The Stacks project

Proof. We see that (1) $\Rightarrow $ (3) as a finite free module is a finite projective $R$-module. Conversely, suppose $P^\bullet $ is a bounded above complex of finite projective $R$-modules. Say $P^ i = 0$ for $i > n_0$. We choose a direct sum decompositions $F^{n_0} = P^{n_0} \oplus C^{n_0}$ with $F^{n_0}$ a finite free $R$-module, and inductively

for $n \leq n_0$ with $F^{n_0}$ a finite free $R$-module. As a complex $F^\bullet $ has maps $F^{n - 1} \to F^ n$ which agree with $P^{n - 1} \to P^ n$, induce the identity $C^ n \to C^ n$, and are zero on $C^{n - 1}$. The map $F^\bullet \to P^\bullet $ is a quasi-isomorphism (even a homotopy equivalence) and hence (3) implies (1).

Assume (1). Let $E^\bullet $ be a bounded above complex of finite free $R$-modules and let $E^\bullet \to K^\bullet $ be a quasi-isomorphism. Then the induced maps $\sigma _{\geq m}E^\bullet \to K^\bullet $ from the stupid truncation of $E^\bullet $ to $K^\bullet $ show that $K^\bullet $ is $m$-pseudo-coherent. Hence (1) implies (2).

Assume (2). Since $K^\bullet $ is $0$-pseudo-coherent we see in particular that $K^\bullet $ is bounded above. Let $b$ be an integer such that $H^ i(K^\bullet ) = 0$ for $i > b$. By descending induction on $n \in \mathbf{Z}$ we are going to construct finite free $R$-modules $F^ i$ for $i \geq n$, differentials $d^ i : F^ i \to F^{i + 1}$ for $i \geq n$, maps $\alpha : F^ i \to K^ i$ compatible with differentials, such that (1) $H^ i(\alpha )$ is an isomorphism for $i > n$ and surjective for $i = n$, and (2) $F^ i = 0$ for $i > b$. Picture

The base case is $n = b + 1$ where we can take $F^ i = 0$ for all $i$. Induction step. Let $C^\bullet $ be the cone on $\alpha $ (Derived Categories, Definition 13.9.1). The long exact sequence of cohomology

shows that $H^ i(C^\bullet ) = 0$ for $i \geq n$. By Lemma 15.65.2 we see that $C^\bullet $ is $(n - 1)$-pseudo-coherent. By Lemma 15.65.3 we see that $H^{n - 1}(C^\bullet )$ is a finite $R$-module. In particular, we see that the kernel of $H^ n(F^\bullet ) \to H^ n(K^\bullet )$ is a finite $R$-module. Choose a finite free $R$-module $F^{n - 1}$ and a map $F^{n - 1} \to \mathop{\mathrm{Ker}}(F^ n \to F^{n - 1})$ such that $F^{n - 1}$ surjects onto the kernel of $H^ n(F^\bullet ) \to H^ n(K^\bullet )$. We extend our of complexes to

Note that $\alpha ^{n - 1}$ exists because the image of $F^{n - 1} \to F^ n \to K^ n$ is in the image of $K^{n - 1} \to K^ n$ by construction. Dnote again $C^\bullet $ the cone of this extended map of complexes. At this point we see that we get an exact sequence

In other words, we see that the cokernel of $H^{n - 1}(F^\bullet ) \to H^{n - 1}(K^\bullet )$ is a finite $R$-module, say generated by the classes of $\xi _1, \ldots , \xi _ r \in \mathop{\mathrm{Ker}}(K^{n - 1} \to K^ n)$. Then we replace $F^{n - 1}$ by $F^{n - 1} \oplus R^{\oplus r}$ where the basis elements in the free summand map to zero in $F^ n$ and the to $\xi _ i$ in $K^{n - 1}$. This finishes the proof of the induction step. $\square$