Lemma 10.38.9. Let $R$ be a ring. Let $S \to S'$ be a faithfully flat map of $R$-algebras. Let $M$ be a module over $S$, and set $M' = S' \otimes _ S M$. Then $M$ is flat over $R$ if and only if $M'$ is flat over $R$.

Proof. Let $N \to N'$ be an injection of $R$-modules. By the faithful flatness of $S \to S'$ we have

$\mathop{\mathrm{Ker}}(N \otimes _ R M \to N' \otimes _ R M) \otimes _ S S' = \mathop{\mathrm{Ker}}(N \otimes _ R M' \to N' \otimes _ R M')$

Hence the equivalence of the lemma follows from the second characterization of flatness in Lemma 10.38.5. $\square$

Comment #5897 by Jack on

Perhaps this is pedantic, but the stated equality uses only the flatness of $S\to S'$. The fact that $S'\to S$ is faithfully flat allows us to conclude that the term on the left is zero iff it were zero before tensoring with $S'$. thanks!

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