Lemma 10.39.14. Let R be a ring. Let M be an R-module. The following are equivalent
M is faithfully flat, and
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.
Lemma 10.39.14. Let R be a ring. Let M be an R-module. The following are equivalent
M is faithfully flat, and
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
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