Lemma 10.39.9 (0584)—The Stacks project

Lemma 10.39.9. Let $R$ be a ring. Let $S \to S'$ be a flat map of $R$-algebras. Let $M$ be a module over $S$, and set $M' = S' \otimes _ S M$.

If $M$ is flat over $R$, then $M'$ is flat over $R$.
If $S \to S'$ is faithfully flat, 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 flatness of $S \to S'$ we have

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

If $M$ is flat over $R$, then the left hand side is zero and we find that $M'$ is flat over $R$ by the second characterization of flatness in Lemma 10.39.5. If $M'$ is flat over $R$ then we have the vanishing of the right hand side and if in addition $S \to S'$ is faithfully flat, this implies that $\mathop{\mathrm{Ker}}(N \otimes _ R M \to N' \otimes _ R M)$ is zero which in turn shows that $M$ is flat over $R$. $\square$

Comments (3)

Comment #5897 by Jack on January 23, 2021 at 18:41

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!

Comment #5963 by Johan on March 08, 2021 at 21:24

Not pedantic at all. In fact, it just so happened that I needed this lemma when $S \to S'$ is just flat and $M$ is flat over $R$ with conclusion $M'$ being flat over $R$ as well. Thanks! (I am editing this now so I can't put the link here yet.)

Comment #6098 by Johan on April 29, 2021 at 19:28

This was changed here. Thanks again.

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