Characterisation of faithfully flat homomorphisms

Question

Let $\phi :R \rightarrow S$ be a ring homomorphism. Then $S$ is faithfully flat iff $\phi$ is injective and $S/\phi (R)$ is $R$ flat.

I have been able to show that $\phi$ is injective if $S$ is faithfully flat.

Answer 1

Since you already have shown that $\phi$ is injective, let’s assume that $R \subseteq S$.

To show that $S$ is faithfully flat, I will use the equivalence that

i) $S$ is faithfully flat;
ii) $S$ is flat and for any nonzero $R$-module $N$, $N \otimes_R S \neq 0$.

Consider the exact sequence $$ 0 \to R \to S \to R/S \to 0. $$ The long exact seqeucne of Tor shows that
a) if $R/S$ is $R$-flat, then $S$ is $R$-flat, and
b) if $S$ if $R$-flat, then $\operatorname{Tor}_i^R (N, R/S) = 0$ for any $R$-module $N$ and $i \ge 2$.

Thus, we may reduce your question as follows,

suppose that $R \subseteq S$ and $S$ is $R$-flat, then $S$ is faithfully flat iff $S/R$ is $R$-flat.

Now, we use Item b), which implies that $R/S$ is flat iff $\operatorname{Tor}_1^R (N, S/R) = 0$ for any $R$-module $N$. Then by Item ii), the equivalence in question follows immediately from the exact sequence $$ 0 \to \operatorname{Tor}_1^R(N, R/S) \to N \otimes_R R \to N \otimes_R S. $$ Remember that $N \otimes_R R \cong N$.