Math Genius: 2 out of 3 property for faithfully flat ring maps

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Proposition 1: Let $f : Ato B$ and $g : Bto C$ be ring maps such that $g$ is faithfully flat. Then the composition $gf$ is flat (resp. faithfully flat) if and only if $f$ is flat (resp. faithfully flat).

Proof: Certainly flatness (resp. faithful flatness) of $f$ implies flatness (resp. faithful flatness) of $gf.$

Conversely, suppose that $gf$ is flat, and let $M’to Mto M”$ be an exact sequence of $A$-modules. Then flatness of $gf$ implies that $M’otimes_A Cto Motimes_A Cto M”otimes_A C$ is exact as well, whence by faithful flatness of $g$ we have that $M’otimes_A Bto Motimes_A Bto M”otimes_A B$ is exact.

If $gf$ is faithfully flat, and $M’to Mto M”$ is a sequence of $A$-modules such that the composition $M’to M”$ is $0,$ then faithful flatness of $gf$ implies that $M’to Mto M”$ is exact if and only if $M’otimes_A Cto Motimes_A Cto M”otimes_A C$ is. Faithful flatness of $g$ implies that this is exact if and only if $M’otimes_A Bto Motimes_A Bto M”otimes_A B$ is. $square$

This result has been surprisingly hard to track down in the literature, despite the simplicity of the proof. (The proof and statement of proposition 1 in the flat case can be found here; I do not know of anywhere that the version of proposition 1 for faithful flatness exists, although I’m sure I just haven’t searched carefully enough.) Note that this proposition is not true only assuming flatness of $g,$ as evidenced by the composition $k[t^2, t^3]to k[t]to k(t).$

I was originally interested in the situation where we assume that the composition is flat, and use that to deduce that $f$ is flat. However, I was wondering whether the following stronger proposition is true.

Proposition 2: Let $f : Ato B$ and $g : Bto C$ be ring maps such that the composition $gf : Ato C$ is faithfully flat. Then $f$ is faithfully flat.

Disclosure: I asked this question in the hopes that this 2 out of 3 property for faithfully flat morphisms will more easily searchable for anyone trying to find a result of this type in the future.

Indeed, Proposition 2 is true.

Proof: Recall that $Ato B$ is faithfully flat if and only if the canonical map $Nto Notimes_A B$ is injective for every $A$-module $N.$ To that end, let $N$ be an $A$-module. We need to prove that $Nto Notimes_A B$ is injective, but we know by faithful flatness of $gf$ that the composition $Nto Notimes_A Bto Notimes_A C$ is injective. Thus, it follows that $Nto Notimes_A B$ must be injective. $square$

Certainly, it need not be true that $Bto C$ be faithfully flat even if both $Ato B$ and $Ato C$ are (consider the composition $kto k[X]to k(X),$ for $k$ a field). So the best we can hope for is Proposition 2.

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