Let $f : X\rightarrow S$ be a flat finite type morphism of schemes with $S$ integral and Noetherian. Let $\eta\in S$ be the generic point.
Let $\{\sigma_i\}$ be a collection of sections of $f$ (possibly infinite), which are Zariski dense in $X_\eta$. I'm interested in additional conditions on $f,\{\sigma_i\}$ under which one or both of the following properties are satisfied:
P1: There exists a point $s\in S – \{\eta\}$ such that $\{\sigma_i\}$ is dense in the fiber $X_s$.
P2: There exists a nonempty open $U\subset S$ such that P1 holds for every $s\in U$.
Clearly a necessary condition is that $S – \{\eta\}$ needs to be “large'' (e.g., P1 will often fail if $S$ is the spec of a discrete valuation ring). From now on lets assume $S – \{\eta\}$ is Zariski dense in $S$.
My intuition is that under mild additional assumptions there should be some kind of semicontinuity result for the dimensions of the Zariski closures inside fibers. In particular, the set of $s\in S$ such that $\{\sigma_i\}$ is not Zariski dense in $X_s$ should be closed in $S$. However I'm not aware of any results in this direction.
Here are some specific questions:
(1) Do P1, P2 hold under the above assumptions?
(2) What if we also assume $f$ has geometrically irreducible fibers?
(3) What if $X$ is a semisimple affine algebraic $S$-group scheme?
Best Answer
Building on Yosemite Stan's example, for any scheme over any number ring, even if you take all the sections, it will still not be Zariski dense in any special fiber, because all sections go through the rational points of the special fiber, which are not dense as the base field is finite. Of course, there are many examples of schemes over number rings with Zariski dense global sections.
Similar examples work for $\mathbb A^1$ over any curve of dimension $1$ over a countable field. There are countably many closed points, order them, and choose the $n$th section to agree with at least one of the previous $n-1$ sections at all of the first $n-1$ closed points. Then in the $n$'th point there will be at most $n$ distinct sections, so they won't be Zariski dense.
Over an uncountable field, the same construction proves that any countable set of points can be the set where the sections aren't dense, so there will not be an open set where the sections are dense. However, one can prove there is at least one closed point where the sections are dense.
First, pick a countable subset of the sections that is Zariski dense at the generic point. Then, for every $d$, by Zariski density we can find some finite set of sections that don't satisfy any nontrivial degree $d$ equation over the generic point, and then the set where those sections do satisfy some nontrivial degree $d$ equation is contained in a proper closed subset. Over un uncountable field, the complement of the union of countably many proper closed subsets will be nonempty, and any point in that nonempty closed set does the trick.