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Let $A \to B$ be a homomorphism of commutative rings. Why are the following conditions equivalent?

  • $A \to B$ is faithfully flat.

  • $A \to B$ is injective, flat and $B/A$ is a flat $A$-module.

This should be elementary, but at the moment I don't see how to prove it. I know the usual characterizations of faithfully flat homomorphisms (which can be found in Atiyah-Macdonald for example).

Martin Brandenburg
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    (From a geometric point of view, a morphism of schemes is faithfully flat if it is flat and surjective. An injective map of rings induces a dominant map on spectra, and a flat map (of finite type) is open, so it implies faithful flatness.) – Watson Nov 13 '18 at 21:15
  • @Watson I don't understand your comment. What about $\mathbb Z\to\mathbb Z[1/n]$, which is flat, injective and of finite type? – Yai0Phah Dec 01 '21 at 10:50

3 Answers3

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Because $\rm A \to B$ is injective you have the following sequence $$ 0 \to \rm A \to B \to B/A \to 0.$$

Then because $\rm B/A$ is flat, we have for any $\rm A$-module $\rm M$, $$ 0 \to \rm M \to B \otimes M \to B/A \otimes M \to 0.$$

Hence $ \rm B \otimes M =0 \Rightarrow M = 0$ and so $\rm A \to \rm B$ is faithfully flat.

That's one part.

Damien L
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I seem to get old because I had already come across this result three years ago. It is Lemma 5.5. in Lurie's paper on Tannaka duality. It even works in arbitrary tame abelian $\otimes$-categories.

Martin Brandenburg
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To go the other way, suppose $f\colon A\to B$ is faithfully flat. The map $f\otimes B\colon A\otimes_AB \to B\otimes_AB$ is injective since it has a section, namely the multiplication map from $B\otimes_AB\to B$ given by $b\otimes b'\mapsto bb'$. Which means it is injective. The original map must therefore be injective, since it is so after a faithfully flat base change.

Pramathanath Sastry
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