Lemma 66.31.4. Let $S$ be a scheme. Let $X \to Y \to Z$ be morphisms of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Let $x \in |X|$ with image $y \in |Y|$.

1. If $\mathcal{F}$ is flat at $x$ over $Y$ and $Y$ is flat at $y$ over $Z$, then $\mathcal{F}$ is flat at $x$ over $Z$.

2. Let $x : \mathop{\mathrm{Spec}}(K) \to X$ be a representative of $x$. If

1. $\mathcal{F}$ is flat at $x$ over $Y$,

2. $x^*\mathcal{F} \not= 0$, and

3. $\mathcal{F}$ is flat at $x$ over $Z$,

then $Y$ is flat at $y$ over $Z$.

3. Let $\overline{x}$ be a geometric point of $X$ lying over $x$ with image $\overline{y}$ in $Y$. If $\mathcal{F}_{\overline{x}}$ is a faithfully flat $\mathcal{O}_{Y, \overline{y}}$-module and $\mathcal{F}$ is flat at $x$ over $Z$, then $Y$ is flat at $y$ over $Z$.

Proof. Pick $\overline{x}$ and $\overline{y}$ as in part (3) and denote $\overline{z}$ the induced geometric point of $Z$. Via the characterization of flatness in Lemmas 66.31.1 and 66.30.8 the lemma reduces to a purely algebraic question on the local ring map $\mathcal{O}_{Z, \overline{z}} \to \mathcal{O}_{Y, \overline{y}}$ and the module $\mathcal{F}_{\overline{x}}$. Part (1) follows from Algebra, Lemma 10.39.4. We remark that condition (2)(b) guarantees that $\mathcal{F}_{\overline{x}}/ \mathfrak m_{\overline{y}} \mathcal{F}_{\overline{x}}$ is nonzero. Hence (2)(a) $+$ (2)(b) imply that $\mathcal{F}_{\overline{x}}$ is a faithfully flat $\mathcal{O}_{Y, \overline{y}}$-module, see Algebra, Lemma 10.39.15. Thus (2) is a special case of (3). Finally, (3) follows from Algebra, Lemma 10.39.10. $\square$

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