Lemma 67.30.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Then $f$ is flat if and only if $f$ is flat at all points of $|X|$.
Proof. Choose a commutative diagram
where $U$ and $V$ are schemes, the vertical arrows are étale, and $a$ is surjective. By definition $f$ is flat if and only if $h$ is flat (Definition 67.22.2). By definition $f$ is flat at $x \in |X|$ if and only if $h$ is flat at some (equivalently any) $u \in U$ which maps to $x$ (Definition 67.22.6). Thus the lemma follows from the fact that a morphism of schemes is flat if and only if it is flat at all points of the source (Morphisms, Definition 29.25.1). $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like
$\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.