Lemma 15.96.9. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. Assume $I_ i(M^\bullet , f)$ is a principal ideal for all $i \in \mathbf{Z}$. Consider the ideal $J(M^\bullet , f) = \sum _ i J_ i(M^\bullet , f)$ of $A/fA$. Consider the set of prime ideals

\begin{align*} E & = \{ f \in \mathfrak p \subset A \mid \mathop{\mathrm{Ker}}(d^ i \bmod f^2)_\mathfrak p \text{ surjects onto } \mathop{\mathrm{Ker}}(d^ i \bmod f)_\mathfrak p \text{ for all }i \in \mathbf{Z}\} \\ & = \{ f \in \mathfrak p \subset A \mid \text{the localizations }\beta _\mathfrak p \text{ of the Bockstein operators are zero}\} \end{align*}

Then we have

1. $J(M^\bullet , f)$ is finitely generated,

2. $A/fA \to C = (A/fA)/J(M^\bullet , f)$ is surjective of finite presentation,

3. $J(M^\bullet , f)_\mathfrak p = 0$ for $\mathfrak p \in E$,

4. if $f \in \mathfrak p$ and $H^ i(M^\bullet )_\mathfrak p$ is free for all $i \in \mathbf{Z}$, then $\mathfrak p \in E$, and

5. the cohomology modules of $\eta _ f M^\bullet \otimes _ A C$ are finite locally free $C$-modules.

Proof. The equality in the definition of $E$ follows from Lemma 15.95.7 and in addition the final statement of that lemma implies part (4).

Part (1) is true because the ideals $J_ i(M^\bullet , f)$ are finitely generated and because $M^\bullet$ is bounded and hence $J_ i(M^\bullet , f)$ is zero for almost all $i$. Part (2) is just a reformulation of part (1).

Proof of (3). By Lemma 15.96.5 we find that $(\eta _ fM)^ i$ is finite locally free of rank $r_ i$ for all $i$. Consider the map

$(1, d^ i) : (\eta _ fM)^ i / f(\eta _ fM)^ i \longrightarrow f^ iM^ i/f^{i + 1}M^ i \oplus f^{i + 1}M^{i + 1}/f^{i + 2}M^{i + 1}$

Pick $\mathfrak p \in E$. By Lemma 15.95.8 and the local freeness of the modules $(\eta _ fM)^ i$ we may write

$\left((\eta _ fM)^ i / f(\eta _ fM)^ i\right)_\mathfrak p = (A/fA)_\mathfrak p^{\oplus m_ i} \oplus (A/fA)_\mathfrak p^{\oplus n_ i}$

compatible with the arrow $(1, d^ i)$ above. By the universal property of the ideal $J_ i(M^\bullet , f)$ we conclude that $J_ i(M^\bullet , f)_\mathfrak p = 0$. Hence $I_\mathfrak p = fA_\mathfrak p$ for $\mathfrak p \in E$.

Proof of (5). Observe that the differential on $\eta _ fM^\bullet$ fits into a commutative diagram

$\xymatrix{ (\eta _ fM)^ i \ar[d] \ar[r] & f^ iM^ i \oplus f^{i + 1}M^{i + 1} \ar[d]^{\left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)} \\ (\eta _ fM)^{i + 1} \ar[r] & f^{i + 1}M^ i \oplus f^{i + 2}M^{i + 2} }$

By construction, after tensoring with $C$, the modules on the left are direct sums of direct summands of the summands on the right. Picture

$\xymatrix{ (\eta _ fM)^ i \otimes _ A C \ar[d] \ar@{=}[r] & K^ i \oplus L^ i \ar[r] \ar[d] & f^ iM^ i \otimes _ A C \oplus f^{i + 1}M^{i + 1} \otimes _ A C \ar[d]^{\left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)} \\ (\eta _ fM)^{i + 1} \otimes _ A C \ar@{=}[r] & K^{i + 1} \oplus L^{i + 1} \ar[r] & f^{i + 1}M^ i \otimes _ A C \oplus f^{i + 2}M^{i + 2} \otimes _ A C }$

where the horizontal arrows are compatible with direct sum decompositions as well as inclusions of direct summands. It follows that the differential identifies $L^ i$ with a direct summand of $K^{i + 1}$ and we conclude that the cohomology of $\eta _ fM^\bullet \otimes _ A C$ in degree $i$ is the module $K^{i + 1}/L^ i$ which is finite projective as desired. $\square$

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