The Stacks project

Lemma 38.28.6. In Situation 38.28.1 let $K$ be as in Lemma 38.28.2. Denote $X_0 \subset X$ the closed subset consisting of points lying over the closed subset $\mathop{\mathrm{Spec}}(A_1) = \mathop{\mathrm{Spec}}(A_2) = \ldots $ of $\mathop{\mathrm{Spec}}(A)$. There exists an open $W \subset X$ containing $X_0$ such that

  1. $H^ i(K)|_ W$ is zero unless $i = 0$,

  2. $\mathcal{F} = H^0(K)|_ W$ is of finite presentation, and

  3. $\mathcal{F}_ n = \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{O}_{X_ n}$.

Proof. Fix $n \geq 1$. By construction there is a canonical map $K \to \mathcal{F}_ n$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ and hence a canonical map $H^0(K) \to \mathcal{F}_ n$ of quasi-coherent sheaves. This explains the meaning of part (3).

Let $x \in X_0$ be a point. We will find an open neighbourhood $W$ of $x$ such that (1), (2), and (3) are true. Since $X_0$ is quasi-compact this will prove the lemma. Let $U \subset X$ be an affine open neighbourhood of $x$. Say $U = \mathop{\mathrm{Spec}}(B)$. Choose a surjection $P \to B$ with $P$ smooth over $A$. By Lemma 38.28.4 and the definition of relative pseudo-coherence there exists a bounded above complex $F^\bullet $ of finite free $P$-modules representing $Ri_*K$ where $i : U \to \mathop{\mathrm{Spec}}(P)$ is the closed immersion induced by the presentation. Let $M_ n$ be the $B$-module corresponding to $\mathcal{F}_ n|_ U$. By Lemma 38.28.5

\[ H^ i(F^\bullet \otimes _ A A_ n) = \left\{ \begin{matrix} 0 & \text{if} & i \not= 0 \\ M_ n & \text{if} & i = 0 \end{matrix} \right. \]

Let $i$ be the maximal index such that $F^ i$ is nonzero. If $i \leq 0$, then (1), (2), and (3) are true. If not, then $i > 0$ and we see that the rank of the map

\[ F^{i - 1} \to F^ i \]

in the point $x$ is maximal. Hence in an open neighbourhood of $x$ inside $\mathop{\mathrm{Spec}}(P)$ the rank is maximal. Thus after replacing $P$ by a principal localization we may assume that the displayed map is surjective. Since $F^ i$ is finite free we may choose a splitting $F^{i - 1} = F' \oplus F^ i$. Then we may replace $F^\bullet $ by the complex

\[ \ldots \to F^{i - 2} \to F' \to 0 \to \ldots \]

and we win by induction on $i$. $\square$


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