The Stacks project

Lemma 15.62.5. Let $R$ be a ring. Let $K^\bullet $ be a complex of $R$-modules. The following are equivalent

  1. $K^\bullet $ is pseudo-coherent,

  2. $K^\bullet $ is $m$-pseudo-coherent for every $m \in \mathbf{Z}$, and

  3. $K^\bullet $ is quasi-isomorphic to a bounded above complex of finite projective $R$-modules.

If (1), (2), and (3) hold and $H^ i(K^\bullet ) = 0$ for $i > b$, then we can find a quasi-isomorphism $F^\bullet \to K^\bullet $ with $F^ i$ finite free $R$-modules and $F^ i = 0$ for $i > b$.

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

\[ F^{n - 1} = P^{n - 1} \oplus C^ n \oplus C^{n - 1} \]

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

\[ \xymatrix{ & F^ n \ar[r] \ar[d]^\alpha & F^{n + 1} \ar[d]^\alpha \ar[r] & \ldots \\ K^{n - 1} \ar[r] & K^ n \ar[r] & K^{n + 1} \ar[r] & \ldots } \]

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 15.62.2 we see that $C^\bullet $ is $(n - 1)$-pseudo-coherent. By Lemma 15.62.3 we see that $H^{n - 1}(C^\bullet )$ is a finite $R$-module. Choose a finite free $R$-module $F^{n - 1}$ and a map $\beta : F^{n - 1} \to C^{n - 1}$ such that the composition $F^{n - 1} \to C^{n - 1} \to C^ n$ is zero and such that $F^{n - 1}$ surjects onto $H^{n - 1}(C^\bullet )$. Since $C^{n - 1} = K^{n - 1} \oplus F^ n$ we can write $\beta = (\alpha ^{n - 1}, -d^{n - 1})$. The vanishing of the composition $F^{n - 1} \to C^{n - 1} \to C^ n$ implies these maps fit into a morphism of complexes

\[ \xymatrix{ & F^{n - 1} \ar[d]^{\alpha ^{n - 1}} \ar[r]_{d^{n - 1}} & F^ n \ar[r] \ar[d]^\alpha & F^{n + 1} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & K^{n - 1} \ar[r] & K^ n \ar[r] & K^{n + 1} \ar[r] & \ldots } \]

Moreover, these maps define a morphism of distinguished triangles

\[ \xymatrix{ (F^ n \to \ldots ) \ar[r] \ar[d] & (F^{n - 1} \to \ldots ) \ar[r] \ar[d] & F^{n - 1} \ar[r] \ar[d]_\beta & (F^ n \to \ldots )[1] \ar[d] \\ (F^ n \to \ldots ) \ar[r] & K^\bullet \ar[r] & C^\bullet \ar[r] & (F^ n \to \ldots )[1] } \]

Hence our choice of $\beta $ implies that the map of complexes $(F^{n - 1} \to \ldots ) \to K^\bullet $ induces an isomorphism on cohomology in degrees $\geq n$ and a surjection in degree $n - 1$. This finishes the proof of the lemma. $\square$


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