Relating the normalization of a variety over $\mathbb{Z}$ to its reduction mod $p$. I'm looking for a proof (or better, a reference) for the following claim.
Claim: Let $X$ be an (irreducible) variety defined over $\mathbb{Z}$. Let $\nu:\tilde{X}\rightarrow X$ be its normalization. Then, for large enough primes $p$, $\tilde{X}/p$ is the normalization of $X/p$.
The way I was trying to do this was by appealing to Serre's "normal = $R1+S2$" criterion. Doing this, it isn't difficult to prove that $X/p$ is $R1$ for all large primes. I'm having more difficulty with the $S2$ part.
So, I'd appreciate either an elementary/direct proof of the above claim or a proof (or counterexample) of "if $X$ is $S_k$ over $\mathbb{Z}$ then $X/p$ is $S_k$ over $\mathbb{F}_p$ for large enough primes $p$''.
 A: As you ask for a reference, look at EGA, IV.12.1.6(iv).  It is not in the form you want. But combined with the fact that the image of constructible subset is constructible (Theorem of Chevalley), it gives a complete answer to your question.
Edit Add some details. 
Let $d=\dim X_{\mathbb Q}=\dim X-1$.
Let $Z\subset X$ be a strict closed subset such that $\tilde{X} \setminus v^{-1}(Z)\to X\setminus Z$ is an isomorphism. Shrinking $S=\mathrm{Spec}(\mathbb Z)$ if necessary, we can suppose that for all $s\in S$, 
$$\dim Z_s = \dim Z_{\mathbb Q}< d =\dim X_s$$ 
(this is one of Chevalley's theorems, EGA IV.13.1.1). As the irreducible components of $X_s$ all have dimension $\ge d$ (loc. cit.), $Z_s$ is nonwhere dense in $X_s$ and $\tilde{X}_s\to X_s$ is birational. 
It remains to show that $\tilde{X}_s$ is normal over a small open subset of $S$.
As $\tilde{X}_{\mathbb Q}$ is normal, hence geometrically normal, shrinking $S$ if necessary, $\tilde{X}_s$ is geometrically normal (hence normal) for all $s\in S$ (EGA IV.12.1.6).
Remark It is not always possible to have $\tilde{X}_s$ irreducible for all but finitely many $s$. As a counterexample, consider $X=\mathbb A^1_K$ over a number field $K\ne \mathbb Q$. 
