Lemma 38.2.3. Let $X \to T \to S$ be morphisms of schemes with $T \to S$ étale. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $x \in X$ be a point. Then
In particular $\mathcal{F}$ is flat over $S$ if and only if $\mathcal{F}$ is flat over $T$.
Proof. As an étale morphism is a flat morphism (see Morphisms, Lemma 29.36.12) the implication “$\Leftarrow $” follows from Algebra, Lemma 10.39.4. For the converse assume that $\mathcal{F}$ is flat at $x$ over $S$. Denote $\tilde x \in X \times _ S T$ the point lying over $x$ in $X$ and over the image of $x$ in $T$ in $T$. Then $(X \times _ S T \to X)^*\mathcal{F}$ is flat at $\tilde x$ over $T$ via $\text{pr}_2 : X \times _ S T \to T$, see Morphisms, Lemma 29.25.7. The diagonal $\Delta _{T/S} : T \to T \times _ S T$ is an open immersion; combine Morphisms, Lemmas 29.35.13 and 29.36.5. So $X$ is identified with open subscheme of $X \times _ S T$, the restriction of $\text{pr}_2$ to this open is the given morphism $X \to T$, the point $\tilde x$ corresponds to the point $x$ in this open, and $(X \times _ S T \to X)^*\mathcal{F}$ restricted to this open is $\mathcal{F}$. Whence we see that $\mathcal{F}$ is flat at $x$ over $T$. $\square$
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