The Stacks project

Proof. We will prove by induction on $-1 \leq m \leq n$ that there exists a differential graded submodule $P' \subset P$ such that

1. $F_ mP \subset P'$,

2. for $i \geq m$ the quotient $F_{i + 1}P \cap P'/(F_ iP \cap P')$ is isomorphic to $\bigoplus _{j \in J'_ i} A[k_{i, j}]$ for some finite subsets $J'_ i \subset J_ i$, and

3. $E \to P$ factors through $P'$.

The base case is $m = n$ where we can take $P' = P$.

Induction step. Assume $P'$ works for $m$. For $i \geq m$ and $j \in J'_ i$ let $x_{i, j} \in F_{i + 1}P \cap P'$ be a homogeneous element of degree $k_{i, j}$ whose image in $F_{i + 1}P \cap P'/(F_ iP \cap P')$ is the generator in the summand corresponding to $j \in J_ i$. The $x_{i, j}$ generate $P'/F_ mP$ as an $A$-module. Write

with $y_{i, j} \in F_ mP$ and $a_{i, j}^{i', j'} \in A$. There exists a finite subset $J'_{m - 1} \subset J_{m - 1}$ such that each $y_{i, j}$ maps to an element of the submodule $\bigoplus _{j \in J'_{m - 1}} A[k_{m - 1, j}]$ of $F_ mP/F_{m - 1}P$. Let $P'' \subset F_ mP$ be the inverse image of $\bigoplus _{j \in J'_{m - 1}} A[k_{m - 1, j}]$ under the map $F_ mP \to F_ mP/F_{m - 1}P$. Then we see that the $A$-submodule

is a differential graded submodule of the type we are looking for. Moreover

Since $E$ is compact, the composition of the given map $E \to P'$ with the quotient map, factors through a finite direct subsum of the module displayed above. Hence after enlarging $J'_{m - 1}$ we may assume $E \to P'$ factors through $P'' + \sum x_{i, j}A$ as desired. $\square$

Proof. Assume $E$ is compact. By Lemma 22.20.4 we may assume that $E$ is represented by a differential graded $A$-module $P$ with property (P). Consider the distinguished triangle

coming from the admissible short exact sequence of Lemma 22.20.1. Since $E$ is compact we have $\delta = \sum _{i = 1, \ldots , n} \delta _ i$ for some $\delta _ i : P \to F_ iP[1]$. Since the composition of $\delta$ with the map $\bigoplus F_ iP[1] \to \bigoplus F_ iP[1]$ is zero (Derived Categories, Lemma 13.4.1) it follows that $\delta = 0$ (follows as $\bigoplus F_ iP \to \bigoplus F_ iP$ maps the summand $F_ iP$ via the difference of $\text{id}$ and the inclusion map into $F_{i - 1}P$). Thus we see that the identity on $E$ factors through $\bigoplus F_ iP$ in $D(A, \text{d})$ (by Derived Categories, Lemma 13.4.11). Next, we use that $P$ is compact again to see that the map $E \to \bigoplus F_ iP$ factors through $\bigoplus _{i = 1, \ldots , n} F_ iP$ for some $n$. In other words, the identity on $E$ factors through $\bigoplus _{i = 1, \ldots , n} F_ iP$. By Lemma 22.36.3 we see that the identity of $E$ factors as $E \to P \to E$ where $P$ is as in part (2) of the statement of the lemma. In other words, we have proven that (1) implies (2).

Assume (2). By Derived Categories, Lemma 13.37.2 it suffices to show that $P$ gives a compact object. Observe that $P$ has property (P), hence we have

for any differential graded module $M$ by Lemma 22.22.3. As direct sums in $D(A, \text{d})$ are given by direct sums of graded modules (Lemma 22.22.4) we reduce to showing that $\mathop{\mathrm{Hom}}\nolimits _{K(A, \text{d})}(P, M)$ commutes with direct sums. Using that $K(A, \text{d})$ is a triangulated category, that $\mathop{\mathrm{Hom}}\nolimits$ is a cohomological functor in the first variable, and the filtration on $P$, we reduce to the case that $P$ is a finite direct sum of shifts of $A$. Thus we reduce to the case $P = A[k]$ which is clear. $\square$

Proof. Observe that the collection of objects of $D(A, \text{d})$ for which such a pair of integers exists is a saturated, strictly full triangulated subcategory of $D(A, \text{d})$. Thus by Proposition 22.36.4 it suffices to prove this when $E$ is represented by a differential graded module $P$ which has a finite filtration $F_\bullet$ by differential graded submodules such that $F_ iP/F_{i - 1}P$ are finite direct sums of shifts of $A$. Using the compatibility with triangles, we see that it suffices to prove it for $P = A$. In this case $\mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(A, M) = H^0(M)$ and the result holds with $a = b = 0$. $\square$

Proof. Let $\mathcal{D} \subset K(A, \text{d})$ be the triangulated subcategory discussed in Remark 22.36.1. Let $P$ be an object of $\mathcal{D}$ which is finite projective as a graded $A$-module. Then $P$ represents a compact object of $D(A, \text{d})$ by Remark 22.36.2.

To prove the converse, let $E$ be a compact object of $D(A, \text{d})$. Fix $a \leq b$ as in Lemma 22.36.5. After decreasing $a$ and increasing $b$ if necessary, we may also assume that $H^ i(E) = 0$ for $i \not\in [a, b]$ (this follows from Proposition 22.36.4 and our assumption on $A$). Moreover, fix an integer $c > 0$ such that $A^ n = 0$ if $|n| \geq c$.

By Proposition 22.36.4 we see that $E$ is a direct summand, in $D(A, \text{d})$, of a differential graded $A$-module $P$ which has a finite filtration $F_\bullet$ by differential graded submodules such that $F_ iP/F_{i - 1}P$ are finite direct sums of shifts of $A$. In particular, $P$ has property (P) and we have $\mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(P, M) = \mathop{\mathrm{Hom}}\nolimits _{K(A, \text{d})}(P, M)$ for any differential graded module $M$ by Lemma 22.22.3. In other words, $P$ is an object of the triangulated subcategory $\mathcal{D} \subset K(A, \text{d})$ discussed in Remark 22.36.1. Note that $P$ is finite free as a graded $A$-module.

Choose $n > 0$ such that $b + 4c - n < a$. Represent the projector onto $E$ by an endomorphism $\varphi : P \to P$ of differential graded $A$-modules. Consider the distinguished triangle

in $K(A, \text{d})$ where $C$ is the cone of the first arrow. Then $C$ is an object of $\mathcal{D}$, we have $C \cong E \oplus E[1]$ in $D(A, \text{d})$, and $C$ is a finite graded free $A$-module. Next, consider a distinguished triangle

in $K(A, \text{d})$ where $C'$ is the cone on a morphism $C[1] \to C$ representing the composition

in $D(A, \text{d})$. Then we see that $C'$ represents $E \oplus E[2]$. Continuing in this manner we see that we can find a differential graded $A$-module $P$ which is an object of $\mathcal{D}$, is a finite free as a graded $A$-module, and represents $E \oplus E[n]$.

Choose a basis $x_ i$, $i \in I$ of homogeneous elements for $P$ as an $A$-module. Let $d_ i = \deg (x_ i)$. Let $P_1$ be the $A$-submodule of $P$ generated by $x_ i$ and $\text{d}(x_ i)$ for $d_ i \leq a - c - 1$. Let $P_2$ be the $A$-submodule of $P$ generated by $x_ i$ and $\text{d}(x_ i)$ for $d_ i \geq b - n + c$. We observe

1. $P_1$ and $P_2$ are differential graded submodules of $P$,

2. $P_1^ t = 0$ for $t \geq a$,

3. $P_1^ t = P^ t$ for $t \leq a - 2c$,

4. $P_2^ t = 0$ for $t \leq b - n$,

5. $P_2^ t = P^ t$ for $t \geq b - n + 2c$.

As $b - n + 2c \geq a - 2c$ by our choice of $n$ we obtain a short exact sequence of differential graded $A$-modules

Since $P$ is projective as a graded $A$-module this is an admissible short exact sequence (Lemma 22.16.1). Hence we obtain a boundary map $\delta : P \to (P_1 \cap P_2)[1]$ in $K(A, \text{d})$, see Lemma 22.7.2. Since $P = E \oplus E[n]$ and since $P_1 \cap P_2$ lives in degrees $(b - n, a)$ we find that $\mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(E \oplus E[n], (P_1 \cap P_2)[1])$ is zero. Therefore $\delta = 0$ as a morphism in $K(A, \text{d})$ as $P$ is an object of $\mathcal{D}$. By Derived Categories, Lemma 13.4.11 we can find a map $s : P \to P_1 \oplus P_2$ such that $\pi \circ s = \text{id}_ P + \text{d}h + h\text{d}$ for some $h : P \to P$ of degree $-1$. Since $P_1 \oplus P_2 \to P$ is surjective and since $P$ is projective as a graded $A$-module we can choose a homogeneous lift $\tilde h : P \to P_1 \oplus P_2$ of $h$. Then we change $s$ into $s + \text{d} \tilde h + \tilde h \text{d}$ to get $\pi \circ s = \text{id}_ P$. This means we obtain a direct sum decomposition $P = s^{-1}(P_1) \oplus s^{-1}(P_2)$. Since $s^{-1}(P_2)$ is equal to $P$ in degrees $\geq b - n + 2c$ we see that $s^{-1}(P_2) \to P \to E$ is a quasi-isomorphism, i.e., an isomorphism in $D(A, \text{d})$. This finishes the proof. $\square$