Lemma 10.59.3. Suppose that $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of finite $R$-modules. Then there exists a submodule $N \subset M'$ with finite colength $l$ and $c \geq 0$ such that
and
for all $n \geq c$.
Proof. Note that $M/I^ nM \to M''/I^ nM''$ is surjective with kernel $M' / M' \cap I^ nM$. By the Artin-Rees Lemma 10.51.2 there exists a constant $c$ such that $M' \cap I^ nM = I^{n - c}(M' \cap I^ cM)$. Denote $N = M' \cap I^ cM$. Note that $I^ c M' \subset N \subset M'$. Hence $\text{length}_ R(M' / M' \cap I^ nM) = \text{length}_ R(M'/N) + \text{length}_ R(N/I^{n - c}N)$ for $n \geq c$. From the short exact sequence
and additivity of lengths (Lemma 10.52.3) we obtain the equality
for $n \geq c$. We have $\varphi _{I, M}(n) = \chi _{I, M}(n) - \chi _{I, M}(n - 1)$ and similarly for the modules $M''$ and $N$. Hence we get $\varphi _{I, M}(n) = \varphi _{I, M''}(n) + \varphi _{I, N}(n-c)$ for $n \geq c$. $\square$
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