Theorem 36.15.3. Let $X$ be a quasi-compact and quasi-separated scheme. The category $D_\mathit{QCoh}(\mathcal{O}_ X)$ can be generated by a single perfect object. More precisely, there exists a perfect object $P$ of $D(\mathcal{O}_ X)$ such that for $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ the following are equivalent

$E = 0$, and

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P[n], E) = 0$ for all $n \in \mathbf{Z}$.

**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

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K'[n], E) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ V)}(K[n], E|_ V) \]

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

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(Q[n], E) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(Q|_ U[n], E|_ U) \]

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.
$\square$

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