The Stacks project

Proof. To see what the statement means, please look at Derived Categories, Definitions 13.6.1 and 13.36.3. It was shown in Lemmas 15.74.4 and 15.74.5 that $D_{perf}(R) \subset D(R)$ is a strictly full, saturated, triangulated subcategory of $D(R)$. Of course $R \in D_{perf}(R)$.

Recall that $\langle R \rangle = \bigcup \langle R \rangle _ n$. To finish the proof we will show that if $M \in D_{perf}(R)$ is represented by

with $M^ i$ finite projective, then $M \in \langle R \rangle _{b - a + 1}$. The proof is by induction on $b - a$. By definition $\langle R \rangle _1$ contains any finite projective $R$-module placed in any degree; this deals with the base case $b - a = 0$ of the induction. In general, we consider the distinguished triangle

By induction the truncated complex $\sigma _{\leq b - 1}M^\bullet $ is in $\langle R \rangle _{b - a}$ and $M_ b[-b]$ is in $\langle R \rangle _1$. Hence $M^\bullet \in \langle R \rangle _{b - a + 1}$ by definition. $\square$

Proof. Consider the short exact sequence of complexes

where the first map is given by $1 - t_ n$ in degree $n$ where $t_ n : L_ n^\bullet \to L_{n + 1}^\bullet $ is the transition map. By Derived Categories, Lemma 13.12.1 this is a distinguished triangle in $D(R)$. Apply the homological functor $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(K, -)$, see Derived Categories, Lemma 13.4.2. Thus a long exact cohomology sequence

Since we have assumed that $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(K, \bigoplus L^\bullet _ n)$ is equal to $\bigoplus \mathop{\mathrm{Hom}}\nolimits _{D(R)}(K, L^\bullet _ n)$ we see that the first map on every row of the diagram is injective (by the explicit description of this map as the sum of the maps induced by $1 - t_ n$). Hence we conclude that $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(K, \mathop{\mathrm{colim}}\nolimits L^\bullet _ n)$ is the cokernel of the first map of the middle row in the diagram above which is what we had to show. $\square$

Proof. Assume $K$ is perfect, i.e., $K$ is quasi-isomorphic to a bounded complex $P^\bullet $ of finite projective modules, see Definition 15.74.1. If $E_ i$ is represented by the complex $E_ i^\bullet $, then $\bigoplus E_ i$ is represented by the complex whose degree $n$ term is $\bigoplus E_ i^ n$. On the other hand, as $P^ n$ is projective for all $n$ we have $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(P^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(P^\bullet , K^\bullet )$ for every complex of $R$-modules $K^\bullet $, see Derived Categories, Lemma 13.19.8. Thus $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(P^\bullet , E^\bullet )$ is the cohomology of the complex

Since $P^\bullet $ is bounded we see that we may replace the $\prod $ signs by $\bigoplus $ signs in the complex above. Since each $P^ n$ is a finite $R$-module we see that $\mathop{\mathrm{Hom}}\nolimits _ R(P^ n, \bigoplus _ i E_ i^ m) = \bigoplus _ i \mathop{\mathrm{Hom}}\nolimits _ R(P^ n, E_ i^ m)$ for all $n, m$. Combining these remarks we see that the map of Derived Categories, Definition 13.37.1 is a bijection.

Conversely, assume $K$ is compact. Represent $K$ by a complex $K^\bullet $ and consider the map

where we have used the canonical truncations, see Homology, Section 12.15. This makes sense as in each degree the direct sum on the right is finite. By assumption this map factors through a finite direct sum. We conclude that $K \to \tau _{\geq n} K$ is zero for at least one $n$, i.e., $K$ is in $D^{-}(R)$.

Since $K \in D^{-}(R)$ and since every $R$-module is a quotient of a free module, we may represent $K$ by a bounded above complex $K^\bullet $ of free $R$-modules, see Derived Categories, Lemma 13.15.4. Note that we have

where we have used the stupid truncations, see Homology, Section 12.15. Hence by Lemma 15.78.2 we see that $1 : K^\bullet \to K^\bullet $ factors through $\sigma _{\geq n}K^\bullet \to K^\bullet $ in $D(R)$. Thus we see that $1 : K^\bullet \to K^\bullet $ factors as

in $D(R)$ for some complex $L^\bullet $ which is bounded and whose terms are free $R$-modules. Say $L^ i = 0$ for $i \not\in [a, b]$. Fix $a, b$ from now on. Let $c$ be the largest integer $\leq b + 1$ such that we can find a factorization of $1_{K^\bullet }$ as above with $L^ i$ is finite free for $i < c$. We will show by induction that $c = b + 1$. Namely, write $L^ c = \bigoplus _{\lambda \in \Lambda } R$. Since $L^{c - 1}$ is finite free we can find a finite subset $\Lambda ' \subset \Lambda $ such that $L^{c - 1} \to L^ c$ factors through $\bigoplus _{\lambda \in \Lambda '} R \subset L^ c$. Consider the map of complexes

given by the projection onto the factors corresponding to $\Lambda \setminus \Lambda '$ in degree $i$. By our assumption on $K$ we see that, after possibly replacing $\Lambda '$ by a larger finite subset, we may assume that $\pi \circ \varphi = 0$ in $D(R)$. Let $(L')^\bullet \subset L^\bullet $ be the kernel of $\pi $. Since $\pi $ is surjective we get a short exact sequence of complexes, which gives a distinguished triangle in $D(R)$ (see Derived Categories, Lemma 13.12.1). Since $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(K, -)$ is homological (see Derived Categories, Lemma 13.4.2) and $\pi \circ \varphi = 0$, we can find a morphism $\varphi ' : K^\bullet \to (L')^\bullet $ in $D(R)$ whose composition with $(L')^\bullet \to L^\bullet $ gives $\varphi $. Setting $\psi '$ equal to the composition of $\psi $ with $(L')^\bullet \to L^\bullet $ we obtain a new factorization. Since $(L')^\bullet $ agrees with $L^\bullet $ except in degree $c$ and since $(L')^ c = \bigoplus _{\lambda \in \Lambda '} R$ the induction step is proved.

The conclusion of the discussion of the preceding paragraph is that $1_ K : K \to K$ factors as

in $D(R)$ where $L$ can be represented by a finite complex of free $R$-modules. In particular we see that $L$ is perfect. Note that $e = \varphi \circ \psi \in \text{End}_{D(R)}(L)$ is an idempotent. By Derived Categories, Lemma 13.4.14 we see that $L = \mathop{\mathrm{Ker}}(e) \oplus \mathop{\mathrm{Ker}}(1 - e)$. The map $\varphi : K \to L$ induces an isomorphism with $\mathop{\mathrm{Ker}}(1 - e)$ in $D(R)$. Hence we finally conclude that $K$ is perfect by Lemma 15.74.5. $\square$

Proof. Choose a finite complex $\overline{P}^\bullet $ of finite projective $R/I$-modules representing $K \otimes _ R^\mathbf {L} R/I$, see Definition 15.74.1. By Lemma 15.75.3 there exists a complex $P^\bullet $ of projective $R$-modules representing $K$ such that $\overline{P}^\bullet = P^\bullet /IP^\bullet $. It follows from Nakayama's lemma (Algebra, Lemma 10.20.1) that $P^\bullet $ is a finite complex of finite projective $R$-modules. $\square$

Proof. It is clear that we may assume replace $R$ by $R/IJ$ and $K$ by $K \otimes _ R^\mathbf {L} R/IJ$. Then $R \to R/(I \cap J)$ is a surjection whose kernel has square zero. Hence by Lemma 15.78.4 it suffices to prove that $K \otimes _ R^\mathbf {L} R/(I \cap J)$ is perfect. Thus we may assume that $I \cap J = 0$.

We prove the lemma in case $I \cap J = 0$. First, we may represent $K$ by a K-flat complex $K^\bullet $ with all $K^ n$ flat, see Lemma 15.59.10. Then we see that we have a short exact sequence of complexes

Note that $K^\bullet /IK^\bullet $ represents $K \otimes ^\mathbf {L}_ R R/I$ by construction of the derived tensor product. Similarly for $K^\bullet /JK^\bullet $ and $K^\bullet /(I + J)K^\bullet $. Note that $K^\bullet /(I + J)K^\bullet $ is a perfect complex of $R/(I + J)$-modules, see Lemma 15.74.9. Hence the complexes $K^\bullet /IK^\bullet $, and $K^\bullet /JK^\bullet $ and $K^\bullet /(I + J)K^\bullet $ have finitely many nonzero cohomology groups (since a perfect complex has finite Tor-amplitude, see Lemma 15.74.2). We conclude that $K \in D^ b(R)$ by the long exact cohomology sequence associated to short exact sequence of complexes displayed above. In particular we assume $K^\bullet $ is a bounded above complex of free $R$-modules (see Derived Categories, Lemma 13.15.4).

We will now show that $K$ is perfect using the criterion of Proposition 15.78.3. Thus we let $E_ j \in D(R)$ be a family of objects parametrized by a set $J$. We choose complexes $E_ j^\bullet $ with flat terms representing $E_ j$, see for example Lemma 15.59.10. It is clear that

is a short exact sequence of complexes. Taking direct sums we obtain a similar short exact sequence

(Note that $- \otimes _ R R/I$ commutes with direct sums.) This short exact sequence determines a distinguished triangle in $D(R)$, see Derived Categories, Lemma 13.12.1. Apply the homological functor $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(K, -)$ (see Derived Categories, Lemma 13.4.2) to get a commutative diagram

with exact columns. It is clear that, for any complex $E^\bullet $ of $R$-modules we have

and similarly for when dividing by $J$ or $I + J$, see Derived Categories, Lemma 13.19.8. Derived Categories. Thus all the horizontal arrows, except for possibly the middle one, are isomorphisms as the complexes $K^\bullet /IK^\bullet $, $K^\bullet /JK^\bullet $, $K^\bullet /(I + J)K^\bullet $ are perfect complexes of $R/I$, $R/J$, $R/(I + J)$-modules, see Proposition 15.78.3. It follows from the $5$-lemma (Homology, Lemma 12.5.20) that the middle map is an isomorphism and the lemma follows by Proposition 15.78.3. $\square$