Lemma 36.15.2. In Situation 36.9.1 denote $j : U \to X$ the open immersion and let $K$ be the perfect object of $D(\mathcal{O}_ X)$ corresponding to the Koszul complex on $f_1, \ldots , f_ r$ over $A$. For $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ the following are equivalent

$E = Rj_*(E|_ U)$, and

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

**Proof.**
Choose a distinguished triangle $E \to Rj_*(E|_ U) \to N \to E[1]$. Observe that

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[n], Rj_*(E|_ U)) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(K|_ U[n], E) = 0 \]

for all $n$ as $K|_ U = 0$. Thus it suffices to prove the result for $N$. In other words, we may assume that $E$ restricts to zero on $U$. Observe that there are distinguished triangles

\[ K^\bullet (f_1^{e_1}, \ldots , f_ i^{e'_ i}, \ldots , f_ r^{e_ r}) \to K^\bullet (f_1^{e_1}, \ldots , f_ i^{e'_ i + e''_ i}, \ldots , f_ r^{e_ r}) \to K^\bullet (f_1^{e_1}, \ldots , f_ i^{e''_ i}, \ldots , f_ r^{e_ r}) \to \ldots \]

of Koszul complexes, see More on Algebra, Lemma 15.28.11. Hence if $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(K[n], E) = 0$ for all $n \in \mathbf{Z}$ then the same thing is true for the $K$ replaced by $K_ e$ as in Lemma 36.9.6. Thus our lemma follows immediately from that one and the fact that $E$ is determined by the complex of $A$-modules $R\Gamma (X, E)$, see Lemma 36.3.5.
$\square$

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