Lemma 15.62.18. Let $R$ be a Noetherian ring. Then

1. A complex of $R$-modules $K^\bullet$ is $m$-pseudo-coherent if and only if $K^\bullet \in D^{-}(R)$ and $H^ i(K^\bullet )$ is a finite $R$-module for $i \geq m$.

2. A complex of $R$-modules $K^\bullet$ is pseudo-coherent if and only if $K^\bullet \in D^{-}(R)$ and $H^ i(K^\bullet )$ is a finite $R$-module for all $i$.

3. An $R$-module is pseudo-coherent if and only if it is finite.

Proof. In Algebra, Lemma 10.70.1 we have seen that any finite $R$-module is pseudo-coherent. On the other hand, a pseudo-coherent module is finite, see Lemma 15.62.4. Hence (3) holds. Suppose that $K^\bullet$ is an $m$-pseudo-coherent complex. Then there exists a bounded complex of finite free $R$-modules $E^\bullet$ such that $H^ i(K^\bullet )$ is isomorphic to $H^ i(E^\bullet )$ for $i > m$ and such that $H^ m(K^\bullet )$ is a quotient of $H^ m(E^\bullet )$. Thus it is clear that each $H^ i(K^\bullet )$, $i \geq m$ is a finite module. The converse implication in (1) follows from Lemma 15.62.11 and part (3). Part (2) follows from (1) and Lemma 15.62.5. $\square$

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