The Stacks project

Proof. By induction on $i$. Base case $i = 1$. The short exact sequence $0 \to I^ n \to A \to A/I^ n \to 0$ determines an injection $\text{Tor}_1^ A(M, A/I^ n) \subset I^ n \otimes _ A M$, see Algebra, Remark 10.75.9. As $M$ is annihilated by $I^ m$ we see that the map $I^ n \otimes _ A M \to I^{n - m} \otimes _ A M$ is zero for $n \geq m$. Hence the result holds with $c = m$.

Induction step. Let $i > 1$ and assume $c$ works for $i - 1$. By More on Algebra, Lemma 15.27.3 applied to $M = A/I^ m$ we can choose $c' \geq 0$ such that $\text{Tor}_ i(A/I^ m, A/I^ n) \to \text{Tor}_ i(A/I^ m, A/I^{n - c'})$ is zero for $n \geq c'$. Let $M$ be annihilated by $I^ m$. Choose a short exact sequence

The corresponding long exact sequence of tors gives an exact sequence

for all integers $n \geq 0$. If $n \geq c + c'$, then the map $\text{Tor}_{i - 1}^ A(S, A/I^ n) \to \text{Tor}_{i - 1}^ A(S, A/I^{n - c})$ is zero and the map $\text{Tor}_ i^ A(A/I^ m, A/I^{n - c}) \to \text{Tor}_ i^ A(A/I^ m, A/I^{n - c - c'})$ is zero. Combined with the short exact sequences this implies the result holds for $i$ with constant $c + c'$. $\square$