Lemma 15.86.1. In Situation 15.84.14. If $A$ is Noetherian, then for every $n$ there exists an $m \geq n$ such that $K_ m^\bullet \to K_ n^\bullet $ factors through the map $K_ m^\bullet \to A/(f_1^ m, \ldots , f_ r^ m)$. In other words, the pro-objects $\{ K_ n^\bullet \} $ and $\{ A/(f_1^ n, \ldots , f_ r^ n)\} $ of $D(A)$ are isomorphic.

**Proof.**
Note that the Koszul complexes have length $r$. Thus the dual of Derived Categories, Lemma 13.12.5 implies it suffices to show that for every $p < 0$ and $n \in \mathbf{N}$ there exists an $m \geq n$ such that $H^ p(K_ m^\bullet ) \to H^ p(K_ n^\bullet )$ is zero. Since $A$ is Noetherian, we see that

is a finite $A$-module. Moreover, the map $K_ m^ p \to K_ n^ p$ is given by a diagonal matrix whose entries are in the ideal $(f_1^{m - n}, \ldots , f_ r^{m - n})$ if $p < 0$ (in fact they are in the $|p|$th power of that ideal). Note that $H^ p(K_ n^\bullet )$ is annihilated by $I = (f_1^ n, \ldots , f_ r^ n)$, see Lemma 15.28.6. Now $I^ t \subset (f_1^{m - n}, \ldots , f_ r^{m - n})$ for $m = n + tr$. Thus by Artin-Rees (Algebra, Lemma 10.50.2) for some $m$ large enough we see that the image of $K_ m^ p \to K_ n^ p$ intersected with $\mathop{\mathrm{Ker}}(K_ n^ p \to K_ n^{p + 1})$ is contained in $I \mathop{\mathrm{Ker}}(K_ n^ p \to K_ n^{p + 1})$. For this $m$ we get the zero map. $\square$

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