Lemma 15.91.1. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$, $N$ be finite $A$-modules with $N$ annihilated by $I$. For each $p > 0$ there exists an $n$ such that the map $\mathop{\mathrm{Ext}}\nolimits _ A^ p(M, N) \to \text{Ext}_ A^ p(I^ nM, N)$ is zero.

**Proof.**
The result is clear for $p = 0$ (with $n = 1$). Choose a short exact sequence $0 \to K \to A^{\oplus t} \to M \to 0$. For $n$ pick a short exact sequence $0 \to L \to A^{\oplus s} \to I^ nM \to 0$. It is clear that we can construct a map of short exact sequences

such that $A^{\oplus s} \to A^{\oplus t}$ has image in $(I^ n)^{\oplus t}$. By Artin-Rees (Algebra, Lemma 10.50.2) we see that $L \to K$ has image contained in $I^{n - c}K$ if $n \geq c$. At this point the exact sequence

and the corresponding sequence for $\mathop{\mathrm{Ext}}\nolimits ^1_ A(I^ nM, N)$ show that the lemma holds for $p = 1$ with $n = c + 1$. Moreover, we see that the result for $p - 1$ and the module $K$ implies the result for $p$ and the module $M$ by the commutativity of the diagram

for $p > 1$. Some details omitted. $\square$

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