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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