## 15.74 Lifting complexes

Let $R$ be a ring. Let $I \subset R$ be an ideal. The lifting problem we will consider is the following. Suppose given an object $K$ of $D(R)$ and a complex $E^\bullet$ of $R/I$-modules such that $E^\bullet$ represents $K \otimes _ R^\mathbf {L} R/I$ in $D(R)$. Question: Does there exist a complex of $R$-modules $P^\bullet$ lifting $E^\bullet$ representing $K$ in $D(R)$? In general the answer to this question is no, but in good cases something can be done. We first discuss lifting acyclic complexes.

Lemma 15.74.1. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $\mathcal{P}$ be a class of $R$-modules. Assume

1. each $P \in \mathcal{P}$ is a projective $R$-module,

2. if $P_1 \in \mathcal{P}$ and $P_1 \oplus P_2 \in \mathcal{P}$, then $P_2 \in \mathcal{P}$, and

3. if $f : P_1 \to P_2$, $P_1, P_2 \in \mathcal{P}$ is surjective modulo $I$, then $f$ is surjective.

Then given any bounded above acyclic complex $E^\bullet$ whose terms are of the form $P/IP$ for $P \in \mathcal{P}$ there exists a bounded above acyclic complex $P^\bullet$ whose terms are in $\mathcal{P}$ lifting $E^\bullet$.

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

$\xymatrix{ & & P^ n \ar[r] \ar[d] & P^{n + 1} \ar[r] \ar[d] & \ldots \ar[r] & P^ b \ar[r] \ar[d] & 0 \ar[r] \ar[d] & \ldots \\ \ldots \ar[r] & E^{n - 1} \ar[r] & E^ n \ar[r] & E^{n + 1} \ar[r] & \ldots \ar[r] & E^ b \ar[r] & 0 \ar[r] & \ldots }$

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 $n$. $\square$

Lemma 15.74.2. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $\mathcal{P}$ be a class of $R$-modules. Let $K \in D(R)$ and let $E^\bullet$ be a complex of $R/I$-modules representing $K \otimes _ R^\mathbf {L} R/I$. Assume

1. each $P \in \mathcal{P}$ is a projective $R$-module,

2. $P_1 \in \mathcal{P}$ and $P_1 \oplus P_2 \in \mathcal{P}$ if and only if $P_1, P_2 \in \mathcal{P}$,

3. if $f : P_1 \to P_2$, $P_1, P_2 \in \mathcal{P}$ is surjective modulo $I$, then $f$ is surjective,

4. $E^\bullet$ is bounded above and $E^ i$ is of the form $P/IP$ for $P \in \mathcal{P}$, and

5. $K$ can be represented by a bounded above complex whose terms are in $\mathcal{P}$.

Then there exists a bounded above complex $P^\bullet$ whose terms are in $\mathcal{P}$ with $P^\bullet /IP^\bullet$ isomorphic to $E^\bullet$ and representing $K$ in $D(R)$.

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 15.74.1 and find a acyclic lift $A^\bullet$ of $C^\bullet$. The complex $A^\bullet$ is bounded above and has terms in $\mathcal{P}$. In

$\xymatrix{ K^\bullet \ar@{..>}[r] \ar[d] & A^\bullet \ar[d] \\ K^\bullet /IK^\bullet \ar[r] & C^\bullet \ar[r] & E^\bullet  }$

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

$P_1 \xrightarrow {f} P_2 \xrightarrow {projection} P'_1$

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. $\square$

Lemma 15.74.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $E^\bullet$ be a complex of $R/I$-modules. Let $K$ be an object of $D(R)$. Assume that

1. $E^\bullet$ is a bounded above complex of projective $R/I$-modules,

2. $K \otimes _ R^\mathbf {L} R/I$ is represented by $E^\bullet$ in $D(R/I)$, and

3. $I$ is a nilpotent ideal.

Then there exists a bounded above complex $P^\bullet$ of projective $R$-modules representing $K$ in $D(R)$ such that $P^\bullet \otimes _ R R/I$ is isomorphic to $E^\bullet$.

Proof. We apply Lemma 15.74.2 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 15.58.12). Then derived tensor products are represented by usual tensor products. Thus we consider the exact sequence

$0 \to K^\bullet \otimes _ R I^ n/I^{n + 1} \to K^\bullet \otimes _ R R/I^{n + 1} \to K^\bullet \otimes _ R R/I^ n \to 0$

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

$K \otimes _ R^\mathbf {L} I^ n/I^{n + 1} = K \otimes _ R^\mathbf {L} R/I \otimes _{R/I}^\mathbf {L} I^ n/I^{n + 1}$

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 $0$. $\square$

Lemma 15.74.4. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $E^\bullet$ be a complex of $R/I$-modules. Let $K$ be an object of $D(R)$. Assume that

1. $E^\bullet$ is a bounded above complex of finite stably free $R/I$-modules,

2. $K \otimes _ R^\mathbf {L} R/I$ is represented by $E^\bullet$ in $D(R/I)$,

3. $K^\bullet$ is pseudo-coherent, and

4. every element of $1 + I$ is invertible.

Then there exists a bounded above complex $P^\bullet$ of finite stably free $R$-modules representing $K$ in $D(R)$ such that $P^\bullet \otimes _ R R/I$ is isomorphic to $E^\bullet$. Moreover, if $E^ i$ is free, then $P^ i$ is free.

Proof. We apply Lemma 15.74.2 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. $\square$

Lemma 15.74.5. Let $(R, \mathfrak m, \kappa )$ be a local ring. Let $K \in D(R)$ be pseudo-coherent. Set $d_ i = \dim _\kappa H^ i(K \otimes _ R^\mathbf {L} \kappa )$. Then $d_ i < \infty$ and for some $b \in \mathbf{Z}$ we have $d_ i = 0$ for $i > b$. Then there exists a complex

$\ldots \to R^{\oplus d_{b - 2}} \to R^{\oplus d_{b - 1}} \to R^{\oplus d_ b} \to 0 \to \ldots$

representing $K$ in $D(R)$. Moreover, this complex is unique up to isomorphism(!).

Proof. Observe that $K \otimes _ R^\mathbf {L} \kappa$ is pseudo-coherent as an object of $D(\kappa )$, see Lemma 15.63.12. 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 15.74.4 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. $\square$

Lemma 15.74.6. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime. Let $K \in D(R)$ be perfect. Set $d_ i = \dim _{\kappa (\mathfrak p)} H^ i(K \otimes _ R^\mathbf {L} \kappa (\mathfrak p))$. Then $d_ i < \infty$ and only a finite number are nonzero. Then there exists an $f \in R$, $f \not\in \mathfrak p$ and a complex

$\ldots \to 0 \to R_ f^{\oplus d_ a} \to R_ f^{\oplus d_{a + 1}} \to \ldots \to R_ f^{\oplus d_{b - 1}} \to R_ f^{\oplus d_ b} \to 0 \to \ldots$

representing $K \otimes _ R^\mathbf {L} R_ f$ in $D(R_ f)$.

Proof. Observe that $K \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ is perfect as an object of $D(\kappa (\mathfrak p))$, see Lemma 15.73.9. Hence only a finite number of $d_ i$ are nonzero and they are all finite. Applying Lemma 15.74.5 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 15.73.17. Some details omitted. $\square$

Lemma 15.74.7. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime. Let $M^\bullet$ and $N^\bullet$ be bounded complexes of finite projective $R$-modules representing the same object of $D(R)$. Then there exists an $f \in R$, $f \not\in \mathfrak p$ such that there is an isomorphism (!) of complexes

$M^\bullet _ f \oplus P^\bullet \cong N^\bullet _ f \oplus Q^\bullet$

where $P^\bullet$ and $Q^\bullet$ are finite direct sums of trivial complexes, i.e., complexes of the form the form $\ldots \to 0 \to R_ f \xrightarrow {1} R_ f \to 0 \to \ldots$ (placed in arbitrary degrees).

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 15.74.5, and the fact that any complex is a direct sum of a trivial complex and a minimal one (Algebra, Lemma 10.102.2). $\square$

Lemma 15.74.8. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $E^\bullet$ be a complex of $R/I$-modules. Let $K$ be an object of $D(R)$. Assume that

1. $E^\bullet$ is a bounded above complex of finite projective $R/I$-modules,

2. $K \otimes _ R^\mathbf {L} R/I$ is represented by $E^\bullet$ in $D(R/I)$,

3. $K$ is pseudo-coherent, and

4. $(R, I)$ is a henselian pair.

Then there exists a bounded above complex $P^\bullet$ of finite projective $R$-modules representing $K$ in $D(R)$ such that $P^\bullet \otimes _ R R/I$ is isomorphic to $E^\bullet$. Moreover, if $E^ i$ is free, then $P^ i$ is free.

Proof. We apply Lemma 15.74.2 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 15.74.2 applies and we find $P^\bullet$ as desired. The final assertion of the lemma follows from Lemma 15.3.5. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).