Let $j:U\rightarrow X$ an open immersion between k-schemes of finite type and $f:X\rightarrow S$ a surjective k-morphism of finite type.
We suppose that $f\circ j:U\rightarrow S$ is faithfully flat, does it imply that f is faithfully flat?
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7$\begingroup$ Why the heck should it imply that $f$ is flat? Did you forget a hypothesis? $\endgroup$– AngeloCommented Apr 14, 2013 at 16:16
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$\begingroup$ There is no reason for $f$ to be flat outside $U$. For example $S=\mathbb{A}^1=U$, $X = \mathbb{A}^1 \sqcup pt$. $\endgroup$– Piotr AchingerCommented Apr 14, 2013 at 16:17
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5$\begingroup$ ...or let $f:X\to S$ be any birational morphism and $U\subseteq X$ the locus where it is an isomorphism. $\endgroup$– Sándor KovácsCommented Apr 14, 2013 at 17:52
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$\begingroup$ Is it true when $U$ is "large" in some sense? $\endgroup$– Martin BrandenburgCommented Apr 14, 2013 at 21:13
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1 Answer
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In comments to the question, Martin Brandenburg asks if the answer should be positive for $U$ large. Here is one positive result in this direction.
Let $k$ be an algebraically closed field. A surjective birational $k$-morphism from a connected normal $k$-scheme of finite type to a regular $k$-scheme of finite type that is flat outside of a set of codimension $\geq 2$ is faithfully flat. To see this, use a purity theorem and the fact that for a birational morphism with a normal target, the flat locus coincides with the etale locus.
This fails if we drop the assumption on the source or on the target. This also fails if you do not assume that the morphism is birational.