Lemma 67.30.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\overline{x}$ be a geometric point of $X$ lying over the point $x \in |X|$. Let $\overline{y} = f \circ \overline{x}$. The following are equivalent
$f$ is flat at $x$, and
the map on étale local rings $\mathcal{O}_{Y, \overline{y}} \to \mathcal{O}_{X, \overline{x}}$ is flat.
Proof. Choose a commutative diagram
where $U$ and $V$ are schemes, $a, b$ are étale, and $u \in U$ mapping to $x$. We can find a geometric point $\overline{u} : \mathop{\mathrm{Spec}}(k) \to U$ lying over $u$ with $\overline{x} = a \circ \overline{u}$, see Properties of Spaces, Lemma 66.19.4. Set $\overline{v} = h \circ \overline{u}$ with image $v \in V$. We know that
see Properties of Spaces, Lemma 66.22.1. We obtain a commutative diagram
of local rings with flat horizontal arrows. We have to show that the left vertical arrow is flat if and only if the right vertical arrow is. Algebra, Lemma 10.39.9 tells us $\mathcal{O}_{U, u}$ is flat over $\mathcal{O}_{V, v}$ if and only if $\mathcal{O}_{X, \overline{x}}$ is flat over $\mathcal{O}_{V, v}$. Hence the result follows from More on Flatness, Lemma 38.2.5. $\square$
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