Lemma 36.16.3. Let $A$ be a ring. Let $X = \mathbf{P}^ n_ A$. Then
\[ E = \mathcal{O}_ X \oplus \mathcal{O}_ X(-1) \oplus \ldots \oplus \mathcal{O}_ X(-n) \]
is a generator (Derived Categories, Definition 13.36.3) of $D_\mathit{QCoh}(X)$.
Proof. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$. Assume $\mathop{\mathrm{Hom}}\nolimits (E, K[p]) = 0$ for all $p \in \mathbf{Z}$. We have to show that $K = 0$. By Derived Categories, Lemma 13.36.4 we see that $\mathop{\mathrm{Hom}}\nolimits (E', K[p])$ is zero for all $E' \in \langle E \rangle $ and $p \in \mathbf{Z}$. By Lemma 36.16.2 applied with $a = -n - 1$ we see that $\mathcal{O}_ X(-n - 1) \in \langle E \rangle $ because it is quasi-isomorphic to a finite complex whose terms are finite direct sums of summands of $E$. Repeating the argument with $a = -n - 2$ we see that $\mathcal{O}_ X(-n - 2) \in \langle E \rangle $. Arguing by induction we find that $\mathcal{O}_ X(-m) \in \langle E \rangle $ for all $m \geq 0$. Since
\[ \mathop{\mathrm{Hom}}\nolimits (\mathcal{O}_ X(-m), K[p]) = H^ p(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ X(m)) = H^ p(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ X(1)^{\otimes m}) \]
we conclude that $K = 0$ by Lemma 36.16.1. (This also uses that $\mathcal{O}_ X(1)$ is an ample invertible sheaf on $X$ which follows from Properties, Lemma 28.26.12.) $\square$
Comments (0)