Lemma 36.34.2. Let $A$ be a ring. Let $n \geq 0$. Let $K \in D_\mathit{QCoh}(\mathcal{O}_{\mathbf{P}^ n_ A})$. The following are equivalent

1. $K$ is pseudo-coherent,

2. $R\Gamma (\mathbf{P}^ n_ A, E \otimes ^\mathbf {L} K)$ is a pseudo-coherent object of $D(A)$ for each pseudo-coherent object $E$ of $D(\mathcal{O}_{\mathbf{P}^ n_ A})$,

3. $R\Gamma (\mathbf{P}^ n_ A, E \otimes ^\mathbf {L} K)$ is a pseudo-coherent object of $D(A)$ for each perfect object $E$ of $D(\mathcal{O}_{\mathbf{P}^ n_ A})$,

4. $R\mathop{\mathrm{Hom}}\nolimits _{\mathbf{P}^ n_ A}(E, K)$ is a pseudo-coherent object of $D(A)$ for each perfect object $E$ of $D(\mathcal{O}_{\mathbf{P}^ n_ A})$,

5. $R\Gamma (\mathbf{P}^ n_ A, K \otimes ^\mathbf {L} \mathcal{O}_{\mathbf{P}^ n_ A}(d))$ is pseudo-coherent object of $D(A)$ for $d = 0, 1, \ldots , n$.

Proof. Recall that

$R\mathop{\mathrm{Hom}}\nolimits _{\mathbf{P}^ n_ A}(E, K) = R\Gamma (\mathbf{P}^ n_ A, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{\mathbf{P}^ n_ A}}(E, K))$

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

$- \otimes ^\mathbf {L}_ R P : D(R) \longrightarrow D_\mathit{QCoh}(\mathcal{O}_{\mathbf{P}^ n_ A})$

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

$R\mathop{\mathrm{Hom}}\nolimits (R, M) = R\mathop{\mathrm{Hom}}\nolimits _{\mathbf{P}^ n_ A}(P, K)$

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

$R\mathop{\mathrm{Hom}}\nolimits _{\mathbf{P}^ n_ A}(P, K) = R\Gamma (\mathbf{P}^ n_ A, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{\mathbf{P}^ n_ A}}(E, K)) = R\Gamma (\mathbf{P}^ n_ A, P^\vee \otimes ^\mathbf {L}_{\mathcal{O}_{\mathbf{P}^ n_ A}} K).$

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. $\square$

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