# On an openess property

Let $S=Spec(A)$ where $A$ is a noetherian integral domain. Let $f:X\rightarrow S$ be a flat, proper morphism of schemes. Let $U\subset X$ be an open and $V=f(U)$ (in particular $V$ is open by flatness). Assume that the fibers $X_s\subset U$ for $s$ in a dense set of points of finite type of $S$. Is it true that I can find a non empty open $V_1\in V$ such that $U\times_S V_1=X\times_S V_1$?

• Let us consider $S=Spec(k)$, where $k$ is a field. Then for any non-empty open $V_1$, $V_1=S=Spec(k)$, therefore you would need $U=X$ in order to have $U\times V_1=X\times V_1$. – Marci Jul 11 '13 at 0:26
• @Marci this is the simple case. Indeed $X_s\subset U$ for $s$ is a dense set of points. – gemi Jul 11 '13 at 8:40

It is enough to suppose $X_s\subseteq U$ for one point $s\in S$ and $f$ proper (flatness is useless).
Indeed, $X\setminus U$ is closed, so $f(X\setminus U)$ is closed and is different from $S$ because it does not contain $s$. Now take $V_1=S\setminus f(X\setminus U)$. It is dense in $S$ because the latter is irreducible. By construction, $f^{-1}(V_1)\subseteq U$, hence $$X\times_S V_1=f^{-1}(V_1)=U\cap f^{-1}(V_1)=U\times_S V_1.$$
• Very nice, QiL'8 . So the result only requires $f$ to be closed: it is purely topological (density of $V_1$ not being required) and has nothing to do with algebraic geometry, nor schemes, nor arithmetic geometry (the tags chosen by gemi) ! It never ceases to amaze me that theorems can become much easier when you forget about most of the hypotheses (here: scheme, noetherian, flat, proper, dense set of $s$) ! – Georges Elencwajg Jul 14 '13 at 12:08