From complex solution to solutions over finite fields There are several ways (Hilbert's Nullstellensatz, model theory, transcendence bases etc.) to prove the following (amazing!) result:
If $f_1,...,f_r$ is a system of polynomials in $n$ variables with integral coefficients, then it has a solution with coordinates in $\mathbb{C}$ if and only if it has solutions with coordinates in $\overline{\mathbb{F}_p}$ for almost all primes $p$.
Question: What are interesting, explicit examples of the implication which yields solutions over finite fields out of a complex solution? Is there a system of polynomials, where the primes $p$ such that there is a solution over $\overline{\mathbb{F}_p}$ are not known, and their existence is only known by the abstract result above? I am not interested in polynomials which somehow artifically encode some undecidable statements of ZFC ;).
 A: I maintain that there's no reasonable way to go in the direction you want. For example, when $r = 1$ the single polynomial $f(x) = x^2 + 1$ has roots in $\mathbb{C}$ and roots in $\overline{ \mathbb{F}_p }$ for all $p$, but for the latter roots half of the time they lie in $\mathbb{F}_p$ and half of the time they lie in $\mathbb{F}_{p^2}$ and I don't see any reasonable way to write them down using the roots over $\mathbb{C}$ somehow.
In the other direction, I doubt it is possible to get more explicit than the following: if the $f_i$ have a solution over almost all $\overline{ \mathbb{F}_p }$, then they have a solution in the ultraproduct $\prod \overline{ \mathbb{F}_p }/U$ where $U$ is a non-principal ultrafilter on the primes. This ultraproduct is an algebraically closed field of characteristic zero with continuum cardinality, hence is abstractly isomorphic to $\mathbb{C}$. Note that there are two non-canonical choices here, both of which (I think) are generally impossible to make without some form of AC and both of which (I think) are unavoidable: the choice of a non-principal ultrafilter, then the choice of an abstract isomorphism with $\mathbb{C}$. 
