Lemma 51.4.11. Let $I$ be an ideal of a Noetherian ring $A$. Set $d = \text{cd}(A, I)$. For $n \geq 1$ choose a finite free resolution

\[ \ldots \to P_ n^{-2} \to P_ n^{-1} \to P_ n^0 \to A/I^ n \to 0 \]

For every $n \geq 1$ there exists an $m > n$ such that $A/I^ m \to A/I^ n$ can be represented by a map of complexes $\alpha ^\bullet : P^\bullet _ m \to P^\bullet _ n$ with $\alpha ^ i = 0$ for $i < -d$.

**Proof.**
We assume $d \geq 1$; the proof in the cases $d = -1, 0$ is omitted. Denote $d_ n^ i : P^ i \to P^{i + 1}$ the differentials. Set $M = \mathop{\mathrm{Im}}(d_ n^{-d - 1} : P_ n^{-d - 1} \to P_ n^{-d})$. Denote $\xi = d_ n^{-d - 1} \in \mathop{\mathrm{Hom}}\nolimits _ A(P_ n^{-d - 1}, M)$. Since $\xi \circ d_ n^{-d - 2} = 0$ we see that $\xi $ defines an element $\overline{\xi }$ of $\mathop{\mathrm{Ext}}\nolimits ^{d + 1}_ A(A/I^ n, M)$. Since $\text{cd}(A, I) = d$ we have $0 = H^{d + 1}_ Z(M) = H^{d + 1}_ I(M) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^{d + 1}(A/I^ m, M)$. Here we have used Dualizing Complexes, Lemmas 47.10.1 and 47.8.2. Thus we may pick $m$ such that $\overline{\xi }$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{d + 1}_ A(A/I^ m, M)$. Choose any map of complexes $\alpha ^\bullet : P_ m^\bullet \to P_ n^\bullet $ representing $A/I^ m \to A/I^ n$. The choice of $m$ implies we can find $\varphi : P_ m^{-d} \to M$ such that

\[ \xi \circ \alpha ^{-d - 1} = \varphi \circ d_ m^{-d - 1} \]

If we think of $\varphi $ as a map $P_ m^{-d} \to P_ n^{-d}$, then we see that $d_ n^{-d} \circ \varphi = 0$ and $\alpha ^{-d} \circ d_ m^{-d - 1} = d_ n^{-d - 1} \circ \alpha ^{-d - 1} = \varphi \circ d_ m^{-d - 1}$. Thus we obtain a map of complexes

\[ \xymatrix{ \ldots \ar[r] & P_ n^{-d - 1} \ar[r] & P_ n^{-d} \ar[r] & P_ n^{-d + 1} \ar[r] & \ldots \\ \ldots \ar[r] & P_ m^{-d - 1} \ar[r] \ar[u]_0 & P_ m^{-d} \ar[r] \ar[u]_{\alpha ^{-d} - \varphi } & P_ m^{-d + 1} \ar[u]_{\alpha ^{-d + 1}} \ar[r] & \ldots } \]

and everything is clear.
$\square$

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