The Stacks project

Proof. The argument preceding the lemma applies for the first case because $Rj_*$ maps $D_\mathit{QCoh}(\mathcal{O}_ U)$ into $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 36.4.1. It is clear that $Rj_*$ maps $D^+_\mathit{QCoh}(\mathcal{O}_ U)$ into $D^+_\mathit{QCoh}(\mathcal{O}_ X)$ which implies the statement on bounded below complexes. Finally, Lemma 36.4.1 guarantees that $Rj_*$ maps $D^-_\mathit{QCoh}(\mathcal{O}_ U)$ into $D^-_\mathit{QCoh}(\mathcal{O}_ X)$ if $X$ is quasi-compact. Combining these two we obtain the last statement. $\square$

Proof. Let $K$ be an object of $D^ b_{\textit{Coh}}(\mathcal{O}_ U)$. By Proposition 36.11.2 we can represent $K$ by a bounded complex $\mathcal{F}^\bullet$ of coherent $\mathcal{O}_ U$-modules. Say $\mathcal{F}^ i = 0$ for $i \not\in [a, b]$ for some $a \leq b$. Since $j$ is quasi-compact and separated, the terms of the bounded complex $j_*\mathcal{F}^\bullet$ are quasi-coherent modules on $X$, see Schemes, Lemma 26.24.1. We inductively pick a coherent submodule $\mathcal{G}^ i \subset j_*\mathcal{F}^ i$ as follows. For $i = a$ we pick any coherent submodule $\mathcal{G}^ a \subset j_*\mathcal{F}^ a$ whose restriction to $U$ is $\mathcal{F}^ a$. This is possible by Properties, Lemma 28.22.2. For $i > a$ we first pick any coherent submodule $\mathcal{H}^ i \subset j_*\mathcal{F}^ i$ whose restriction to $U$ is $\mathcal{F}^ i$ and then we set $\mathcal{G}^ i = \mathop{\mathrm{Im}}(\mathcal{H}^ i \oplus \mathcal{G}^{i - 1} \to j_*\mathcal{F}^ i)$. It is clear that $\mathcal{G}^\bullet \subset j_*\mathcal{F}^\bullet$ is a bounded complex of coherent $\mathcal{O}_ X$-modules whose restriction to $U$ is $\mathcal{F}^\bullet$ as desired. $\square$

Proof. By Lemma 36.10.1 we see that $E$ is an object of $D_\mathit{QCoh}(\mathcal{O}_ U)$. By Lemma 36.13.1 we may assume $E = E'|U$ for some object $E'$ of $D_\mathit{QCoh}(\mathcal{O}_ X)$. Write $X = \mathop{\mathrm{Spec}}(A)$. By Lemma 36.3.5 we can find a complex $M^\bullet$ of $A$-modules whose associated complex of $\mathcal{O}_ X$-modules is a representative of $E'$.

Choose $f_1, \ldots , f_ r \in A$ such that $U = D(f_1) \cup \ldots \cup D(f_ r)$. By Lemma 36.10.2 the complexes $M^\bullet _{f_ j}$ are pseudo-coherent complexes of $A_{f_ j}$-modules. Let $n$ be an integer. Assume we have a map of complexes $\alpha : F^\bullet \to M^\bullet$ where $F^\bullet$ is bounded above, $F^ i = 0$ for $i < n$, each $F^ i$ is a finite free $R$-module, such that

is an isomorphism for $i > n$ and surjective for $i = n$. Picture

Since each $M^\bullet _{f_ j}$ has vanishing cohomology in large degrees we can find such a map for $n \gg 0$. By induction on $n$ we are going to extend this to a map of complexes $F^\bullet \to M^\bullet$ such that $H^ i(\alpha _{f_ j})$ is an isomorphism for all $i$. The lemma will follow by taking $\widetilde{F^\bullet }$.

The induction step will be to extend the diagram above by adding $F^{n - 1}$. Let $C^\bullet$ be the cone on $\alpha$ (Derived Categories, Definition 13.9.1). The long exact sequence of cohomology shows that $H^ i(C^\bullet _{f_ j}) = 0$ for $i \geq n$. By More on Algebra, Lemma 15.64.2 we see that $C^\bullet _{f_ j}$ is $(n - 1)$-pseudo-coherent. By More on Algebra, Lemma 15.64.3 we see that $H^{n - 1}(C^\bullet _{f_ j})$ is a finite $A_{f_ j}$-module. Choose a finite free $A$-module $F^{n - 1}$ and an $A$-module $\beta : F^{n - 1} \to C^{n - 1}$ such that the composition $F^{n - 1} \to C^{n - 1} \to C^ n$ is zero and such that $F^{n - 1}_{f_ j}$ surjects onto $H^{n - 1}(C^\bullet _{f_ j})$. (Some details omitted; hint: clear denominators.) Since $C^{n - 1} = M^{n - 1} \oplus F^ n$ we can write $\beta = (\alpha ^{n - 1}, -d^{n - 1})$. The vanishing of the composition $F^{n - 1} \to C^{n - 1} \to C^ n$ implies these maps fit into a morphism of complexes

Moreover, these maps define a morphism of distinguished triangles

Hence our choice of $\beta$ implies that the map of complexes $(F^{-1} \to \ldots ) \to M^\bullet$ induces an isomorphism on cohomology localized at $f_ j$ in degrees $\geq n$ and a surjection in degree $n - 1$. This finishes the proof of the lemma. $\square$

Proof. Recall that $\mathop{\mathrm{Ext}}\nolimits ^ n_{D(\mathcal{O}_ X)}(\mathcal{E}, E) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(\mathcal{E}, E[n])$. We have Mayer-Vietoris for morphisms in the derived category, see Cohomology, Lemma 20.33.3. Thus if $X = U \cup V$ and the result of the lemma holds for $E|_ U$, $E|_ V$, and $E|_{U \cap V}$ for some bound $n_0$, then the result holds for $E$ with bound $n_0 + 1$. Thus it suffices to prove the lemma when $X$ is affine, see Cohomology of Schemes, Lemma 30.4.1.

Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Choose a complex of $A$-modules $M^\bullet$ whose associated complex of quasi-coherent modules represents $E$, see Lemma 36.3.5. Write $\mathcal{E} = \widetilde{P}$ for some $A$-module $P$. Since $\mathcal{E}$ is finite locally free, we see that $P$ is a finite projective $A$-module. We have

The first equality by Lemma 36.3.5, the second equality by Derived Categories, Lemma 13.19.8, and the final equality because $\mathop{\mathrm{Hom}}\nolimits _ A(P, -)$ is an exact functor. As $E$ and hence $M^\bullet$ is bounded we get zero for all sufficiently large $n$. $\square$

Proof. Part (2) follows from (1) as $\mathop{\mathrm{Ext}}\nolimits ^ n_{D(\mathcal{O}_ X)}(K, L) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K, L[n])$. We prove (1). Since $K$ is perfect we have

where $K^\vee$ is the “dual” perfect complex to $K$, see Cohomology, Lemma 20.50.5. Note that $K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} L$ is in $D_\mathit{QCoh}(X)$ by Lemmas 36.3.9 and 36.10.1 (to see that a perfect complex has quasi-coherent cohomology sheaves). Say $K^\vee$ has tor amplitude in $[a, b]$. Then the spectral sequence

shows that $H^ j(K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} L)$ is zero if $H^ q(L) = 0$ for $q \in [j - b, j - a]$. Let $N$ be the integer $d$ of Cohomology of Schemes, Lemma 30.4.4. Then $H^0(X, K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} L)$ vanishes if the cohomology sheaves

are zero. Namely, by the lemma cited and Lemma 36.3.4, we have

and by the vanishing of cohomology sheaves, this is equal to $H^0(X, \tau _{\geq 1}(K^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} L))$ which is zero by Derived Categories, Lemma 13.16.1. It follows that $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K, L)$ is zero if $H^ i(L) = 0$ for $i \in [-b - N, -a]$. $\square$

Proof. Say $X = \mathop{\mathrm{Spec}}(A)$. Recall that a perfect complex is pseudo-coherent, see Cohomology, Lemma 20.49.5. By Lemma 36.13.3 we can find a bounded above complex $\mathcal{F}^\bullet$ of finite free $A$-modules such that $E$ is isomorphic to $\mathcal{F}^\bullet |_ U$ in $D(\mathcal{O}_ U)$. By Cohomology, Lemma 20.49.5 and since $U$ is quasi-compact, we see that $E$ has finite tor dimension, say $E$ has tor amplitude in $[a, b]$. Pick $r < a$ and set

Since $E$ has tor amplitude in $[a, b]$ we see that $\mathcal{F} = \mathcal{K}|_ U$ is flat (Cohomology, Lemma 20.48.2). Hence $\mathcal{F}$ is flat and of finite presentation, thus finite locally free (Properties, Lemma 28.20.2). It follows that

is a strictly perfect complex on $U$ representing $E$. On the other hand, the complex $P = (\mathcal{F}^ r \to \mathcal{F}^{r + 1} \to \ldots )$ is a perfect complex on $X$. Using stupid truncations we obtain a distinguished triangle

If the map $E \to \mathcal{F}[-r - 1]$ is zero in $D(\mathcal{O}_ U)$, then $P|_ U = \mathcal{F}[-r - 2] \oplus E$, see Derived Categories, Lemma 13.4.11. This will be true for $r \ll 0$ for example by Lemma 36.13.5. $\square$

Proof. Write $X = \mathop{\mathrm{Spec}}(A)$. Write $U = D(f_1) \cup \ldots \cup D(f_ r)$. Choose finite complex of finite projective $A$-modules $M^\bullet$ representing $E$ (Lemma 36.10.7). Choose a complex of $A$-modules $(M')^\bullet$ representing $E'$ (Lemma 36.3.5). In this case the complex $H^\bullet = \mathop{\mathrm{Hom}}\nolimits _ A(M^\bullet , (M')^\bullet )$ is a complex of $A$-modules whose associated complex of quasi-coherent $\mathcal{O}_ X$-modules represents $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, E')$, see Cohomology, Lemma 20.46.9. Then $\alpha$ determines an element $s$ of $H^0(U, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, E'))$, see Cohomology, Lemma 20.42.1. There exists an $e$ and a map

corresponding to $s$, see Proposition 36.9.5. Letting $E_1$ be the object corresponding to complex of quasi-coherent $\mathcal{O}_ X$-modules associated to

we obtain $E_1 \to E$ using the canonical map $I^\bullet (f_1^ e, \ldots , f_ r^ e) \to A$ and $E_1 \to E'$ using $\xi$ and Cohomology, Lemma 20.42.1. $\square$

Proof. By Lemma 36.13.6 we can find a perfect object $E$ of $D(\mathcal{O}_ X)$ such that $E|_ U = \mathcal{F}[r] \oplus F$ for some finite locally free $\mathcal{O}_ U$-module $\mathcal{F}$. By Lemma 36.13.7 we can find a morphism of perfect complexes $\alpha : E_1 \to E$ such that $(E_1)|_ U \cong E|_ U$ and such that $\alpha |_ U$ is the map

Then the cone on $\alpha$ is a solution. $\square$

Proof. We have Mayer-Vietoris for morphisms in the derived category, see Cohomology, Lemma 20.33.3. Thus if $X = U \cup V$ and the result of the lemma holds for $f|_ U$, $f|_ V$, and $f|_{U \cap V}$, then the result holds for $f$. Thus it suffices to prove the lemma when $X$ is affine, see Cohomology of Schemes, Lemma 30.4.1.

Let $X = \mathop{\mathrm{Spec}}(A)$. Then $f \in A$. We will use the equivalence $D(A) = D_\mathit{QCoh}(X)$ of Lemma 36.3.5 without further mention. Represent $E$ by a finite complex of finite projective $A$-modules $P^\bullet$. This is possible by Lemma 36.10.7. Let $t$ be the largest integer such that $P^ t$ is nonzero. The distinguished triangle

shows that by induction on the length of the complex $P^\bullet$ we can reduce to the case where $P^\bullet$ has a single nonzero term. This and the shift functor reduces us to the case where $P^\bullet$ consists of a single finite projective $A$-module $P$ in degree $0$. Represent $E'$ by a complex $M^\bullet$ of $A$-modules. Then $\alpha$ corresponds to a map $P \to H^0(M^\bullet )$. Since the module $H^0(M^\bullet )$ is supported on $V(f)$ by assumption (2) we see that every element of $H^0(M^\bullet )$ is annihilated by a power of $f$. Since $P$ is a finite $A$-module the map $f^ n\alpha : P \to H^0(M^\bullet )$ is zero for some $n$ as desired. $\square$

Proof. Say $T = V(g_1, \ldots , g_ s)$. After replacing $g_ j$ by a power we may assume multiplication by $g_ j$ is zero on $F$, see Lemma 36.13.9. Choose $E$ as in Lemma 36.13.8. Note that $g_ j : E \to E$ restricts to zero on $U$. Choose a distinguished triangle

By Derived Categories, Lemma 13.4.11 the object $C_1$ restricts to $F \oplus F[1] \oplus F[1] \oplus F[2]$ on $U$. Moreover, $g_1 : C_1 \to C_1$ has square zero by Derived Categories, Lemma 13.4.5. Namely, the diagram

is commutative since the compositions $E \xrightarrow {g_1} E \to C_1$ and $C_1 \to E[1] \xrightarrow {g_1} E[1]$ are zero. Continuing, setting $C_{i + 1}$ equal to the cone of the map $g_ i : C_ i \to C_ i$ we obtain a perfect complex $C_ s$ on $X$ supported on $T$ whose restriction to $U$ gives

Choose morphisms of perfect complexes $\beta : C' \to C_ s$ and $\gamma : C' \to C_ s$ as in Lemma 36.13.7 such that $\beta |_ U$ is an isomorphism and such that $\gamma |_ U \circ \beta |_ U^{-1}$ is the morphism

which is the identity on all summands except for $F$ where it is zero. By Lemma 36.13.7 we also have $C' = C_ s \otimes ^\mathbf {L} I$ for some perfect complex $I$ on $X$. Hence the nullity of $g_ j^2\text{id}_{C_ s}$ implies the same thing for $C'$. Thus $C'$ is supported on $T$ as well. Then $\text{Cone}(\gamma )$ is a solution. $\square$

Proof. Since $X$ is quasi-compact there exists an integer $m$ such that $X = U \cup V_1 \cup \ldots \cup V_ m$ for some affine opens $V_ j$ of $X$. Arguing by induction on $m$ we see that we may assume $m = 1$. In other words, we may assume that $X = U \cup V$ with $V$ affine. By Lemma 36.13.10 we can choose a perfect object $Q$ in $D(\mathcal{O}_ V)$ supported on $T \cap V$ and an isomorphism $Q|_{U \cap V} \to (P \oplus P[1])|_{U \cap V}$. By Lemma 36.13.7 we can replace $Q$ by $Q \otimes ^\mathbf {L} I$ (still supported on $T \cap V$) and assume that the map

lifts to $Q \to E|_ V$. By Cohomology, Lemma 20.45.1 we find an morphism $a : R \to E$ of $D(\mathcal{O}_ X)$ such that $a|_ U$ is isomorphic to $P \oplus P[1] \to E|_ U$ and $a|_ V$ isomorphic to $Q \to E|_ V$. Thus $R$ is perfect and supported on $T$ as desired. $\square$