Relative generic flatness. - MathOverflow most recent 30 from http://mathoverflow.net2013-06-20T03:20:33Zhttp://mathoverflow.net/feeds/question/88001http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/88001/relative-generic-flatnessRelative generic flatness.Rami2012-02-09T15:57:36Z2012-02-09T21:03:52Z
<p>It is known that any morphism is flat at an open set of points. I'd like to know if there is a relative version of this fact. </p>
<p>Let $f: X \rightarrow Y$ and $g:Y \rightarrow S$ be morphisms of varieties, and let $h=g \circ f$ be their composition. Suppose that $h$ and $g$ are flat. Is it true that the set
$\{x\in X| f|_{h ^{-1}(h(x))} : h ^{-1}(h(x)) \to g ^{-1}(h(x)) \text{ is flat at } x \}$
is open?</p>
http://mathoverflow.net/questions/88001/relative-generic-flatness/88007#88007Answer by Laurent Moret-Bailly for Relative generic flatness.Laurent Moret-Bailly2012-02-09T16:41:44Z2012-02-09T16:41:44Z<p>With these assumptions, the set in question is the set of points in $X$ where $f$ is flat; hence it is open. This is "flatness by fibers", EGA IV (11.3.10) (applied with $\mathcal{F}=\mathcal{O}_X$).</p>