# When must a set of sections which is Zariski dense in the generic fiber also be dense in some special fiber?

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?

• Hmmm, perhaps the following would work? (Hopefully I didn't misunderstand the setup.) Take $X=\mathbb{G}_m$ over $S=Spec(\mathbb{Z}[1/2])$. Then it looks like $2^n\in \mathbb{G}_m(\mathbb{Q})$ gives $\mathbb{Q}$-points (i.e. $\eta$-points) for any $n\in \mathbb{Z}$. This countable set of points is Zariski dense. But their reductions mod $p$ are all contained in the finite set $\mathbb{G}_m(\mathbb{F}_p)$, so they will not be Zariski dense. Dec 9, 2021 at 1:48

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.
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.
• Thanks for your answer! In your final paragraph, what do you mean by a "degree $d$ equation"? Are you still considering $\mathbb{A}^1$ over a curve? Also for the final sentence, presumably you wanted to say that the union of countably many proper closed subsets can't be the whole thing? Dec 9, 2021 at 4:43
• @stupid_question_bot I'm fixing an embedding to either $\mathbb A^n$ or $\mathbb P^n$. And yes. Dec 9, 2021 at 4:59
• To fix notation, suppose $k$ is an uncountable field, $S$ a $k$-scheme, and $X = \mathbb{A}^n_S$. Given $d\ge 1$, as you say you can find $n_d$ sections $\sigma_1,\ldots,\sigma_{n_d}$ which don't satisfy any nontrivial degree $d$ equations over the generic point of $S$. Why must the set $\{s\in S \;|\;\;(\sigma_1,\ldots,\sigma_{n_d})\cap X_s \text{ satisfies a nontrivial degree$d$equation in$X_s$}\}$ be closed? Dec 9, 2021 at 5:54