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).
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$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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.
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