Lemma 15.101.4. 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 $\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, I^ nM) \to \mathop{\mathrm{Ext}}\nolimits _ A^ p(K, M)$ is contained in $I^{n - c}\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, M)$ for $n \geq c$.
Proof. Choose a bounded above complex $P^\bullet $ of finite free $A$-modules representing $K$. Then $\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, M)$ 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 the result by Lemma 15.101.1 (and much more is true). $\square$
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