Lemma 15.96.4. Let $A$ be a ring, let $\mathfrak p \subset A$ be a prime ideal, and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. If $H^ i(M^\bullet )_\mathfrak p$ is free for all $i$, then $I_ i(M^\bullet , f)_\mathfrak p$ is a principal ideal and in fact generated by a power of $f$ for all $i$.

Proof. We may assume $A$ is local with maximal ideal $\mathfrak p$ by Lemma 15.96.3. We may also replace $M^\bullet$ with a quasi-isomorphic complex by Lemma 15.96.1. By our assumption on the freeness of cohomology modules we see that $M^\bullet$ is quasi-isomorphic to the complex whose term in degree $i$ is $H^ i(M^\bullet )$ with vanishing differentials, see for example Derived Categories, Lemma 13.27.9. In other words, we may assume the differentials in the complex $M^\bullet$ are all zero. In this case it is clear that $I_ i(M^\bullet , f) = (f^{r_ i})$ is principal. $\square$

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