Lemma 15.4.2. Assumptions as in Lemma 15.4.1. Let $Q = \mathop{\mathrm{Coker}}(g)$ and $Q' = \mathop{\mathrm{Coker}}(g')$. Then $\text{Gr}_ I(Q) \cong \text{Gr}_ I(Q')$ as graded $\text{Gr}_ I(A)$-modules.

Proof. In degree $n$ we have $\text{Gr}_ I(Q)_ n = I^ nN/(I^{n + 1}N + g(M) \cap I^ nN)$ and similarly for $Q'$. We claim that

$g(M) \cap I^ nN \subset I^{n + 1}N + g'(M) \cap I^ nN.$

By symmetry (the proof of the claim will only use that $c$ works for $g$ which also holds for $g'$ by the lemma) this will imply that

$I^{n + 1}N + g(M) \cap I^ nN = I^{n + 1}N + g'(M) \cap I^ nN$

whence $\text{Gr}_ I(Q)_ n$ and $\text{Gr}_ I(Q')_ n$ agree as subquotients of $N$, implying the lemma. Observe that the claim is clear for $n \leq c$ as $f = f' \bmod I^{c + 1}N$. If $n > c$, then suppose $b \in g(M) \cap I^ nN$. Write $b = g(a)$ for $a \in I^{n - c}M$. Set $b' = g'(a)$. We have $b - b' = (g - g')(a) \in I^{n + 1}N$ as desired. $\square$

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