Lemma 15.101.7. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. There exists an integer $n > 0$ such that $I^ nM \to M$ factors through the map $I \otimes _ A^\mathbf {L} M \to M$ in $D(A)$.

**Proof.**
This follows from Lemma 15.101.6. It can also been seen directly as follows. Consider the distinguished triangle

By the axioms of a triangulated category it suffices to prove that $I^ nM \to A/I \otimes _ A^\mathbf {L} M$ is zero in $D(A)$ for some $n$. Choose generators $f_1, \ldots , f_ r$ of $I$ and let $K = K_\bullet (A, f_1, \ldots , f_ r)$ be the Koszul complex and consider the factorization $A \to K \to A/I$ of the quotient map. Then we see that it suffices to show that $I^ nM \to K \otimes _ A M$ is zero in $D(A)$ for some $n > 0$. Suppose that we have found an $n > 0$ such that $I^ nM \to K \otimes _ A M$ factors through $\tau _{\geq t}(K \otimes _ A M)$ in $D(A)$. Then the obstruction to factoring through $\tau _{\geq t + 1}(K \otimes _ A M)$ is an element in $\mathop{\mathrm{Ext}}\nolimits ^ t(I^ nM, H_ t(K \otimes _ A M))$. The finite $A$-module $H_ t(K \otimes _ A M)$ is annihilated by $I$. Then by Lemma 15.101.3 we can after increasing $n$ assume this obstruction element is zero. Repeating this a finite number of times we find $n$ such that $I^ nM \to K \otimes _ A M$ factors through $0 = \tau _{\geq r + 1}(K \otimes _ A M)$ in $D(A)$ and we win. $\square$

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