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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 ;).

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    Indicate the nature of the coefficients of the f_i (e.g., they can't be random complex numbers or it wouldn't make sense to speak of solns in char. p) and what "almost all" means.2011-04-24
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    I assume the $f_i$ have integer coefficients and "almost all" means "all but finitely many." But I am not sure what to make of the question. Even when $r = 1$ how are you supposed to produce a root of an integer polynomial over $\overline{\mathbb{F}_p}$ out of a complex root? I don't think you can say anything more than "the two fields are both algebraically closed, therefore this polynomial admits a root over both of them."2011-04-24
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    Qiaochu: Yes, I too presumed that Martin meant to specify those condition just as you wrote them. But I still think he should include them in his question.2011-04-25
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    I can't give an example, but I can give a very interesting application of the "if" direction (which I think can also be proved using complex analysis): if $f: \mathbb{C}^n \to \mathbb{C}^n$ is an injective polynomial map (with integer coefficients), then $f$ is surjective.2011-04-25
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    @KCd: I've edited the question. In such obvious cases, feel free to edit it by yourself. @Justin: Nice application. Grothendieck has generalized it to schemes, when I remember correctly. To which polynomial system do you apply the result?2011-04-26
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    http://en.wikipedia.org/wiki/Ax%E2%80%93Grothendieck_theorem2011-04-26

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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 $\mathbb{C}$.

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    Yes I know that. My question was not about constructing roots, but rather if (see my question) there are proofs that certain equations are solvable by using this procedure, whereas no other proof is known.2011-05-29
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    @Martin: I see. I was confused by your use of the word "yield." I thought you wanted an algorithm which, given a complex root, would construct a root over finite fields. Perhaps you might find an answer to your question in some moduli space.2011-05-29