Lemma 15.76.4. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $i \in \mathbf{Z}$. Let $K^\bullet$ be a pseudo-coherent complex of $R$-modules such that $H^ i(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak p)) = 0$. Then there exists an $f \in R$, $f \not\in \mathfrak p$ and a canonical direct sum decomposition

$K^\bullet \otimes _ R R_ f = \tau _{\geq i + 1}(K^\bullet \otimes _ R R_ f) \oplus \tau _{\leq i - 1}(K^\bullet \otimes _ R R_ f)$

in $D(R_ f)$ with $\tau _{\geq i + 1}(K^\bullet \otimes _ R R_ f)$ a perfect complex with tor-amplitude in $[i + 1, \infty ]$.

Proof. This is an often used special case of Lemma 15.76.2. A direct proof is as follows. We may assume that $K^\bullet$ is a bounded above complex of finite free $R$-modules. Let us inspect what is happening in degree $i$:

$\ldots \to K^{i - 2} \to R^{\oplus l} \to R^{\oplus m} \to R^{\oplus n} \to K^{i + 2} \to \ldots$

Let $A$ be the $m \times l$ matrix corresponding to $K^{i - 1} \to K^ i$ and let $B$ be the $n \times m$ matrix corresponding to $K^ i \to K^{i + 1}$. The assumption is that $A \bmod \mathfrak p$ has rank $r$ and that $B \bmod \mathfrak p$ has rank $m - r$. In other words, there is some $r \times r$ minor $a$ of $A$ which is not in $\mathfrak p$ and there is some $(m - r) \times (m - r)$-minor $b$ of $B$ which is not in $\mathfrak p$. Set $f = ab$. Then after inverting $f$ we can find direct sum decompositions $K^{i - 1} = R^{\oplus l - r} \oplus R^{\oplus r}$, $K^ i = R^{\oplus r} \oplus R^{\oplus m - r}$, $K^{i + 1} = R^{\oplus m - r} \oplus R^{\oplus n - m + r}$ such that the module map $K^{i - 1} \to K^ i$ kills of $R^{\oplus l - r}$ and induces an isomorphism of $R^{\oplus r}$ onto the corresponding summand of $K^ i$ and such that the module map $K^ i \to K^{i + 1}$ kills of $R^{\oplus r}$ and induces an isomorphism of $R^{\oplus m - r}$ onto the corresponding summand of $K^{i + 1}$. Thus $K^\bullet$ becomes quasi-isomorphic to

$\ldots \to K^{i - 2} \to R^{\oplus l - r} \to 0 \to R^{\oplus n - m + r} \to K^{i + 2} \to \ldots$

and everything is clear. $\square$

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