Lemma 36.34.1. Let $A$ be a ring. Let $R$ be a (possibly noncommutative) $A$-algebra which is finite free as an $A$-module. Then any object $M$ of $D(R)$ which is pseudo-coherent in $D(A)$ can be represented by a bounded above complex of finite free (right) $R$-modules.
Proof. Choose a complex $M^\bullet $ of right $R$-modules representing $M$. Since $M$ is pseudo-coherent we have $H^ i(M) = 0$ for large enough $i$. Let $m$ be the smallest index such that $H^ m(M)$ is nonzero. Then $H^ m(M)$ is a finite $A$-module by More on Algebra, Lemma 15.64.3. Thus we can choose a finite free $R$-module $F^ m$ and a map $F^ m \to M^ m$ such that $F^ m \to M^ m \to M^{m + 1}$ is zero and such that $F^ m \to H^ m(M)$ is surjective. Picture:
\[ \xymatrix{ & F^ m \ar[d]^\alpha \ar[r] & 0 \ar[d] \ar[r] & \ldots \\ M^{m - 1} \ar[r] & M^ m \ar[r] & M^{m + 1} \ar[r] & \ldots } \]
By descending induction on $n \leq m$ 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 $i > m$. Picture
\[ \xymatrix{ & F^ n \ar[r] \ar[d]^\alpha & F^{n + 1} \ar[d]^\alpha \ar[r] & \ldots \ar[r] & F^ i \ar[d]^\alpha \ar[r] & 0 \ar[d] \ar[r] & \ldots \\ M^{n - 1} \ar[r] & M^ n \ar[r] & M^{n + 1} \ar[r] & \ldots \ar[r] & M^ i \ar[r] & M^{i + 1} \ar[r] & \ldots } \]
The base case is $n = m$ which we've done above. 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 $i \geq n$. Observe that $F^\bullet $ is pseudo-coherent as a complex of $A$-modules because $R$ is finite free as an $A$-module. Hence by More on Algebra, Lemma 15.64.2 we see that $C^\bullet $ is $(n - 1)$-pseudo-coherent as a complex of $A$-modules. By More on Algebra, Lemma 15.64.3 we see that $H^{n - 1}(C^\bullet )$ is a finite $A$-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 $C^{n - 1} = M^{n - 1} \oplus F^ n$ 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
\[ \xymatrix{ & F^{n - 1} \ar[d]^{\alpha ^{n - 1}} \ar[r]_{d^{n - 1}} & F^ n \ar[r] \ar[d]^\alpha & F^{n + 1} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & M^{n - 1} \ar[r] & M^ n \ar[r] & M^{n + 1} \ar[r] & \ldots } \]
Moreover, these maps define a morphism of distinguished triangles
\[ \xymatrix{ (F^ n \to \ldots ) \ar[r] \ar[d] & (F^{n - 1} \to \ldots ) \ar[r] \ar[d] & F^{n - 1} \ar[r] \ar[d]_\beta & (F^ n \to \ldots )[1] \ar[d] \\ (F^ n \to \ldots ) \ar[r] & M^\bullet \ar[r] & C^\bullet \ar[r] & (F^ n \to \ldots )[1] } \]
Hence our choice of $\beta $ implies that the map of complexes $(F^{n - 1} \to \ldots ) \to M^\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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