Lemma 15.66.4. Let $R$ be a ring. Let $a \in \mathbf{Z}$ and let $K$ be an object of $D(R)$. The following are equivalent

1. $K$ has tor-amplitude in $[a, \infty ]$, and

2. $K$ is quasi-isomorphic to a K-flat complex $E^\bullet$ whose terms are flat $R$-modules with $E^ i = 0$ for $i \not\in [a, \infty ]$.

Proof. The implication (2) $\Rightarrow$ (1) is immediate. Assume (1) holds. First we choose a K-flat complex $K^\bullet$ with flat terms representing $K$, see Lemma 15.59.10. For any $R$-module $M$ the cohomology of

$K^{n - 1} \otimes _ R M \to K^ n \otimes _ R M \to K^{n + 1} \otimes _ R M$

computes $H^ n(K \otimes _ R^\mathbf {L} M)$. This is always zero for $n < a$. Hence if we apply Lemma 15.66.2 to the complex $\ldots \to K^{a - 1} \to K^ a \to K^{a + 1}$ we conclude that $N = \mathop{\mathrm{Coker}}(K^{a - 1} \to K^ a)$ is a flat $R$-module. We set

$E^\bullet = \tau _{\geq a}K^\bullet = (\ldots \to 0 \to N \to K^{a + 1} \to \ldots )$

The kernel $L^\bullet$ of $K^\bullet \to E^\bullet$ is the complex

$L^\bullet = (\ldots \to K^{a - 1} \to I \to 0 \to \ldots )$

where $I \subset K^ a$ is the image of $K^{a - 1} \to K^ a$. Since we have the short exact sequence $0 \to I \to K^ a \to N \to 0$ we see that $I$ is a flat $R$-module. Thus $L^\bullet$ is a bounded above complex of flat modules, hence K-flat by Lemma 15.59.7. It follows that $E^\bullet$ is K-flat by Lemma 15.59.6. $\square$

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