The Stacks project

Lemma 36.13.9. Let $X$ be an affine scheme. Let $T \subset X$ be a closed subset such that $X \setminus T$ is quasi-compact. Let $U \subset X$ be a quasi-compact open. For every perfect object $F$ of $D(\mathcal{O}_ U)$ supported on $T \cap U$ the object $F \oplus F[1]$ is the restriction of a perfect object $E$ of $D(\mathcal{O}_ X)$ supported in $T$.

Proof. Say $T = V(g_1, \ldots , g_ s)$. After replacing $g_ j$ by a power we may assume multiplication by $g_ j$ is zero on $F$, see Lemma 36.13.8. Choose $E$ as in Lemma 36.13.7. Note that $g_ j : E \to E$ restricts to zero on $U$. Choose a distinguished triangle

\[ E \xrightarrow {g_1} E \to C_1 \to E[1] \]

By Derived Categories, Lemma 13.4.11 the object $C_1$ restricts to $F \oplus F[1] \oplus F[1] \oplus F[2]$ on $U$. Moreover, $g_1 : C_1 \to C_1$ has square zero by Derived Categories, Lemma 13.4.5. Namely, the diagram

\[ \xymatrix{ E \ar[r] \ar[d]_0 & C_1 \ar[d]_{g_1} \ar[r] & E[1] \ar[d]_0 \\ E \ar[r] & C_1 \ar[r] & E[1] } \]

is commutative since the compositions $E \xrightarrow {g_1} E \to C_1$ and $C_1 \to E[1] \xrightarrow {g_1} E[1]$ are zero. Continuing, setting $C_{i + 1}$ equal to the cone of the map $g_ i : C_ i \to C_ i$ we obtain a perfect complex $C_ s$ on $X$ supported on $T$ whose restriction to $U$ gives

\[ F \oplus F[1]^{\oplus s} \oplus F[2]^{\oplus {s \choose 2}} \oplus \ldots \oplus F[s] \]

Choose morphisms of perfect complexes $\beta : C' \to C_ s$ and $\gamma : C' \to C_ s$ as in Lemma 36.13.6 such that $\beta |_ U$ is an isomorphism and such that $\gamma |_ U \circ \beta |_ U^{-1}$ is the morphism

\[ F \oplus F[1]^{\oplus s} \oplus F[2]^{\oplus {s \choose 2}} \oplus \ldots \oplus F[s] \to F \oplus F[1]^{\oplus s} \oplus F[2]^{\oplus {s \choose 2}} \oplus \ldots \oplus F[s] \]

which is the identity on all summands except for $F$ where it is zero. By Lemma 36.13.6 we also have $C' = C_ s \otimes ^\mathbf {L} I$ for some perfect complex $I$ on $X$. Hence the nullity of $g_ j^2\text{id}_{C_ s}$ implies the same thing for $C'$. Thus $C'$ is supported on $T$ as well. Then $\text{Cone}(\gamma )$ is a solution. $\square$

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Your email address will not be published.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 08EK. Beware of the difference between the letter 'O' and the digit '0'.