Lemma 38.13.11 (Finite type version of Theorem 38.13.6). Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume

1. $X \to S$ is locally of finite type,

2. $\mathcal{F}$ is an $\mathcal{O}_ X$-module of finite type, and

3. the set of weakly associated points of $S$ is locally finite in $S$.

Then $U = \{ x \in X \mid \mathcal{F}\text{ flat at }x\text{ over }S\}$ is open in $X$ and $\mathcal{F}|_ U$ is flat over $S$ and locally finitely presented relative to $S$ (see More on Morphisms, Definition 37.58.1).

Proof. The question is local on $X$ and $S$. Thus we may assume $X$ and $S$ are affine. Then we may choose a closed immersion $i : X \to \mathbf{A}^ n_ S$. We apply Theorem 38.13.6 to $X' = \mathbf{A}^ n_ S \to S$ and the quasi-coherent module $\mathcal{F}' = i_*\mathcal{F}$ of finite type and we find that

$U' = \{ x' \in X' \mid \mathcal{F}'\text{ flat at }x'\text{ over }S\}$

is open in $X'$ and that $\mathcal{F}'|_{U'}$ is of finite presentation. Since $\mathcal{F}'$ restricts to zero on $X' \setminus i(X)$ and since $\mathcal{F}'_{i(x)} \cong \mathcal{F}_ x$ for all $x \in X$ we see that

$U' = i(U) \amalg (X' \setminus i(X))$

Hence $U = i^{-1}(U')$ is open. Moreover, it is clear that $\mathcal{F}'|_{U'} = (i|_ U)_*(\mathcal{F}|_ U)$. Hence we conclude that $\mathcal{F}|_ U$ is finitely presented relative to $S$ by More on Morphisms, Lemmas 37.58.3 and 37.58.4. $\square$

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