The Stacks project

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:

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

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

Moreover, these maps define a morphism of distinguished triangles

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. $\square$

Proof. Recall that

by definition, see Cohomology, Section 20.44. Thus parts (4) and (3) are equivalent by Cohomology, Lemma 20.50.5.

Since every perfect complex is pseudo-coherent, it is clear that (2) implies (3).

Assume (1) holds. Then $E \otimes ^\mathbf {L} K$ is pseudo-coherent for every pseudo-coherent $E$, see Cohomology, Lemma 20.47.5. By Lemma 36.30.5 the direct image of such a pseudo-coherent complex is pseudo-coherent and we see that (2) is true.

Part (3) implies (5) because we can take $E = \mathcal{O}_{\mathbf{P}^ n_ A}(d)$ for $d = 0, 1, \ldots , n$.

To finish the proof we have to show that (5) implies (1). Let $P$ be as in (36.20.0.1) and $R$ as in (36.20.0.2). By Lemma 36.20.1 we have an equivalence

Let $M \in D(R)$ be an object such that $M \otimes ^\mathbf {L} P = K$. By Differential Graded Algebra, Lemma 22.35.4 there is an isomorphism

in $D(A)$. Arguing as above we obtain

Using that $P^\vee $ is the direct sum of $\mathcal{O}_{\mathbf{P}^ n_ A}(d)$ for $d = 0, 1, \ldots , n$ and (5) we conclude $R\mathop{\mathrm{Hom}}\nolimits (R, M)$ is pseudo-coherent as a complex of $A$-modules. Of course $M = R\mathop{\mathrm{Hom}}\nolimits (R, M)$ in $D(A)$. Thus $M$ is pseudo-coherent as a complex of $A$-modules. By Lemma 36.34.1 we may represent $M$ by a bounded above complex $F^\bullet $ of finite free $R$-modules. Then $F^\bullet = \bigcup _{p \geq 0} \sigma _{\geq p}F^\bullet $ is a filtration which shows that $F^\bullet $ is a differential graded $R$-module with property (P), see Differential Graded Algebra, Section 22.20. Hence $K = M \otimes ^\mathbf {L}_ R P$ is represented by $F^\bullet \otimes _ R P$ (follows from the construction of the derived tensor functor, see for example the proof of Differential Graded Algebra, Lemma 22.35.3). Since $F^\bullet \otimes _ R P$ is a bounded above complex whose terms are direct sums of copies of $P$ we conclude that the lemma is true. $\square$

Proof. There exists an integer $N$ such that $R\Gamma (X, -) : D_\mathit{QCoh}(\mathcal{O}_ X) \to D(A)$ has cohomological dimension $N$ as explained in Lemma 36.4.1. Let $b \in \mathbf{Z}$ be such that $H^ i(K) = 0$ for $i > b$. Let $E$ be pseudo-coherent on $X$. It suffices to show that $R\Gamma (X, E \otimes ^\mathbf {L} K)$ is $m$-pseudo-coherent for every $m$. Choose an approximation $P \to E$ by a perfect complex $P$ of $(X, E, m - N - 1 - b)$. This is possible by Theorem 36.14.6. Choose a distinguished triangle

in $D_\mathit{QCoh}(\mathcal{O}_ X)$. The cohomology sheaves of $C$ are zero in degrees $\geq m - N - 1 - b$. Hence the cohomology sheaves of $C \otimes ^\mathbf {L} K$ are zero in degrees $\geq m - N - 1$. Thus the cohomology of $R\Gamma (X, C \otimes ^\mathbf {L} K)$ are zero in degrees $\geq m - 1$. Hence

is an isomorphism on cohomology in degrees $\geq m$. By assumption the source is pseudo-coherent. We conclude that $R\Gamma (X, E \otimes ^\mathbf {L} K)$ is $m$-pseudo-coherent as desired. $\square$