Lemma 15.64.4 (064T)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.64: Pseudo-coherent modules, I
Lemma 15.64.4 (cite)

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Lemma 15.64.4. Let $R$ be a ring. Let $M$ be an $R$-module. Then

$M$ is $0$-pseudo-coherent if and only if $M$ is a finite $R$-module,
$M$ is $(-1)$-pseudo-coherent if and only if $M$ is a finitely presented $R$-module,
$M$ is $(-d)$-pseudo-coherent if and only if there exists a resolution

\[ R^{\oplus a_ d} \to R^{\oplus a_{d - 1}} \to \ldots \to R^{\oplus a_0} \to M \to 0 \]

of length $d$, and
$M$ is pseudo-coherent if and only if there exists an infinite resolution

\[ \ldots \to R^{\oplus a_1} \to R^{\oplus a_0} \to M \to 0 \]

by finite free $R$-modules.

Proof. If $M$ is of finite type (resp. of finite presentation), then $M$ is $0$-pseudo-coherent (resp. $(-1)$-pseudo-coherent) as follows from the discussion preceding Definition 15.64.1. Conversely, if $M$ is $0$-pseudo-coherent, then $M = H^0(M[0])$ is of finite type by Lemma 15.64.3. If $M$ is $(-1)$-pseudo-coherent, then it is $0$-pseudo-coherent hence of finite type. Choose a surjection $R^{\oplus a} \to M$ and denote $K = \mathop{\mathrm{Ker}}(R^{\oplus a} \to M)$. By Lemma 15.64.2 we see that $K$ is $0$-pseudo-coherent, hence of finite type, whence $M$ is of finite presentation.

To prove the third and fourth statement use induction and an argument similar to the above (details omitted). $\square$

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