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

$X \to S$ is locally of finite presentation,

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

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 an $\mathcal{O}_ U$-module of finite presentation and flat over $S$.

**Proof.**
Let $x \in X$ be such that $\mathcal{F}$ is flat at $x$ over $S$. We have to find an open neighbourhood of $x$ such that $\mathcal{F}$ restricts to a $S$-flat finitely presented module on this neighbourhood. The problem is local on $X$ and $S$, hence we may assume that $X$ and $S$ are affine. As $\mathcal{F}_ x$ is a finitely presented $\mathcal{O}_{X, x}$-module by Lemma 38.10.9 we conclude from Algebra, Lemma 10.126.5 there exists a finitely presented $\mathcal{O}_ X$-module $\mathcal{F}'$ and a map $\varphi : \mathcal{F}' \to \mathcal{F}$ which induces an isomorphism $\varphi _ x : \mathcal{F}'_ x \to \mathcal{F}_ x$. In particular we see that $\mathcal{F}'$ is flat over $S$ at $x$, hence by openness of flatness More on Morphisms, Theorem 37.15.1 we see that after shrinking $X$ we may assume that $\mathcal{F}'$ is flat over $S$. As $\mathcal{F}$ is of finite type after shrinking $X$ we may assume that $\varphi $ is surjective, see Modules, Lemma 17.9.4 or alternatively use Nakayama's lemma (Algebra, Lemma 10.20.1). By Lemma 38.13.5 we have

\[ \text{WeakAss}_ X(\mathcal{F}') \subset \bigcup \nolimits _{s \in \text{WeakAss}(S)} \text{Ass}_{X_ s}(\mathcal{F}'_ s) \]

As $\text{WeakAss}(S)$ is finite by assumption and since $\text{Ass}_{X_ s}(\mathcal{F}'_ s)$ is finite by Divisors, Lemma 31.2.5 we conclude that $\text{WeakAss}_ X(\mathcal{F}')$ is finite. Using Algebra, Lemma 10.15.2 we may, after shrinking $X$ once more, assume that $\text{WeakAss}_ X(\mathcal{F}')$ is contained in the generalization of $x$. Now consider $\mathcal{K} = \mathop{\mathrm{Ker}}(\varphi )$. We have $\text{WeakAss}_ X(\mathcal{K}) \subset \text{WeakAss}_ X(\mathcal{F}')$ (by Divisors, Lemma 31.5.4) but on the other hand, $\varphi _ x$ is an isomorphism, also $\varphi _{x'}$ is an isomorphism for all $x' \leadsto x$. We conclude that $\text{WeakAss}_ X(\mathcal{K}) = \emptyset $ whence $\mathcal{K} = 0$ by Divisors, Lemma 31.5.5.
$\square$

## Comments (1)

Comment #1114 by Simon Pepin Lehalleur on

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