The Stacks project

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. $\square$

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

is the desired resolution. $\square$

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

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

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

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

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. $\square$

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. $\square$

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. $\square$

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 $A = B = R$). $\square$

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. $\square$

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. $\square$

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

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

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 $\square$

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

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

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

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

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. $\square$

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

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

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. $\square$