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Let $X$ and $S$ $k$-schemes of finite type . ($k$ a field) and $U$ an open subset of $X$

Let $f:X\rightarrow S$ a $k$-morphism of finite type.

We assume that for any closed point $s\in S(\bar{k})$, $ U_{s}\neq\emptyset$, do we have that $U_{s}$ is non-empty for any $s\in S$?

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Sure, just use Chevalley's theorem that the image of any finite type map between noetherian schemes is constructible (in the space of the entire target scheme, not just on its subspace of closed points!). Since a constructible subset of $S$ containing all closed points is the entire space (proof: reformulate via emptiness of its constructible complement!), that's it. More generally, for any Jacobson scheme $S$ and its subspace of closed points $S^0$, $A \mapsto A\cap S^0$ is an inclusion-preserving bijection between the sets of constructible subsets of $S$ and $S^0$. –  user61789 Jun 29 '13 at 14:54
so in fact if $S$ is a Jacobson scheme and $f:X\rightarrow S$ a finitely presented morphism then the statement is true? –  prochet Jun 29 '13 at 15:27
Yeah, 10.1--10.4 in IV$_3$. –  user61789 Jun 29 '13 at 22:41
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