Lemma 51.20.3. Let $I$ be an ideal of a Noetherian ring $A$. For all $n \geq 1$ there exists an $m > n$ such that the map $A/I^ m \to A/I^ n$ satisfies the equivalent conditions of Lemma 51.20.2 with $e = \text{cd}(A, I)$.

**Proof.**
Let $\xi \in \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ A(A/I^ n, N)$ be the element constructed in Lemma 51.20.2 part (5). Since $e = \text{cd}(A, I)$ we have $0 = H^{e + 1}_ Z(N) = H^{e + 1}_ I(N) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^{e + 1}(A/I^ m, N)$ by Dualizing Complexes, Lemmas 47.10.1 and 47.8.2. Thus we may pick $m \geq n$ such that $\xi $ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ A(A/I^ m, N)$ as desired.
$\square$

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