Lemma 15.96.8. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ and $N^\bullet$ be two bounded complexes of finite free $A$-modules representing the same object in $D(A)$. Assume $I_ i(M^\bullet , f)$ is a principal ideal for all $i \in \mathbf{Z}$. Then $J_ i(M^\bullet , f) = J_ i(N^\bullet , f)$ as ideals in $A/fA$.

Proof. Observe that the fact that $I_ i(M^\bullet , f)$ is a principal ideal implies that $I_ i(M^\bullet , f)$ is a principal ideal by Lemma 15.96.1 and hence the statement makes sense. As in the proof of Lemma 15.96.1 we may assume $N^\bullet = M^\bullet \oplus Q^\bullet$ for some trivial complex $Q^\bullet$, i.e.,

$Q^\bullet = \ldots \to 0 \to A \xrightarrow {1} A \to 0 \to \ldots$

where $A$ is placed in degree $j$ and $j + 1$. Since $\eta _ f$ is compatible with direct sums, we see that the map

$(1, d^ i) : (\eta _ fN)^ i / f(\eta _ fN)^ i \longrightarrow f^ iN^ i/f^{i + 1}N^ i \oplus f^{i + 1}N^{i + 1}/f^{i + 2}N^{i + 1}$

is the direct sum of the corresponding map for $M^\bullet$ and for $Q^\bullet$. By the universal property defining the ideals in question, we conclude that $J_ i(N^\bullet , f) = J_ i(M^\bullet , f) + J_ i(Q^\bullet , f)$. Hence it suffices to show that $J_ i(Q^\bullet , f) = 0$ for all $i$. This is a computation that we omit. $\square$

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