## 15.74 Perfect complexes

A perfect complex is a pseudo-coherent complex of finite tor dimension. We will not use this as the definition, but define perfect complexes over a ring directly as follows.

Definition 15.74.1. Let $R$ be a ring. Denote $D(R)$ the derived category of the abelian category of $R$-modules.

An object $K$ of $D(R)$ is *perfect* if it is quasi-isomorphic to a bounded complex of finite projective $R$-modules.

An $R$-module $M$ is *perfect* if $M[0]$ is a perfect object in $D(R)$.

For example, over a Noetherian ring a finite module is perfect if and only if it has finite projective dimension, see Lemma 15.74.3 and Algebra, Definition 10.109.2.

Lemma 15.74.2. Let $K^\bullet $ be an object of $D(R)$. The following are equivalent

$K^\bullet $ is perfect, and

$K^\bullet $ is pseudo-coherent and has finite tor dimension.

If (1) and (2) hold and $K^\bullet $ has tor-amplitude in $[a, b]$, then $K^\bullet $ is quasi-isomorphic to a complex $E^\bullet $ of finite projective $R$-modules with $E^ i = 0$ for $i \not\in [a, b]$.

**Proof.**
It is clear that (1) implies (2), see Lemmas 15.64.5 and 15.66.3. Assume (2) holds and that $K^\bullet $ has tor-amplitude in $[a, b]$. In particular, $H^ i(K^\bullet ) = 0$ for $i > b$. Choose a complex $F^\bullet $ of finite free $R$-modules with $F^ i = 0$ for $i > b$ and a quasi-isomorphism $F^\bullet \to K^\bullet $ (Lemma 15.64.5). Set $E^\bullet = \tau _{\geq a}F^\bullet $. Note that $E^ i$ is finite free except $E^ a$ which is a finitely presented $R$-module. By Lemma 15.66.2 $E^ a$ is flat. Hence by Algebra, Lemma 10.78.2 we see that $E^ a$ is finite projective.
$\square$

Lemma 15.74.3. Let $M$ be a module over a ring $R$. The following are equivalent

$M$ is a perfect module, and

there exists a resolution

\[ 0 \to F_ d \to \ldots \to F_1 \to F_0 \to M \to 0 \]

with each $F_ i$ a finite projective $R$-module.

**Proof.**
Assume (2). Then the complex $E^\bullet $ with $E^{-i} = F_ i$ is quasi-isomorphic to $M[0]$. Hence $M$ is perfect. Conversely, assume (1). By Lemmas 15.74.2 and 15.64.4 we can find resolution $E^\bullet \to M$ with $E^{-i}$ a finite free $R$-module. By Lemma 15.66.2 we see that $F_ d = \mathop{\mathrm{Coker}}(E^{d - 1} \to E^ d)$ is flat for some $d$ sufficiently large. By Algebra, Lemma 10.78.2 we see that $F_ d$ is finite projective. Hence

\[ 0 \to F_ d \to E^{-d+1} \to \ldots \to E^0 \to M \to 0 \]

is the desired resolution.
$\square$

Lemma 15.74.4. Let $R$ be a ring. Let $(K^\bullet , L^\bullet , M^\bullet , f, g, h)$ be a distinguished triangle in $D(R)$. If two out of three of $K^\bullet , L^\bullet , M^\bullet $ are perfect then the third is also perfect.

**Proof.**
Combine Lemmas 15.74.2, 15.64.6, and 15.66.5.
$\square$

Lemma 15.74.5. Let $R$ be a ring. If $K^\bullet \oplus L^\bullet $ is perfect, then so are $K^\bullet $ and $L^\bullet $.

**Proof.**
Follows from Lemmas 15.74.2, 15.64.8, and 15.66.7.
$\square$

Lemma 15.74.6. Let $R$ be a ring. Let $K^\bullet $ be a bounded complex of perfect $R$-modules. Then $K^\bullet $ is a perfect complex.

**Proof.**
Follows by induction on the length of the finite complex: use Lemma 15.74.4 and the stupid truncations.
$\square$

Lemma 15.74.7. Let $R$ be a ring. If $K^\bullet \in D^ b(R)$ and all its cohomology modules are perfect, then $K^\bullet $ is perfect.

**Proof.**
Follows by induction on the length of the finite complex: use Lemma 15.74.4 and the canonical truncations.
$\square$

Lemma 15.74.8. Let $A \to B$ be a ring map. Assume that $B$ is perfect as an $A$-module. Let $K^\bullet $ be a perfect complex of $B$-modules. Then $K^\bullet $ is perfect as a complex of $A$-modules.

**Proof.**
Using Lemma 15.74.2 this translates into the corresponding results for pseudo-coherent modules and modules of finite tor dimension. See Lemma 15.66.12 and Lemma 15.64.11 for those results.
$\square$

Lemma 15.74.9. Let $A \to B$ be a ring map. Let $K^\bullet $ be a perfect complex of $A$-modules. Then $K^\bullet \otimes _ A^{\mathbf{L}} B$ is a perfect complex of $B$-modules.

**Proof.**
Using Lemma 15.74.2 this translates into the corresponding results for pseudo-coherent modules and modules of finite tor dimension. See Lemma 15.66.13 and Lemma 15.64.12 for those results.
$\square$

Lemma 15.74.10. Let $A \to B$ be a flat ring map. Let $M$ be a perfect $A$-module. Then $M \otimes _ A B$ is a perfect $B$-module.

**Proof.**
By Lemma 15.74.3 the assumption implies that $M$ has a finite resolution $F_\bullet $ by finite projective $R$-modules. As $A \to B$ is flat the complex $F_\bullet \otimes _ A B$ is a finite length resolution of $M \otimes _ A B$ by finite projective modules over $B$. Hence $M \otimes _ A B$ is perfect.
$\square$

Lemma 15.74.11. Let $R$ be a ring. If $K$ and $L$ are perfect objects of $D(R)$, then $K \otimes _ R^\mathbf {L} L$ is a perfect object too.

**Proof.**
We can prove this using the definition as follows. We may represent $K$, resp. $L$ by a bounded complex $K^\bullet $, resp. $L^\bullet $ of finite projective $R$-modules. Then $K \otimes _ R^\mathbf {L} L$ is represented by the bounded complex $\text{Tot}(K^\bullet \otimes _ R L^\bullet )$. The terms of this complex are direct sums of the modules $M^ a \otimes _ R L^ b$. Since $M^ a$ and $L^ b$ are direct summands of finite free $R$-modules, so is $M^ a \otimes _ R L^ b$. Hence we conclude the terms of the complex $\text{Tot}(K^\bullet \otimes _ R L^\bullet )$ are finite projective.

Another proof can be given using the characterization of perfect complexes in Lemma 15.74.2 and the corresponding lemmas for pseudo-coherent complexes (Lemma 15.64.16) and for tor amplitude (Lemma 15.66.10 used with $A = B = R$).
$\square$

Lemma 15.74.12. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$ be elements which generate the unit ideal. Let $K^\bullet $ be a complex of $R$-modules. If for each $i$ the complex $K^\bullet \otimes _ R R_{f_ i}$ is perfect, then $K^\bullet $ is perfect.

**Proof.**
Using Lemma 15.74.2 this translates into the corresponding results for pseudo-coherent modules and modules of finite tor dimension. See Lemma 15.66.16 and Lemma 15.64.14 for those results.
$\square$

Lemma 15.74.13. Let $R$ be a ring. Let $K^\bullet $ be a complex of $R$-modules. Let $R \to R'$ be a faithfully flat ring map. If the complex $K^\bullet \otimes _ R R'$ is perfect, then $K^\bullet $ is perfect.

**Proof.**
Using Lemma 15.74.2 this translates into the corresponding results for pseudo-coherent modules and modules of finite tor dimension. See Lemma 15.66.17 and Lemma 15.64.15 for those results.
$\square$

Lemma 15.74.14. Let $R$ be a regular ring. Then

an $R$-module is perfect if and only if it is a finite $R$-module, and

a complex of $R$-modules $K^\bullet $ is perfect if and only if $K^\bullet \in D^ b(R)$ and each $H^ i(K^\bullet )$ is a finite $R$-module.

**Proof.**
Any perfect $R$-module is finite by definition. Conversely, let $M$ be a finite $R$-module. Choose a resolution

\[ \ldots \to F_2 \xrightarrow {d_2} F_1 \xrightarrow {d_1} F_0 \to M \to 0 \]

with $F_ i$ finite free $R$-modules (Algebra, Lemma 10.71.1). Set $M_ i = \mathop{\mathrm{Ker}}(d_ i)$. Denote $U_ i \subset \mathop{\mathrm{Spec}}(R)$ the set of primes $\mathfrak p$ such that $M_{i, \mathfrak p}$ is free; $U_ i$ is open by Algebra, Lemma 10.79.3. We have a exact sequence $0 \to M_{i + 1} \to F_{i + 1} \to M_ i \to 0$. If $\mathfrak p \in U_ i$, then $0 \to M_{i + 1, \mathfrak p} \to F_{i + 1, \mathfrak p} \to M_{i, \mathfrak p} \to 0$ splits. Thus $M_{i + 1, \mathfrak p}$ is finite projective, hence free (Algebra, Lemma 10.78.2). This shows that $U_ i \subset U_{i + 1}$. We claim that $\mathop{\mathrm{Spec}}(R) = \bigcup U_ i$. Namely, for every prime ideal $\mathfrak p$ the regular local ring $R_\mathfrak p$ has finite global dimension by Algebra, Proposition 10.110.1. It follows that $M_{i, \mathfrak p}$ is finite projective (hence free) for $i \gg 0$ for example by Algebra, Lemma 10.109.3. Since the spectrum of $R$ is Noetherian (Algebra, Lemma 10.31.5) we conclude that $U_ n = \mathop{\mathrm{Spec}}(R)$ for some $n$. Then $M_ n$ is a projective $R$-module by Algebra, Lemma 10.78.2. Thus

\[ 0 \to M_ n \to F_ n \to \ldots \to F_1 \to M \to 0 \]

is a bounded resolution by finite projective modules and hence $M$ is perfect. This proves part (1).

Let $K^\bullet $ be a complex of $R$-modules. If $K^\bullet $ is perfect, then it is in $D^ b(R)$ and it is quasi-isomorphic to a finite complex of finite projective $R$-modules so certainly each $H^ i(K^\bullet )$ is a finite $R$-module (as $R$ is Noetherian). Conversely, suppose that $K^\bullet $ is in $D^ b(R)$ and each $H^ i(K^\bullet )$ is a finite $R$-module. Then by (1) each $H^ i(K^\bullet )$ is a perfect $R$-module, whence $K^\bullet $ is perfect by Lemma 15.74.7
$\square$

Lemma 15.74.15. Let $A$ be a ring. Let $K \in D(A)$ be perfect. Then $K^\vee = R\mathop{\mathrm{Hom}}\nolimits _ A(K, A)$ is a perfect complex and $K \cong (K^\vee )^\vee $. There are functorial isomorphisms

\[ L \otimes _ A^\mathbf {L} K^\vee = R\mathop{\mathrm{Hom}}\nolimits _ A(K, L) \quad \text{and}\quad H^0(L \otimes _ A^\mathbf {L} K^\vee ) = \mathop{\mathrm{Ext}}\nolimits _ A^0(K, L) \]

for $L \in D(A)$.

**Proof.**
We can represent $K$ by a complex $K^\bullet $ of finite projective $A$-modules. By Lemma 15.73.2 the object $K^\vee $ is represented by the complex $E^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , A)$. Note that $E^ n = \mathop{\mathrm{Hom}}\nolimits _ A(K^{-n}, A)$ and the differentials of $E^\bullet $ are the transpose of the differentials of $K^\bullet $ up to sign. Observe that $E^\bullet $ is the left dual of $K^\bullet $ in the symmetric monoidal category of complexes of $R$-modules, see Lemma 15.72.2. There is a canonical map

\[ K^\bullet = \text{Tot}(\mathop{\mathrm{Hom}}\nolimits ^\bullet (A, A) \otimes _ A K^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , A), A) \]

which up to sign uses the evaluation map in each degree, see Lemma 15.71.6. (For sign rules see Section 15.72.) Thus this map defines a canonical isomorphism $(K^\vee )^\vee \cong K$ as the double dual of a finite projective module is itself.

The second equality follows from the first by Lemma 15.73.1 and Derived Categories, Lemma 13.19.8 as well as the definition of Ext groups, see Derived Categories, Section 13.27. Let $L^\bullet $ be a complex of $A$-modules representing $L$. By Section 15.72 item (11) there is a canonical isomorphism

\[ \text{Tot}(L^\bullet \otimes _ A E^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , L^\bullet ) \]

of complexes of $A$-modules. This proves the first displayed equality and the proof is complete.
$\square$

slogan
Lemma 15.74.16. Let $A$ be a ring. Let $(K_ n)_{n \in \mathbf{N}}$ be a system of perfect objects of $D(A)$. Let $K = \text{hocolim} K_ n$ be the derived colimit (Derived Categories, Definition 13.33.1). Then for any object $E$ of $D(A)$ we have

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(K, E) = R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_ A K_ n^\vee \]

where $(K_ n^\vee )$ is the inverse system of dual perfect complexes.

**Proof.**
By Lemma 15.74.15 we have $R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_ A K_ n^\vee = R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathrm{Hom}}\nolimits _ A(K_ n, E)$ which fits into the distinguished triangle

\[ R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathrm{Hom}}\nolimits _ A(K_ n, E) \to \prod R\mathop{\mathrm{Hom}}\nolimits _ A(K_ n, E) \to \prod R\mathop{\mathrm{Hom}}\nolimits _ A(K_ n, E) \]

Because $K$ similarly fits into the distinguished triangle $\bigoplus K_ n \to \bigoplus K_ n \to K$ it suffices to show that $\prod R\mathop{\mathrm{Hom}}\nolimits _ A(K_ n, E) = R\mathop{\mathrm{Hom}}\nolimits _ A(\bigoplus K_ n, E)$. This is a formal consequence of (15.73.0.1) and the fact that derived tensor product commutes with direct sums.
$\square$

Lemma 15.74.17. Let $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ be a filtered colimit of rings.

Given a perfect $K$ in $D(R)$ there exists an $i \in I$ and a perfect $K_ i$ in $D(R_ i)$ such that $K \cong K_ i \otimes _{R_ i}^\mathbf {L} R$ in $D(R)$.

Given $0 \in I$ and $K_0, L_0 \in D(R_0)$ with $K_0$ perfect, we have

\[ \mathop{\mathrm{Hom}}\nolimits _{D(R)}(K_0 \otimes _{R_0}^\mathbf {L} R, L_0 \otimes _{R_0}^\mathbf {L} R) = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{D(R_ i)}(K_0 \otimes _{R_0}^\mathbf {L} R_ i, L_0 \otimes _{R_0}^\mathbf {L} R_ i) \]

In other words, the triangulated category of perfect complexes over $R$ is the colimit of the triangulated categories of perfect complexes over $R_ i$.

**Proof.**
We will use the results of Algebra, Lemmas 10.127.5 and 10.127.6 without further mention. These lemmas in particular say that the category of finitely presented $R$-modules is the colimit of the categories of finitely presented $R_ i$-modules. Since finite projective modules can be characterized as summands of finite free modules (Algebra, Lemma 10.78.2) we see that the same is true for the category of finite projective modules. This proves (1) by our definition of perfect objects of $D(R)$.

To prove (2) we may represent $K_0$ by a bounded complex $K_0^\bullet $ of finite projective $R_0$-modules. We may represent $L_0$ by a K-flat complex $L_0^\bullet $ (Lemma 15.59.10). Then we have

\[ \mathop{\mathrm{Hom}}\nolimits _{D(R)}(K_0 \otimes _{R_0}^\mathbf {L} R, L_0 \otimes _{R_0}^\mathbf {L} R) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(K_0^\bullet \otimes _{R_0} R, L_0^\bullet \otimes _{R_0} R) \]

by Derived Categories, Lemma 13.19.8. Similarly for the $\mathop{\mathrm{Hom}}\nolimits $ with $R$ replaced by $R_ i$. Since in the right hand side only a finite number of terms are involved, since

\[ \mathop{\mathrm{Hom}}\nolimits _ R(K_0^ p \otimes _{R_0} R, L_0^ q \otimes _{R_0} R) = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{R_ i}(K_0^ p \otimes _{R_0} R_ i, L_0^ q \otimes _{R_0} R_ i) \]

by the lemmas cited at the beginning of the proof, and since filtered colimits are exact (Algebra, Lemma 10.8.8) we conclude that (2) holds as well.
$\square$

## Comments (6)

Comment #1160 by Matthieu Romagny on

Comment #1177 by Johan on

Comment #5977 by Sandeep on

Comment #5979 by Johan on

Comment #6459 by Owen on

Comment #6467 by Johan on