Lemma 10.99.16. Let $A$ be a ring, let $M$ be an $A$-module, and let $f \in A$. If
$f$ is a nonzerodivisor on $A$ and $M$,
$M_ f$ is a flat $A_ f$-module, and
$M/fM$ is a flat $A/fA$-module,
Then $M$ is a flat $A$-module. Same with “flat” replaced by “faithfully flat”.
Proof. Since $0 \to A \to A \to A/fA \to 0$ and $0 \to M \to M \to M/fM \to 0$ are exact, we find that $\text{Tor}_ i^ A(M, A/fA) = 0$ for $i = 1$ and $i = 2$. By Lemma 10.99.8 we conclude that $\text{Tor}_1^ A(M, N) = 0$ for all $A$-modules $N$ annihilated by $f$ (this uses the flatness of $M/fM$ over $A/fA$). Given an $A$-module $N$ annihilated by $f$ we may choose a short exact sequence $0 \to N' \to F \to N \to 0$ of $A$-modules where $F$ is a direct sum of copies of $A/fA$. From the exact sequence
we conclude that $\text{Tor}_2^ A(M, N) = 0$ for all $A$-modules $N$ annihilated by $f$. Next, let $K$ be an arbitrary $A$-module. We may break the map $f : K \to K$ into two short exact sequences
where $K' = K/K[f] \cong fK$. Applying the exact sequences of Tor we obtain exact sequences
and
Using the vanishing of $\text{Tor}_1^ A(K[f], M)$ and $\text{Tor}_2^ A(K/fK, M)$ we conclude that $f : K \to K$ induces an injective map on $\text{Tor}^ A_1(M, K)$. In other words, we see that $\text{Tor}^ A_1(M, K)$ is zero if and only if $\text{Tor}^ A_1(M, K) \otimes _ A A_ f$ is zero. By Lemma 10.76.1 we have
The vanishing by the flatness of $M_ f$ over $A_ f$ (Lemma 10.75.8). We conclude that $\text{Tor}_1^ A(M, K) = 0$ for all $A$-modules $K$. Hence $M$ is flat over $A$ by Lemma 10.75.8.
We omit the argument for the case of faithfully flat modules. $\square$
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