Lemma 15.65.5. Let $R$ be a ring. Let $K \in D^-(R)$. The following are equivalent:

1. $K$ is pseudo-coherent,

2. for every family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules, the canonical map

$\alpha : K \otimes _ R^\mathbf {L} \left( \prod \nolimits _\alpha Q_{\alpha } \right) \longrightarrow \prod \nolimits _\alpha (K \otimes _ R^\mathbf {L} Q_{\alpha })$

is an isomorphism in $D(R)$,

3. for every $R$-module $Q$ and every set $A$, the canonical map

$\beta : K \otimes _ R^\mathbf {L} Q^ A \longrightarrow (K \otimes _ R^\mathbf {L} Q)^ A$

is an isomorphism in $D(R)$, and

4. for every set $A$, the canonical map

$\gamma : K \otimes _ R^\mathbf {L} R^ A \longrightarrow K^ A$

is an isomorphism in $D(R)$.

Given $m \in \mathbf{Z}$ the following are equivalent

1. $K$ is $m$-pseudo-coherent,

2. for every family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules, with $\alpha$ as above $H^ i(\alpha )$ is an isomorphism for $i > m$ and surjective for $i = m$,

3. for every $R$-module $Q$ and every set $A$, with $\beta$ as above $H^ i(\beta )$ is an isomorphism for $i > m$ and surjective for $i = m$,

4. for every set $A$, with $\gamma$ as above $H^ i(\gamma )$ is an isomorphism for $i > m$ and surjective for $i = m$.

Proof. If $K$ is pseudo-coherent, then $K$ can be represented by a bounded above complex of finite free $R$-modules. Then the derived tensor products are computed by tensoring with this complex. Also, products in $D(R)$ are given by taking products of any choices of representative complexes. Hence (1) implies (2), (3), (4) by the corresponding fact for modules, see Algebra, Proposition 10.89.3.

In the same way (using the tensor product is right exact) the reader shows that (a) implies (b), (c), and (d).

Assume (4) holds. To show that $K$ is pseudo-coherent it suffices to show that $K$ is $m$-pseudo-coherent for all $m$ (Lemma 15.64.5). Hence to finish then proof it suffices to prove that (d) implies (a).

Assume (d). Let $i$ be the largest integer such that $H^ i(K)$ is nonzero. If $i < m$, then we are done. If not, then from (d) and the description of products in $D(R)$ given above we find that $H^ i(K) \otimes _ R R^ A \to H^ i(K)^ A$ is surjective. Hence $H^ i(K)$ is a finitely generated $R$-module by Algebra, Proposition 10.89.2. Thus we may choose a complex $L$ consisting of a single finite free module sitting in degree $i$ and a map of complexes $L \to K$ such that $H^ i(L) \to H^ i(K)$ is surjective. In particular $L$ satisfies (1), (2), (3), and (4). Choose a distinguished triangle

$L \to K \to M \to L[1]$

Then we see that $H^ j(M) = 0$ for $j \geq i$. On the other hand, $M$ still has property (d) by a small argument which we omit. By induction on $i$ we find that $M$ is $m$-pseudo-coherent. Hence $K$ is $m$-pseudo-coherent by Lemma 15.64.2. $\square$

Comment #5095 by 羽山籍真 on

A typo in the proof: "Assume (d)... If $i", here $d$ should be $m$.

Comment #7419 by Torsten Wedhorn on

In several places (in (2), (3), (4) and two times in the proof) it should be $D(R)$ instead of $D(A)$. There is a (small) risk for confusion here since the index set is also denoted by $A$.

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