The Stacks project

Proof. Special case of Lemma 36.13.11. $\square$

Proof. Choose a distinguished triangle $E \to Rj_*(E|_ U) \to N \to E[1]$. Observe that

for all $n$ as $K|_ U = 0$. Thus it suffices to prove the result for $N$. In other words, we may assume that $E$ restricts to zero on $U$. Observe that there are distinguished triangles

of Koszul complexes, see More on Algebra, Lemma 15.28.11. Hence if $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[n], E) = 0$ for all $n \in \mathbf{Z}$ then the same thing is true for the $K$ replaced by $K_ e$ as in Lemma 36.9.6. Thus our lemma follows immediately from that one and the fact that $E$ is determined by the complex of $A$-modules $R\Gamma (X, E)$, see Lemma 36.3.5. $\square$

Proof. We will prove this using the induction principle of Cohomology of Schemes, Lemma 30.4.1.

If $X$ is affine, then $\mathcal{O}_ X$ is a perfect generator. This follows from Lemma 36.3.5.

Assume that $X = U \cup V$ is an open covering with $U$ quasi-compact such that the theorem holds for $U$ and $V$ is an affine open. Let $P$ be a perfect object of $D(\mathcal{O}_ U)$ which is a generator for $D_\mathit{QCoh}(\mathcal{O}_ U)$. Using Lemma 36.15.1 we may choose a perfect object $Q$ of $D(\mathcal{O}_ X)$ whose restriction to $U$ is a direct sum one of whose summands is $P$. Say $V = \mathop{\mathrm{Spec}}(A)$. Let $Z = X \setminus U$. This is a closed subset of $V$ with $V \setminus Z$ quasi-compact. Choose $f_1, \ldots , f_ r \in A$ such that $Z = V(f_1, \ldots , f_ r)$. Let $K \in D(\mathcal{O}_ V)$ be the perfect object corresponding to the Koszul complex on $f_1, \ldots , f_ r$ over $A$. Note that since $K$ is supported on $Z \subset V$ closed, the pushforward $K' = R(V \to X)_*K$ is a perfect object of $D(\mathcal{O}_ X)$ whose restriction to $V$ is $K$ (see Cohomology, Lemma 20.49.10). We claim that $Q \oplus K'$ is a generator for $D_\mathit{QCoh}(\mathcal{O}_ X)$.

Let $E$ be an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$ such that there are no nontrivial maps from any shift of $Q \oplus K'$ into $E$. By Cohomology, Lemma 20.33.6 we have $K' = R(V \to X)_! K$ and hence

Thus by Lemma 36.15.2 the vanishing of these groups implies that $E|_ V$ is isomorphic to $R(U \cap V \to V)_*E|_{U \cap V}$. This implies that $E = R(U \to X)_*E|_ U$ (small detail omitted). If this is the case then

which contains $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(P[n], E|_ U)$ as a direct summand. Thus by our choice of $P$ the vanishing of these groups implies that $E|_ U$ is zero. Whence $E$ is zero. $\square$

Proof. We will prove this using the induction principle of Cohomology of Schemes, Lemma 30.4.1.

Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. In this case there exist $f_1, \ldots , f_ r \in A$ such that $T = V(f_1, \ldots , f_ r)$. Let $K$ be the Koszul complex on $f_1, \ldots , f_ r$ as in Lemma 36.15.2. Then $K$ is a perfect object with cohomology supported on $T$ and hence a perfect object of $D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$. On the other hand, if $E \in D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$ and $\mathop{\mathrm{Hom}}\nolimits (K, E[n]) = 0$ for all $n$, then Lemma 36.15.2 tells us that $E = Rj_*(E|_{X \setminus T}) = 0$. Hence $K$ generates $D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$, (by our definition of generators of triangulated categories in Derived Categories, Definition 13.36.3).

Assume that $X = U \cup V$ is an open covering with $V$ affine and $U$ quasi-compact such that the lemma holds for $U$. Let $P$ be a perfect object of $D(\mathcal{O}_ U)$ supported on $T \cap U$ which is a generator for $D_{\mathit{QCoh}, T \cap U}(\mathcal{O}_ U)$. Using Lemma 36.13.11 we may choose a perfect object $Q$ of $D(\mathcal{O}_ X)$ supported on $T$ whose restriction to $U$ is a direct sum one of whose summands is $P$. Write $V = \mathop{\mathrm{Spec}}(B)$. Let $Z = X \setminus U$. Then $Z$ is a closed subset of $V$ such that $V \setminus Z$ is quasi-compact. As $X$ is quasi-separated, it follows that $Z \cap T$ is a closed subset of $V$ such that $W = V \setminus (Z \cap T)$ is quasi-compact. Thus we can choose $g_1, \ldots , g_ s \in B$ such that $Z \cap T = V(g_1, \ldots , g_ r)$. Let $K \in D(\mathcal{O}_ V)$ be the perfect object corresponding to the Koszul complex on $g_1, \ldots , g_ s$ over $B$. Note that since $K$ is supported on $(Z \cap T) \subset V$ closed, the pushforward $K' = R(V \to X)_*K$ is a perfect object of $D(\mathcal{O}_ X)$ whose restriction to $V$ is $K$ (see Cohomology, Lemma 20.49.10). We claim that $Q \oplus K'$ is a generator for $D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$.

Let $E$ be an object of $D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$ such that there are no nontrivial maps from any shift of $Q \oplus K'$ into $E$. By Cohomology, Lemma 20.33.6 we have $K' = R(V \to X)_! K$ and hence

Thus by Lemma 36.15.2 we have $E|_ V = Rj_*E|_ W$ where $j : W \to V$ is the inclusion. Picture

Since $E$ is supported on $T$ we see that $E|_ W$ is supported on $T \cap W = T \cap U \cap V$ which is closed in $W$. We conclude that

where the second equality is part (1) of Cohomology, Lemma 20.33.6. This implies that $E = R(U \to X)_*E|_ U$ (small detail omitted). If this is the case then

which contains $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(P[n], E|_ U)$ as a direct summand. Thus by our choice of $P$ the vanishing of these groups implies that $E|_ U$ is zero. Whence $E$ is zero. $\square$