Lemma 15.64.18. Let $R$ be a coherent ring (Algebra, Definition 10.90.1). Let $K \in D^-(R)$. The following are equivalent

$K$ is $m$-pseudo-coherent,

$H^ m(K)$ is a finite $R$-module and $H^ i(K)$ is coherent for $i > m$, and

$H^ m(K)$ is a finite $R$-module and $H^ i(K)$ is finitely presented for $i > m$.

Thus $K$ is pseudo-coherent if and only if $H^ i(K)$ is a coherent module for all $i$.

**Proof.**
Recall that an $R$-module $M$ is coherent if and only if it is of finite presentation (Algebra, Lemma 10.90.4). This explains the equivalence of (2) and (3). If so and if we choose an exact sequence $0 \to N \to R^{\oplus m} \to M \to 0$, then $N$ is coherent by Algebra, Lemma 10.90.3. Thus in this case, repeating this procedure with $N$ we find a resolution

\[ \ldots \to R^{\oplus n} \to R^{\oplus m} \to M \to 0 \]

by finite free $R$-modules. In other words, $M$ is pseudo-coherent. The equivalence of (1) and (2) follows from this and Lemmas 15.64.10 and 15.64.4. The final assertion follows from the equivalence of (1) and (2) combined with Lemma 15.64.5.
$\square$

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