
    – Johannes Hahn
    Sep 20, 2021 at 8:47
  • $\begingroup$ I guess my question is if the commutative proof extends to the noncommutative setting. $\endgroup$
    – Tim Montegue
    Sep 20, 2021 at 8:49
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    $\begingroup$ 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. $\endgroup$
    – Johannes Hahn
    Sep 20, 2021 at 8:52

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