The Stacks project

Proof. We will prove this using the induction principle of Lemma 75.9.3.

If $X$ is affine, then $\mathcal{O}_ X$ is a perfect generator. This follows from Lemma 75.4.2 and Derived Categories of Schemes, Lemma 36.3.5.

Assume that $(U \subset X, f : V \to X)$ is an elementary distinguished square with $U$ quasi-compact such that the theorem holds for $U$ and $V$ is an affine scheme. Let $P$ be a perfect object of $D(\mathcal{O}_ U)$ which is a generator for $D_\mathit{QCoh}(\mathcal{O}_ U)$. Using Lemma 75.15.3 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 $Z \subset V$ be the reduced closed subscheme which is the inverse image of $X \setminus U$ and maps isomorphically to it (see Definition 75.9.1). This is a retrocompact closed subset of $V$. 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 $Z$, the pushforward $K' = Rf_*K$ is a perfect object of $D(\mathcal{O}_ X)$ whose restriction to $V$ is $K$ (see Lemmas 75.14.3 and 75.10.7). 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 Lemma 75.10.7 we have $K' = f_! K$ and hence

Thus by Derived Categories of Schemes, Lemma 36.15.2 (using also Lemma 75.4.2) the vanishing of these groups implies that $E|_ V$ is isomorphic to $R(U \times _ X V \to V)_*E|_{U \times _ X V}$. 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. $\square$