Theorem 37.16.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume

1. $X$ is locally of finite presentation over $S$,

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

3. $Y$ is locally of finite type over $S$.

Let $x \in X$. Set $y = f(x)$ and let $s \in S$ be the image of $x$ in $S$. If $\mathcal{F}_ x \not= 0$, then the following are equivalent:

1. $\mathcal{F}$ is flat over $S$ at $x$, and $\mathcal{F}_ s$ is flat over $Y_ s$ at $x$, and

2. $Y$ is flat over $S$ at $y$ and $\mathcal{F}$ is flat over $Y$ at $x$.

Moreover, the set of points $x$ where (1) and (2) hold is open in $\text{Supp}(\mathcal{F})$.

Proof. Consider the ring maps

$\mathcal{O}_{S, s} \longrightarrow \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, x}$

and the module $\mathcal{F}_ x$. The stalk of $\mathcal{F}_ s$ at $x$ is the module $\mathcal{F}_ x/\mathfrak m_ s \mathcal{F}_ x$ and the local ring of $Y_ s$ at $y$ is $\mathcal{O}_{Y, y}/\mathfrak m_ s \mathcal{O}_{Y, y}$. Thus the implication (1) $\Rightarrow$ (2) is Algebra, Lemma 10.128.9. If (2) holds, then the first ring map is faithfully flat and $\mathcal{F}_ x$ is flat over $\mathcal{O}_{Y, y}$ so by Algebra, Lemma 10.39.4 we see that $\mathcal{F}_ x$ is flat over $\mathcal{O}_{S, s}$. Moreover, $\mathcal{F}_ x/\mathfrak m_ s \mathcal{F}_ x$ is the base change of the flat module $\mathcal{F}_ x$ by $\mathcal{O}_{Y, y} \to \mathcal{O}_{Y, y}/\mathfrak m_ s \mathcal{O}_{Y, y}$, hence flat by Algebra, Lemma 10.39.7.

By Morphisms, Lemma 29.21.11 the morphism $f$ is locally of finite presentation. Consider the set

37.16.2.1
$$\label{more-morphisms-equation-open} U = \{ x \in X \mid \mathcal{F} \text{ flat at }x \text{ over both }Y\text{ and }S\} .$$

This set is open in $X$ by Theorem 37.15.1. Note that if $x \in U$, then $\mathcal{F}_ s$ is flat at $x$ over $Y_ s$ as a base change of a flat module under the morphism $Y_ s \to Y$, see Morphisms, Lemma 29.25.7. Hence at every point of $U \cap \text{Supp}(\mathcal{F})$ condition (1) is satisfied. On the other hand, it is clear that if $x \in \text{Supp}(\mathcal{F})$ satisfies (1) and (2), then $x \in U$. Thus the open set we are looking for is $U \cap \text{Supp}(\mathcal{F})$. $\square$

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