Lemma 15.62.8. Let $R$ be a ring. Let $K^\bullet$ be a complex of $R$-modules. Let $m \in \mathbf{Z}$.

1. If $H^ i(K^\bullet ) = 0$ for all $i \geq m$, then $K^\bullet$ is $m$-pseudo-coherent.

2. If $H^ i(K^\bullet ) = 0$ for $i > m$ and $H^ m(K^\bullet )$ is a finite $R$-module, then $K^\bullet$ is $m$-pseudo-coherent.

3. If $H^ i(K^\bullet ) = 0$ for $i > m + 1$, the module $H^{m + 1}(K^\bullet )$ is of finite presentation, and $H^ m(K^\bullet )$ is of finite type, then $K^\bullet$ is $m$-pseudo-coherent.

Proof. It suffices to prove (3). Set $M = H^{m + 1}(K^\bullet )$. Note that $\tau _{\geq m + 1}K^\bullet$ is quasi-isomorphic to $M[- m - 1]$. By Lemma 15.62.4 we see that $M[- m - 1]$ is $m$-pseudo-coherent. Since we have the distinguished triangle

$(\tau _{\leq m}K^\bullet , K^\bullet , \tau _{\geq m + 1}K^\bullet )$

(Derived Categories, Remark 13.12.4) by Lemma 15.62.2 it suffices to prove that $\tau _{\leq m}K^\bullet$ is pseudo-coherent. By assumption $H^ m(\tau _{\leq m}K^\bullet )$ is a finite type $R$-module. Hence we can find a finite free $R$-module $E$ and a map $E \to \mathop{\mathrm{Ker}}(d_ K^ m)$ such that the composition $E \to \mathop{\mathrm{Ker}}(d_ K^ m) \to H^ m(\tau _{\leq m}K^\bullet )$ is surjective. Then $E[-m] \to \tau _{\leq m}K^\bullet$ witnesses the fact that $\tau _{\leq m}K^\bullet$ is $m$-pseudo-coherent. $\square$

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