**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

\[ 0 \to K \to \bigoplus \nolimits _{i \in I} R/I \to N \to 0 \]

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

\[ 0 \to \text{Tor}_1^ R(N, M) \to K \otimes _ R M \to (\bigoplus \nolimits _{i \in I} R/I) \otimes _ R M \to N \otimes _ R M \to 0 \]

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

\begin{align*} K \otimes _ R M & = K \otimes _{R/I} M/IM, \\ (\bigoplus \nolimits _{i \in I} R/I) \otimes _ R M & = (\bigoplus \nolimits _{i \in I} R/I) \otimes _{R/I} M/IM, \\ N \otimes _ R M & = N \otimes _{R/I} M/IM. \end{align*}

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

\[ 0 \to K \otimes _{R/I} M/IM \to (\bigoplus \nolimits _{i \in I} R/I) \otimes _{R/I} M/IM \to N \otimes _{R/} M/IM \to 0 \]

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

In particular, if we apply this to the module $I/I^2$, then we conclude that the sequence

\[ 0 \to I^2 \otimes _ R M \to I \otimes _ R M \to I/I^2 \otimes _ R M \to 0 \]

is short exact. This implies that $I^2 \otimes _ R M \to M$ is injective and it implies that $I/I^2 \otimes _{R/I} M/IM = IM/I^2M$.

Let us prove that $M/I^2M$ is flat over $R/I^2$. Let $I^2 \subset J$ be an ideal. We have to show that $J/I^2 \otimes _{R/I^2} M/I^2M \to M/I^2M$ is injective, see Lemma 10.39.5. As $M/IM$ is flat over $R/I$ we know that the map $(I + J)/I \otimes _{R/I} M/IM \to M/IM$ is injective. The sequence

\[ (I \cap J)/I^2 \otimes _{R/I^2} M/I^2M \to J/I^2 \otimes _{R/I^2} M/I^2M \to (I + J)/I \otimes _{R/I} M/IM \to 0 \]

is exact, as you get it by tensoring the exact sequence $0 \to (I \cap J) \to J \to (I + J)/I \to 0$ by $M/I^2M$. Hence suffices to prove the injectivity of the map $(I \cap J)/I^2 \otimes _{R/I} M/IM \to IM/I^2M$. However, the map $(I \cap J)/I^2 \to I/I^2$ is injective and as $M/IM$ is flat over $R/I$ the map $(I \cap J)/I^2 \otimes _{R/I} M/IM \to I/I^2 \otimes _{R/I} M/IM$ is injective. Since we have previously seen that $I/I^2 \otimes _{R/I} M/IM = IM/I^2M$ we obtain the desired injectivity.

Hence we have proven that the assumptions imply: (a) $\text{Tor}_1^ R(N, M) = 0$ for all $N$ annihilated by $I$, (b) $I^2 \otimes _ R M \to M$ is injective, and (c) $M/I^2M$ is flat over $R/I^2$. Thus we can continue by induction to get the same results for $I^ n$ for all $n \geq 1$.
$\square$

## Comments (2)

Comment #8472 by Et on

Comment #8730 by Kentaro Inoue on