The Stacks project

Proof. From Lemma 10.99.8 we see that $\text{Tor}_1^ A(M, K) = 0$ for all $A$-modules $K$ annihilated by $I$. Given an $A$-module $K$ annihilated by $I$ we may choose a short exact sequence $0 \to N \to F \to K \to 0$ of $A$-modules where $F$ is a direct sum of copies of $A/fA$. We obtain an exact sequence

Thus using assumption (3) and induction on $i$ we conclude that $\text{Tor}_ i^ A(M, K) = 0$ for all $A$-modules $K$ annihilated by $I$ and $i = 1, \ldots , r + 1$.

Suppose that for some $1 \leq j \leq r$ we have shown that $\text{Tor}_ i^ A(M, K) = 0$ for all $A$-modules $K$ annihilated by $f_1, \ldots , f_ j$ and $i = 1, \ldots , j + 1$. Let $K$ be an $A$-module annihilated by $f_1, \ldots , f_{j - 1}$. We may break the map $f_ j : K \to K$ into two short exact sequences

where $K' = K/K[f_ j] \cong f_ jK$. Let $1 \leq i \leq j$. Applying the exact sequences of Tor we obtain exact sequences

and

Using the vanishing of $\text{Tor}_ i^ A(K[f_ j], M)$ and $\text{Tor}_{i + 1}^ A(K/f_ jK, M)$ we conclude that $f_ j : K \to K$ induces an injective map on $\text{Tor}^ A_ i(M, K)$. In other words, we see that $\text{Tor}^ A_ i(M, K)$ is zero if and only if $\text{Tor}^ A_ i(M, K) \otimes _ A A_{f_ j}$ is zero. By Lemma 10.76.1 we have

The vanishing by the flatness of $M_{f_ j}$ over $A_{f_ j}$ (Lemma 10.75.8). We conclude that $\text{Tor}_ i^ A(M, K) = 0$ for all $A$-modules $K$ annihilated by $f_1, \ldots , f_{j - 1}$ and $i = 1, \ldots , j$. By descending induction on $j$, we conclude that this holds for $j = 0$, i.e., we see 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. $\square$