Proposition 15.78.3. Let $R$ be a ring. For an object $K$ of $D(R)$ the following are equivalent

$K$ is perfect, and

$K$ is a compact object of $D(R)$.

Proposition 15.78.3. Let $R$ be a ring. For an object $K$ of $D(R)$ the following are equivalent

$K$ is perfect, and

$K$ is a compact object of $D(R)$.

**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

\[ \prod \mathop{\mathrm{Hom}}\nolimits _ R(P^ n, E^{n - 1}) \to \prod \mathop{\mathrm{Hom}}\nolimits _ R(P^ n, E^ n) \to \prod \mathop{\mathrm{Hom}}\nolimits _ R(P^ n, E^{n + 1}). \]

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

\[ K^\bullet \longrightarrow \bigoplus \nolimits _{n \geq 0} \tau _{\geq n} K^\bullet \]

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

\[ K^\bullet = \bigcup \nolimits _{n \leq 0} \sigma _{\geq n}K^\bullet \]

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

\[ K^\bullet \xrightarrow {\varphi } L^\bullet \xrightarrow {\psi } K^\bullet \]

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

\[ \pi : L^\bullet \longrightarrow (\bigoplus \nolimits _{\lambda \in \Lambda \setminus \Lambda '} R)[-i] \]

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

\[ K \xrightarrow {\varphi } L \xrightarrow {\psi } K \]

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. $\square$

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## Comments (1)

Comment #7810 by Raffaele Lamagna on