The Stacks project

Proof. Omitted. Hints: The associativity constraint is the isomorphism of Lemma 15.59.15 and the commutativity constraint is the isomorphism of Lemma 15.59.14. Having said this the commutativity of various diagrams follows from the corresponding result for the category of complexes of $R$-modules, see Section 15.58. $\square$

Proof. By descending induction on $a = \min (n_ i; i = 1, \ldots , N)$. If $F^ n = 0$ for $n \geq a$, then the result is true with $G^\bullet $ equal to the zero complex. In general, after renumbering we may assume there exists an $1 \leq r \leq N$ such that $n_1 = \ldots = n_ r = a$ and $n_ i > a$ for $i > r$. Choose a basis $b_ j, j \in J$ for $F^ a$. We can choose a finite subset $J' \subset J$ such that $f_ i \in \bigoplus _{j \in J'} Rb_ j$ for $i = 1, \ldots , r$. Choose a basis $c_ k, k \in K$ for $F^{a + 1}$. We can choose a finite subset $K' \subset K$ such that $\text{d}_ F^ a(b_ j) \in \bigoplus _{k \in K'} Rc_ k$ for $j \in J'$. Then we can apply the induction hypothesis to find a subcomplex $H^\bullet \subset F^\bullet $ containing $c_ k \in F^{a + 1}$ for $k \in K'$ and $f_ i \in F^{n_ i}$ for $i > r$. Take $G^\bullet $ equal to $H^\bullet $ in degrees $> a$ and equal to $\bigoplus _{j \in J'} Rb_ j$ in degree $a$. $\square$

Proof. If $M$ is perfect, then we can represent $M$ by a bounded complex $M^\bullet $ of finite projective $R$-modules. In this case $M^\bullet $ has a left dual in the category of complexes by Lemma 15.72.2 which is a fortiori a left dual in $D(R)$.

Assume (1). Say $N$, $\eta : R \to M \otimes _ R^\mathbf {L} N$, and $\epsilon : M \otimes _ R^\mathbf {L} N \to R$ is a left dual as in Categories, Definition 4.43.5. Choose a complex $M^\bullet $ representing $M$. Choose a K-flat complexes $N^\bullet $ with flat terms representing $N$, see Lemma 15.59.10. Then $\eta $ is given by a map of complexes

We can write the image of $1$ as a finite sum

with $m_{n, i} \in M^ n$ and $n_{-n, i} \in N^{-n}$. Let $K^\bullet \subset M^\bullet $ be the subcomplex generated by all the elements $m_{n, i}$ and $\text{d}(m_{n, i})$. By our choice of $N^\bullet $ we find that $\text{Tot}(K^\bullet \otimes _ R N^\bullet ) \subset \text{Tot}(M^\bullet \otimes _ R N^\bullet )$ and $\eta (1)$ is in the subcomplex by our choice above. Denote $K$ the object of $D(R)$ represented by $K^\bullet $. Then we see that $\eta $ factors over a map $\tilde\eta : R \longrightarrow K \otimes _ R^\mathbf {L} N$. Since $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_ M$ we conclude that the identity on $M$ factors through $K$ by the commutative diagram

Since $K$ is bounded above it follows that $M \in D^-(R)$. Thus we can represent $M$ by a bounded above complex $M^\bullet $ of free $R$-modules, see for example Derived Categories, Lemma 13.15.4. Write $\eta (1) = \sum \nolimits _ n \sum \nolimits _ i m_{n, i} \otimes n_{-n, i}$ as before. By Lemma 15.126.2 we can find a subcomplex $K^\bullet \subset M^\bullet $ containing all the elements $m_{n, i}$ which is bounded and consists of finite free $R$-modules. As above we find that the identity on $M$ factors through $K$. Since $K$ is perfect we conclude $M$ is perfect too, see Lemma 15.74.5. $\square$

Proof. Assume (2). Consider the object $R\mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ and the composition map

Checking locally we see that this is an isomorphism; we omit the details. Because $D(R)$ is symmetric monoidal we see that $M$ is invertible.

Assume (1). Observe that an invertible object of a monoidal category has a left dual, namely, its inverse. Thus $M$ is perfect by Lemma 15.126.3. Consider a prime ideal $\mathfrak p \subset R$ with residue field $\kappa $. Then we see that $M \otimes _ R^\mathbf {L} \kappa $ is an invertible object of $D(\kappa )$. Clearly this implies that $\dim H^ i(M \otimes _ R^\mathbf {L} \kappa )$ is nonzero exactly for one $i$ and equal to $1$ in that case. By Lemma 15.75.7 this gives (2).

In the proof above we have seen that (a) holds. Let $U_ n \subset \mathop{\mathrm{Spec}}(R)$ be the union of the opens of the form $D(f)$ such that $M_ f \cong R_ f[-n]$. Clearly, $U_ n \cap U_{n'} = \emptyset $ if $n \not= n'$. If $M$ has tor amplitude in $[a, b]$, then $U_ n = \emptyset $ if $n \not\in [a, b]$. Hence we see that we have a product decomposition $R = \prod _{a \leq n \leq b} R_ n$ as in (d) such that $U_ n$ corresponds to $\mathop{\mathrm{Spec}}(R_ n)$, see Algebra, Lemma 10.24.3. Since $D(R) = \prod _{a \leq n \leq b} D(R_ n)$ and similarly for the category of modules parts (b), (c), and (d) follow immediately. $\square$