Dimension of fibers of scheme over $\mathbb{Z}$

Question

Let $X$ be a reduced scheme of finite type over $\mathbb{Z}$: given a prime $p \in \mathbb{Z}$,I'm going to denote as $X_p$ its base change over $\mathbb{F}_p$. I would be interested in knowing the following:is it true that the dimension of $X_p$ is eventually costant?(i.e there exists an $n \in \mathbb{N}$ such that for every $p \geq n$, we have $\dim X_p=n$).

I tried to reduce to the affine and irreducible case, so that $X=\operatorname{Spec}(A)$ with $A$ a finitely generated domain. By a stronger version of Noether normalization lemma, there exists an $f \in \mathbb{Z}$ such that there exists a finite injective morphism $\mathbb{Z}_f[x_1,x_2, \dots, x_r] \subseteq A_f$.Now, we have $\dim X_f=r+1$ and I would like that to imply $\dim X_p = r$ for every $p $ such that $p$ does not divide $f$. The only thing that might be useful that comes to my mind here is the standard result about fiber dimension:

$$\dim \mathcal{O}_{X_y,x} \geq \dim \mathcal{O}_{X,x}-\dim \mathcal{O}_{Y,y} $$

However, we do not have a real control over $\dim \mathcal{O}_{X,x}$ so I do not know how to conclude. Is there any known statement in the literature? Are there any mild conditions under which this fact is true?

I would also be interested in the following fact: suppose that eventually $\dim X_p=0$ so that every fiber is a finite collection of points. I would like to have $X_p(\overline{\mathbb{F}_p})$ to be also eventually constant. I imagine that one could actually prove something like that the number of irreducible components of maximal dimension of $X_p$ should be eventually constant, but I would know how to prove this.

Answer 1

This is a simple application of some genericness results.

To handle the dimension side of things, use Stacks 05F7:

Lemma: Let $f:X\to Y$ be a morphism of schemes. Assume $Y$ is irreducible with generic point $\eta$ and $f$ of finite type. If $X_\eta$ has dimension $n$, then there exists a nonempty open set $V\subset Y$ so that for all $y\in V$, the fiber $X_y$ has dimension $n$.

Applied to our situation at hand, the open subset $V$ contains all but finitely many primes, so there is a largest prime $p$ not in $V$. Then $\dim X_q$ is constant for all $q>p$.

For the eventually-dimension-zero case, we note that by our previous work, there's an affine open $U=\operatorname{Spec} S^{-1}\Bbb Z \subset \operatorname{Spec} \Bbb Z$ so that $X_U=\operatorname{Spec} S^{-1}A$ is dimension zero. By your noether-normalization type lemma and an application of generic freeness, we may shrink $U$ so that $S^{-1}A$ is actually finite free over $S^{-1}\Bbb Z$.

As $X\to \operatorname{Spec}\Bbb Z$ is finite type and the generic fiber is geometrically reduced (because the localization of a domain is a domain and reduced implies geometrically reduced in characteristic zero), Stacks 0578 applies and we can again shrink $U$ a bit so that on this open set, $X_U\to U$ has geometrically reduced fibers.

At this point, we have that the fiber over every point in $U$ is geometrically reduced and module-free of the same finite rank, so the only obstruction to saying that it has the number of $\overline{\Bbb F_p}$-points equal to the rank is that it be a product of separable field extensions of $\Bbb F_p$. But this is not hard: a non-separable field extension has degree divisible by $p$, so all we need to do to guarantee separability is to restrict to primes larger than the finite, constant rank of our fiber. (This last bit was unnecessary because finite fields are perfect and therefore all algebraic extensions are separable.) So the result is proven.

Your idea about irreducible components is also correct, and can be handled by mostly the same sorts of logic as above: the situation at the generic point will be the situation on a dense open set. I leave the details to you.