Lemma 10.39.4. A composition of (faithfully) flat ring maps is (faithfully) flat. If $R \to R'$ is (faithfully) flat, and $M'$ is a (faithfully) flat $R'$-module, then $M'$ is a (faithfully) flat $R$-module.

Proof. The first statement of the lemma is a particular case of the second, so it is clearly enough to prove the latter. Let $R \to R'$ be a flat ring map, and $M'$ a flat $R'$-module. We need to prove that $M'$ is a flat $R$-module. Let $N_1 \to N_2 \to N_3$ be an exact complex of $R$-modules. Then, the complex $R' \otimes _ R N_1 \to R' \otimes _ R N_2 \to R' \otimes _ R N_3$ is exact (since $R'$ is flat as an $R$-module), and so the complex $M' \otimes _{R'} \left(R' \otimes _ R N_1\right) \to M' \otimes _{R'} \left(R' \otimes _ R N_2\right) \to M' \otimes _{R'} \left(R' \otimes _ R N_3\right)$ is exact (since $M'$ is a flat $R'$-module). Since $M' \otimes _{R'} \left(R' \otimes _ R N\right) \cong \left(M' \otimes _{R'} R'\right) \otimes _ R N \cong M' \otimes _ R N$ for any $R$-module $N$ functorially (by Lemmas 10.12.7 and 10.12.3), this complex is isomorphic to the complex $M' \otimes _ R N_1 \to M' \otimes _ R N_2 \to M' \otimes _ R N_3$, which is therefore also exact. This shows that $M'$ is a flat $R$-module. Tracing this argument backwards, we can show that if $R \to R'$ is faithfully flat, and if $M'$ is faithfully flat as an $R'$-module, then $M'$ is faithfully flat as an $R$-module. $\square$

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