The Stacks project

[Theorem 6.8, Rouquier-dimensions]

Lemma 36.15.4. Let $X$ be a quasi-compact and quasi-separated scheme. Let $T \subset X$ be a closed subset such that $X \setminus T$ is quasi-compact. With notation as above, the category $D_{\mathit{QCoh}, T}(\mathcal{O}_ X)$ is generated by a single perfect object.

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 36.13.10 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 20.46.10). 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

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

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

\[ \xymatrix{ W \ar[r]_ j & V & Z \cap T \ar[l] \ar[d] \\ U \cap V \ar[u]^{j'} \ar[ru]_{j''} & & Z \ar[lu] } \]

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

\[ E|_ V = Rj_*(E|_ W) = Rj_*(Rj'_*(E|_{U \cap V})) = Rj''_*(E|_{U \cap V}) \]

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

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

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 0A9A. Beware of the difference between the letter 'O' and the digit '0'.