# on flat morphisms

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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Why the heck should it imply that $f$ is flat? Did you forget a hypothesis? –  Angelo Apr 14 '13 at 16:16
There is no reason for $f$ to be flat outside $U$. For example $S=\mathbb{A}^1=U$, $X = \mathbb{A}^1 \sqcup pt$. –  Piotr Achinger Apr 14 '13 at 16:17
...or let $f:X\to S$ be any birational morphism and $U\subseteq X$ the locus where it is an isomorphism. –  Sándor Kovács Apr 14 '13 at 17:52
Is it true when $U$ is "large" in some sense? –  Martin Brandenburg Apr 14 '13 at 21:13