Lemma 10.99.8. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. If $M/IM$ is flat over $R/I$ and $\text{Tor}_1^ R(R/I, M) = 0$ then

1. $M/I^ nM$ is flat over $R/I^ n$ for all $n \geq 1$, and

2. for any module $N$ which is annihilated by $I^ m$ for some $m \geq 0$ we have $\text{Tor}_1^ R(N, M) = 0$.

In particular, if $I$ is nilpotent, then $M$ is flat over $R$.

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 $I$.

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

$\text{Tor}_1^ R(N', M) \to \text{Tor}_1^ R(N, M) \to \text{Tor}_1^ R(N'', M)$

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. $\square$

Comment #8472 by Et on

Maybe note in the almost last paragraph that you are using the snake lemma? Took me a moment to figure out what you were trying to do

Comment #8730 by Kentaro Inoue on

I think that this lemma can be proved more easily. In the first paragraph, (2) is proved in the case $m=1$. In the general case, (2) can be reduced to the case $m=1$ by considering the $I$-adic filtration. Then (1) follows from (2) immediately.

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