The Stacks project

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

  1. $E = 0$, and

  2. $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(P[n], E) = 0$ 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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09IS. Beware of the difference between the letter 'O' and the digit '0'.