Lemma 10.39.14. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent

1. $M$ is faithfully flat, and

2. $M$ is flat and for all $R$-module homomorphisms $\alpha : N \to N'$ we have $\alpha = 0$ if and only if $\alpha \otimes \text{id}_ M = 0$.

Proof. If $M$ is faithfully flat, then $0 \to \mathop{\mathrm{Ker}}(\alpha ) \to N \to N'$ is exact if and only if the same holds after tensoring with $M$. This proves (1) implies (2). For the other, assume (2). Let $N_1 \to N_2 \to N_3$ be a complex, and assume the complex $N_1 \otimes _ R M \to N_2 \otimes _ R M \to N_3\otimes _ R M$ is exact. Take $x \in \mathop{\mathrm{Ker}}(N_2 \to N_3)$, and consider the map $\alpha : R \to N_2/\mathop{\mathrm{Im}}(N_1)$, $r \mapsto rx + \mathop{\mathrm{Im}}(N_1)$. By the exactness of the complex $-\otimes _ R M$ we see that $\alpha \otimes \text{id}_ M$ is zero. By assumption we get that $\alpha$ is zero. Hence $x$ is in the image of $N_1 \to N_2$. $\square$

Comment #2340 by Federico Scavia on

Typo in the first line of the proof, an $N'$ is missing in the exact sequence.

Comment #4941 by yogesh on

the exact sequence in the first line of the proof is always exact. maybe it should be N in place of $\ker(\alpha)$

Comment #4942 by yogesh on

oops, nevermind. disregard my previous comment

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