The Stacks project

35.16 Compact and perfect objects

Let $X$ be a Noetherian scheme of finite dimension. By Cohomology, Proposition 20.21.7 and Cohomology on Sites, Lemma 21.48.3 the sheaves of modules $j_!\mathcal{O}_ U$ are compact objects of $D(\mathcal{O}_ X)$ for all opens $U \subset X$. These sheaves are typically not quasi-coherent, hence these do not give perfect objects of the derived category $D(\mathcal{O}_ X)$. However, if we restrict ourselves to complexes with quasi-coherent cohomology sheaves, then this does not happen. Here is the precise statement.

Proposition 35.16.1. Let $X$ be a quasi-compact and quasi-separated scheme. An object of $D_\mathit{QCoh}(\mathcal{O}_ X)$ is compact if and only if it is perfect.

Proof. If $K$ is a perfect object of $D(\mathcal{O}_ X)$ with dual $K^\vee $ (Cohomology, Lemma 20.43.11) we have

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K, M) = H^0(X, K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} M) \]

functorially in $M$. Since $K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} -$ commutes with direct sums and since $H^0(X, -)$ commutes with direct sums on $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 35.4.2 we conclude that $K$ is compact in $D_\mathit{QCoh}(\mathcal{O}_ X)$.

Conversely, let $K$ be a compact object of $D_\mathit{QCoh}(\mathcal{O}_ X)$. To show that $K$ is perfect, it suffices to show that $K|_ U$ is perfect for every affine open $U \subset X$, see Cohomology, Lemma 20.43.2. Observe that $j : U \to X$ is a quasi-compact and separated morphism. Hence $Rj_* : D_\mathit{QCoh}(\mathcal{O}_ U) \to D_\mathit{QCoh}(\mathcal{O}_ X)$ commutes with direct sums, see Lemma 35.4.2. Thus the adjointness of restriction to $U$ and $Rj_*$ implies that $K|_ U$ is a compact object of $D_\mathit{QCoh}(\mathcal{O}_ U)$. Hence we reduce to the case that $X$ is affine.

Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. By Lemma 35.3.5 the problem is translated into the same problem for $D(A)$. For $D(A)$ the result is More on Algebra, Proposition 15.72.3. $\square$

The following result is a strengthening of Proposition 35.16.1. Let $T \subset X$ be a closed subset of a scheme $X$. As before $D_ T(\mathcal{O}_ X)$ denotes the strictly full, saturated, triangulated subcategory consisting of complexes whose cohomology sheaves are supported on $T$. Since taking direct sums commutes with taking cohomology sheaves, it follows that $D_ T(\mathcal{O}_ X)$ has direct sums and that they are equal to direct sums in $D(\mathcal{O}_ X)$.

Lemma 35.16.2. Let $X$ be a quasi-compact and quasi-separated scheme. Let $T \subset X$ be a closed subset such that $X \setminus T$ is quasi-compact. An object of $D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$ is compact if and only if it is perfect as an object of $D(\mathcal{O}_ X)$.

Proof. We observe that $D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$ is a triangulated category with direct sums by the remark preceding the lemma. By Proposition 35.16.1 the perfect objects define compact objects of $D(\mathcal{O}_ X)$ hence a fortiori of any subcategory preserved under taking direct sums. For the converse we will use there exists a generator $E \in D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$ which is a perfect complex of $\mathcal{O}_ X$-modules, see Lemma 35.14.4. Hence by the above, $E$ is compact. Then it follows from Derived Categories, Proposition 13.34.6 that $E$ is a classical generator of the full subcategory of compact objects of $D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$. Thus any compact object can be constructed out of $E$ by a finite sequence of operations consisting of (a) taking shifts, (b) taking finite direct sums, (c) taking cones, and (d) taking direct summands. Each of these operations preserves the property of being perfect and the result follows. $\square$

The following lemma is an application of the ideas that go into the proof of the preceding lemma.

Lemma 35.16.3. Let $X$ be a quasi-compact and quasi-separated scheme. Let $T \subset X$ be a closed subset such that $U = X \setminus T$ is quasi-compact. Let $\alpha : P \to E$ be a morphism of $D_\mathit{QCoh}(\mathcal{O}_ X)$ with either

  1. $P$ is perfect and $E$ supported on $T$, or

  2. $P$ pseudo-coherent, $E$ supported on $T$, and $E$ bounded below.

Then there exists a perfect complex of $\mathcal{O}_ X$-modules $I$ and a map $I \to \mathcal{O}_ X[0]$ such that $I \otimes ^\mathbf {L} P \to E$ is zero and such that $I|_ U \to \mathcal{O}_ U[0]$ is an isomorphism.

Proof. Set $\mathcal{D} = D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$. In both cases the complex $K = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (P, E)$ is an object of $\mathcal{D}$. See Lemma 35.9.8 for quasi-coherence. It is clear that $K$ is supported on $T$ as formation of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits $ commutes with restriction to opens. The map $\alpha $ defines an element of $H^0(K) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\mathcal{O}_ X[0], K)$. Then it suffices to prove the result for the map $\alpha : \mathcal{O}_ X[0] \to K$.

Let $E \in \mathcal{D}$ be a perfect generator, see Lemma 35.14.4. Write

\[ K = \text{hocolim} K_ n \]

as in Derived Categories, Lemma 13.34.3 using the generator $E$. Since the functor $\mathcal{D} \to D(\mathcal{O}_ X)$ commutes with direct sums, we see that $K = \text{hocolim} K_ n$ holds in $D(\mathcal{O}_ X)$. Since $\mathcal{O}_ X$ is a compact object of $D(\mathcal{O}_ X)$ we find an $n$ and a morphism $\alpha _ n : \mathcal{O}_ X \to K_ n$ which gives rise to $\alpha $, see Derived Categories, Lemma 13.31.9. By Derived Categories, Lemma 13.34.4 applied to the morphism $\mathcal{O}_ X[0] \to K_ n$ in the ambient category $D(\mathcal{O}_ X)$ we see that $\alpha _ n$ factors as $\mathcal{O}_ X[0] \to Q \to K_ n$ where $Q$ is an object of $\langle E \rangle $. We conclude that $Q$ is a perfect complex supported on $T$.

Choose a distinguished triangle

\[ I \to \mathcal{O}_ X[0] \to Q \to I[1] \]

By construction $I$ is perfect, the map $I \to \mathcal{O}_ X[0]$ restricts to an isomorphism over $U$, and the composition $I \to K$ is zero as $\alpha $ factors through $Q$. This proves the lemma. $\square$

Comments (2)

Comment #3853 by Matthieu Romagny on

Missing 's' in third sentence: hence these do not give perfect objects

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