The Stacks project

Lemma 10.39.8. Let $R \to R'$ be a faithfully flat ring map. Let $M$ be a module over $R$, and set $M' = R' \otimes _ R M$. Then $M$ is flat over $R$ if and only if $M'$ is flat over $R'$.

Proof. By Lemma 10.39.7 we see that if $M$ is flat then $M'$ is flat. For the converse, suppose that $M'$ is flat. Let $N_1 \to N_2 \to N_3$ be an exact sequence of $R$-modules. We want to show that $N_1 \otimes _ R M \to N_2 \otimes _ R M \to N_3 \otimes _ R M$ is exact. We know that $N_1 \otimes _ R R' \to N_2 \otimes _ R R' \to N_3 \otimes _ R R'$ is exact, because $R \to R'$ is flat. Flatness of $M'$ implies that $N_1 \otimes _ R R' \otimes _{R'} M' \to N_2 \otimes _ R R' \otimes _{R'} M' \to N_3 \otimes _ R R' \otimes _{R'} M'$ is exact. We may write this as $N_1 \otimes _ R M \otimes _ R R' \to N_2 \otimes _ R M \otimes _ R R' \to N_3 \otimes _ R M \otimes _ R R'$. Finally, faithful flatness implies that $N_1 \otimes _ R M \to N_2 \otimes _ R M \to N_3 \otimes _ R M$ is exact. $\square$

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