The Stacks project

Proof. Assume $M/IM$ is flat over $R/I$ and $\text{Tor}_1^ R(R/I, M) = 0$. Let $N$ be an $R/I$-module. Choose a short exact sequence

By the long exact sequence of $\text{Tor}$ and the vanishing of $\text{Tor}_1^ R(R/I, M)$ we get

But since $K$, $\bigoplus _{i \in I} R/I$, and $N$ are all annihilated by $I$ we see that

As $M/IM$ is flat over $R/I$ we conclude that

is exact. Combining this with the above we conclude that $\text{Tor}_1^ R(N, M) = 0$ for any $R$-module $N$ annihilated by $I$.

Let us prove (2) by induction on $m$. The case $m = 1$ was done in the previous paragraph. For $N$ annihilated by $I^ m$ for $m > 1$ we may choose an exact sequence $0 \to N' \to N \to N'' \to 0$ with $N'$ and $N''$ annihilated by $I^{m - 1}$. For example one can take $N' = IN$ and $N'' = N/IN$. Then the exact sequence

and induction prove the vanishing we want.

Finally, we prove (1). Given $n \geq 1$ we have to show that $M/I^ nM$ is flat over $R/I^ n$. In other words, we have to show that the functor $N \mapsto N \otimes _{R/I^ n} M/I^ nM$ is exact on the category of $R$-modules $N$ annihilated by $I^ n$. However, for such $N$ we have $N \otimes _{R/I^ n} M/I^ nM = N \otimes _ R M$. By the vanishing of $\text{Tor}_1$ in (2) we see that the functor $N \mapsto N \otimes _ R M$ is exact on the category of $N$ annihilated by some power of $I$ and we conclude. $\square$