Lemma 36.31.3 (0BDK)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 36: Derived Categories of Schemes
Section 36.31: Perfect complexes
Lemma 36.31.3 (cite)

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Lemma 36.31.3. Let $X$ be a scheme. Let $E \in D(\mathcal{O}_ X)$ be perfect. Given $i, r \in \mathbf{Z}$ , there exists an open subscheme $U \subset X$ characterized by the following

$E|_ U \cong H^ i(E|_ U)[-i]$ and $H^ i(E|_ U)$ is a locally free $\mathcal{O}_ U$ -module of rank $r$ ,
a morphism $f : Y \to X$ factors through $U$ if and only if $Lf^*E$ is isomorphic to a locally free module of rank $r$ placed in degree $i$ .

Proof. Let $\beta _ j : X \to \{ 0, 1, 2, \ldots \}$ for $j \in \mathbf{Z}$ be the functions of Lemma 36.31.1. Then the set

$W = \{ x \in X \mid \beta _ j(x) \leq 0\text{ for all }j \not= i\}$

is open in $X$ and its formation commutes with pullback to any $Y$ over $X$ . This follows from the lemma using that apriori in a neighbourhood of any point only a finite number of the $\beta _ j$ are nonzero. Thus we may replace $X$ by $W$ and assume that $\beta _ j(x) = 0$ for all $x \in X$ and all $j \not= i$ . In this case $H^ i(E)$ is a finite locally free module and $E \cong H^ i(E)[-i]$ , see for example More on Algebra, Lemma 15.75.7. Thus $X$ is the disjoint union of the open subschemes where the rank of $H^ i(E)$ is fixed and we win. $\square$

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