Lemma 15.89.2. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $K \in D(A)$ be pseudo-coherent and let $M$ be a finite $A$-module. For each $p \in \mathbf{Z}$ there exists an $c$ such that the image of $\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, I^ nM) \to \text{Ext}_ A^ p(K, M)$ is contained in $I^{n - c}\text{Ext}_ A^ p(K, M)$ for $n \geq c$.

Proof. Choose a bounded above complex $P^\bullet$ of finite free $A$-modules representing $K$. Then $\text{Ext}_ A^ p(K, M)$ is the cohomology of

$\mathop{\mathrm{Hom}}\nolimits _ A(F^{-p + 1}, M) \xrightarrow {a} \mathop{\mathrm{Hom}}\nolimits _ A(F^{-p}, M) \xrightarrow {b} \mathop{\mathrm{Hom}}\nolimits _ A(F^{-p - 1}, M)$

and $\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, I^ nM)$ is computed by replacing these finite $A$-modules by $I^ n$ times themselves. Thus we want to prove $\mathop{\mathrm{Ker}}(b) \cap I^ n\mathop{\mathrm{Hom}}\nolimits _ A(F^{-p}, M) \subset I^{n - c}\mathop{\mathrm{Ker}}(b)$. This follows from Artin-Rees (Algebra, Lemma 10.50.2). $\square$

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