The Stacks project

Proof. Say $E^ i = 0$ for $i > b$. Assume given $n$ and a morphism of complexes

with $P^ i \in \mathcal{P}$, with $P^ n \to P^{n + 1} \to \ldots \to P^ b$ acyclic in degrees $\geq n + 1$, and with vertical maps inducing isomorphisms $P^ i/IP^ i \to E^ i$. In this situation one can inductively choose isomorphisms $P^ i = Z^ i \oplus Z^{i + 1}$ such that the maps $P^ i \to P^{i + 1}$ are given by $Z^ i \oplus Z^{i + 1} \to Z^{i + 1} \to Z^{i + 1} \oplus Z^{i + 2}$. By property (2) and arguing inductively we see that $Z^ i \in \mathcal{P}$. Choose $P^{n - 1} \in \mathcal{P}$ and an isomorphism $P^{n - 1}/IP^{n - 1} \to E^{n - 1}$. Since $P^{n - 1}$ is projective and since $Z^ n/IZ^ n = \mathop{\mathrm{Im}}(E^{n - 1} \to E^ n)$, we can lift the map $P^{n - 1} \to E^{n - 1} \to E^ n$ to a map $P^{n - 1} \to Z^ n$. By property (3) the map $P^{n - 1} \to Z^ n$ is surjective. Thus we obtain an extension of the diagram by adding $P^{n - 1}$ and the maps just constructed to the left of $P^ n$. Since a diagram of the desired form exists for $n > b$ we conclude by induction on $n$. $\square$

Proof. By assumption (5) we can represent $K$ by a bounded above complex $K^\bullet$ whose terms are in $\mathcal{P}$. Then $K \otimes _ R^\mathbf {L} R/I$ is represented by $K^\bullet /IK^\bullet$. Since $E^\bullet$ is a bounded above complex of projective $R/I$-modules by (4), we can choose a quasi-isomorphism $\delta : E^\bullet \to K^\bullet /IK^\bullet$ (Derived Categories, Lemma 13.19.8). Let $C^\bullet$ be cone on $\delta$ (Derived Categories, Definition 13.9.1). The module $C^ i$ is the direct sum $K^ i/IK^ i \oplus E^{i + 1}$ hence is of the form $P/IP$ for some $P \in \mathcal{P}$ as (2) says in particular that $\mathcal{P}$ is preserved under taking sums. Since $C^\bullet$ is acyclic, we can apply Lemma 15.75.1 and find a acyclic lift $A^\bullet$ of $C^\bullet$. The complex $A^\bullet$ is bounded above and has terms in $\mathcal{P}$. In

we can find the dotted arrow making the diagram commute by Derived Categories, Lemma 13.19.6. We will show below that it follows from (1), (2), (3) that $K^ i \to A^ i$ is the inclusion of a direct summand for every $i$. By property (2) we see that $P^ i = \mathop{\mathrm{Coker}}(K^ i \to A^ i)$ is in $\mathcal{P}$. Thus we can take $P^\bullet = \mathop{\mathrm{Coker}}(K^\bullet \to A^\bullet )[-1]$ to conclude.

To finish the proof we have to show the following: Let $f : P_1 \to P_2$, $P_1, P_2 \in \mathcal{P}$ and $P_1/IP_1 \to P_2/IP_2$ is split injective with cokernel of the form $P_3/IP_3$ for some $P_3 \in \mathcal{P}$, then $f$ is split injective. Write $E_ i = P_ i/IP_ i$. Then $E_2 = E_1 \oplus E_3$. Since $P_2$ is projective we can choose a map $g : P_2 \to P_3$ lifting the map $E_2 \to E_3$. By condition (3) the map $g$ is surjective, hence split as $P_3$ is projective. Set $P_1' = \mathop{\mathrm{Ker}}(g)$ and choose a splitting $P_2 = P'_1 \oplus P_3$. Then $P'_1 \in \mathcal{P}$ by (2). We do not know that $g \circ f = 0$, but we can consider the map

The composition modulo $I$ is an isomorphism. Since $P'_1$ is projective we can split $P_1 = T \oplus P'_1$. If $T = 0$, then we are done, because then $P_2 \to P'_1$ is a splitting of $f$. We see that $T \in \mathcal{P}$ by (2). Calculating modulo $I$ we see that $T/IT = 0$. Since $0 \in \mathcal{P}$ (as the summand of any $P$ in $\mathcal{P}$) we see the map $0 \to T$ is surjective and we conclude that $T = 0$ as desired. $\square$

Proof. We apply Lemma 15.75.2 using the class $\mathcal{P}$ of all projective $R$-modules. Properties (1) and (2) of the lemma are immediate. Property (3) follows from Nakayama's lemma (Algebra, Lemma 10.20.1). Property (4) follows from the fact that we can lift projective $R/I$-modules to projective $R$-modules, see Algebra, Lemma 10.77.5. To see that (5) holds it suffices to show that $K$ is in $D^{-}(R)$. We are given that $K \otimes _ R^\mathbf {L} R/I$ is in $D^{-}(R/I)$ (because $E^\bullet$ is bounded above). We will show by induction on $n$ that $K \otimes _ R^\mathbf {L} R/I^ n$ is in $D^{-}(R/I^ n)$. This will finish the proof because $I$ being nilpotent exactly means that $I^ n = 0$ for some $n$. We may represent $K$ by a K-flat complex $K^\bullet$ with flat terms (Lemma 15.59.10). Then derived tensor products are represented by usual tensor products. Thus we consider the exact sequence

Thus the cohomology of $K \otimes _ R^\mathbf {L} R/I^{n + 1}$ sits in a long exact sequence with the cohomology of $K \otimes _ R^\mathbf {L} R/I^ n$ and the cohomology of

The first cohomologies vanish above a certain degree by induction assumption and the second cohomologies vanish above a certain degree because $K^\bullet \otimes _ R^\mathbf {L} R/I$ is bounded above and $I^ n/I^{n + 1}$ is in degree $0$. $\square$

Proof. One direction follows from Lemma 15.64.12. For the other direction, assume $K$ is pseudo-coherent. Then by Lemma 15.64.5 we can represent $K$ by a bounded above complex $E^\bullet$ of finite free $R$-modules. By Lemma 15.75.3 we can represent $K'$ by a bounded above complex $P^\bullet$ of projective $R'$-modules such that $P^ n \otimes _{R'} R = E^ n$. By Nakayama's lemma we see that $P^ n$ is finite free and we conclude that $K'$ is pseudo-coherent as well. $\square$

Proof. We apply Lemma 15.75.2 using the class $\mathcal{P}$ of all finite stably free $R$-modules. Property (1) of the lemma is immediate. Property (2) follows from Lemma 15.3.2. Property (3) follows from Nakayama's lemma (Algebra, Lemma 10.20.1). Property (4) follows from the fact that we can lift finite stably free $R/I$-modules to finite stably free $R$-modules, see Lemma 15.3.3. Part (5) holds because a pseudo-coherent complex can be represented by a bounded above complex of finite free $R$-modules. The final assertion of the lemma follows from Lemma 15.3.5. $\square$

Proof. Observe that $K \otimes _ R^\mathbf {L} \kappa$ is pseudo-coherent as an object of $D(\kappa )$, see Lemma 15.64.12. Hence the cohomology spaces are finite dimensional and vanish above some cutoff. Every object of $D(\kappa )$ is isomorphic in $D(\kappa )$ to a complex $E^\bullet$ with zero differentials. In particular $E^ i \cong \kappa ^{\oplus d_ i}$ is finite free. Applying Lemma 15.75.5 we obtain the existence.

If we have two complexes $F^\bullet$ and $G^\bullet$ with $F^ i$ and $G^ i$ free of rank $d_ i$ representing $K$. Then we may choose a map of complexes $\beta : F^\bullet \to G^\bullet$ representing the isomorphism $F^\bullet \cong K \cong G^\bullet$, see Derived Categories, Lemma 13.19.8. The induced map of complexes $\beta \otimes 1 : F^\bullet \otimes _ R^\mathbf {L} \kappa \to G^\bullet \otimes _ R^\mathbf {L} \kappa$ must be an isomorphism of complexes as the differentials in $F^\bullet \otimes _ R^\mathbf {L} \kappa$ and $G^\bullet \otimes _ R^\mathbf {L} \kappa$ are zero. Thus $\beta ^ i : F^ i \to G^ i$ is a map of finite free $R$-modules whose reduction modulo $\mathfrak m$ is an isomorphism. Hence $\beta ^ i$ is an isomorphism and we win. $\square$

Proof. Observe that $K \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ is perfect as an object of $D(\kappa (\mathfrak p))$, see Lemma 15.74.9. Hence only a finite number of $d_ i$ are nonzero and they are all finite. Applying Lemma 15.75.6 we get a complex representing $K$ having the desired shape over the local ring $R_\mathfrak p$. We have $R_\mathfrak p = \mathop{\mathrm{colim}}\nolimits R_ f$ for $f \in R$, $f \not\in \mathfrak p$ (Algebra, Lemma 10.9.9). We conclude by Lemma 15.74.17. Some details omitted. $\square$

Proof. If we have an isomorphism of the type described over the localization $R_\mathfrak p$, then using that $R_\mathfrak p = \mathop{\mathrm{colim}}\nolimits R_ f$ (Algebra, Lemma 10.9.9) we can descend the isomorphism to an isomorphism over $R_ f$ for some $f$. Thus we may assume $R$ is local and $\mathfrak p$ is the maximal ideal. In this case the result follows from the uniqueness of a “minimal” complex representing a perfect object, see Lemma 15.75.6, and the fact that any complex is a direct sum of a trivial complex and a minimal one (Algebra, Lemma 10.102.2). $\square$

Proof. We apply Lemma 15.75.2 using the class $\mathcal{P}$ of all finite projective $R$-modules. Properties (1) and (2) of the lemma are immediate. Property (3) follows from Nakayama's lemma (Algebra, Lemma 10.20.1). Property (4) follows from the fact that we can lift finite projective $R/I$-modules to finite projective $R$-modules, see Lemma 15.13.1. Property (5) holds because a pseudo-coherent complex can be represented by a bounded above complex of finite free $R$-modules. Thus Lemma 15.75.2 applies and we find $P^\bullet$ as desired. The final assertion of the lemma follows from Lemma 15.3.5. $\square$