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The Stacks project

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


Comments (4)

Comment #2340 by Federico Scavia on

Typo in the first line of the proof, an 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

Comment #4942 by yogesh on

oops, nevermind. disregard my previous comment

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  • 3 comment(s) on Section 10.39: Flat modules and flat ring maps

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