Lemma 65.37.7. A smooth morphism of algebraic spaces is flat.

**Proof.**
Let $X \to Y$ be a smooth morphism of algebraic spaces. By definition this means there exists a diagram as in Lemma 65.22.1 with $h$ smooth and surjective vertical arrow $a$. By Morphisms, Lemma 29.34.8 $h$ is flat. Hence $X \to Y$ is flat by definition.
$\square$

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