Let $S$ be some (fixed) subset of $\mathbb{Z} [X_1, \dots , X_n]$ which contains only homogeneous polynomials, and if $F$ is a field, let $X(F)$ be the set of $ x \in P^{n-1}(F)$ such that $f (x) = 0$ for any $f \in S$.

Now consider the assertion $(E_p) : X(F)$ is empty for any field $F$ of characteristic $p$. From basic valuation theory $^{*}$ we know that $(E_p)$ implies $(E_0)$ for any $p \geq 0$ (this is analogous to the fact that a diophantine equation which has solutions in $\mathbb{Z}$ has (obviously) solutions in each $\mathbb{F}_p$).
I'm interested in the converse implication. It is false in general that $(E_0)$ implies $(E_p)$ (for a fixed $p>0$) : just consider the set $S$ which contains only the constant polynomial $p$. But in this counterexample $E_l$ fails for only one prime $l$.

Hence the question : is it true that if $(E_0)$ holds, then $(E_p)$ also holds for all but finitely many primes ? If yes, is there some effective upper bound (in term of $S$) for the greatest prime for which $(E_p)$ fails ?

$^{*}$ Here is the argument : Let $F$ be a field of characteristic $0$, and suppose $X(F)$ contains a line $x$ of $F^n$. Since $\mathbb{Q} \subset F$, some non-archimedean norm $ |\cdot|$ on $F$ extends the $p$-adic norm $ |\cdot|_p$ (and has the same range). If $R = \overline{B}(0,1)$ and $ \mathfrak{m} = B(0,1)$ then $K = R/\mathfrak{m}$ is a field of characteristic $p$. Then the image of $x \cap R^n$ in $K^n$ is itself a line, therefore an element of $X(K)$ : contradiction.

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    $\begingroup$ I probably should have also mentioned that in the simplest case $F_1(X,Y)=F_2(X,Y)=0$, the existence of a nontrivial solution in an algebraically closed field $k$ is exactly the condition that the resultant $R(F_1,F_2)$ vanishes in $k$. $\endgroup$ – Joe Silverman Apr 22 '12 at 15:38

No. For example, $3x^3+4y^3+5z^3=0$ has nontrivial solutions mod $p$ for every prime $p$, but it has no nontrivial solutions in $\mathbb{Q}$. Indeed, it has solutions in every $p$-adic field $\mathbb{Q}_p$ and also solutions in $\mathbb{R}$. This is a famous example of Selmer. One says that the "Hasse principle" fails for this example. There is a subtler obstruction to "Local Solutions for all $p$ Implies Global Solutions" called the Brauer-Manin obstruction. The vanishing of the Brauer-Manin obstruction is necessary for the existence of a solution in $\mathbb{Q}$, but it is not always a sufficient condition. The study of such obstructions is an important area of current research.

EDIT: Sorry, I misinterpreted your question. I guess you're asking whether "no solutions in $\mathbb{C}$ implies there is some prime $p$ such that no solutions in $\overline{\mathbb{F}}_p$." The answer to that should be yes. Let $F_1,\ldots,F_r$ be polynomials in your set $S$ that generate the homogeneous ideal generated by the elements of $S$. Then elimination theory says that there is a list of polynomials $E_1,\ldots,E_s$ in the coefficients of $F_1,\ldots,F_r$ such that $F_1,\ldots,F_r$ have a common solution in $\mathbb{P}^{n-1}(k)$ if and only if every $E_i$ vanishes at $F_1,\ldots,F_r$. (This is for algebraically closed fields $k$.) So if $F_1,\ldots,F_r$ have no common zeros in $\mathbb{P}^{n-1}(\mathbb{C})$, then some $E_i(F_1,\ldots,F_r)$ is non-zero, and hence it is nonzero in characteristic $p$ for all but finitely many $p$, and hence $F_1,\ldots,F_r$ have no common zeros in $\mathbb{P}^{n-1}(\overline{\mathbb{F}}_p)$ for all but finitely many $p$. For effective estimates, the term you want to search for is "effective nullstellensatz".

  • $\begingroup$ But your example doesn't satisfy $(E_0)$ : it certainly has solutions in $\mathbb{C}$ for example. $\endgroup$ – js21 Apr 22 '12 at 12:09
  • $\begingroup$ Thanks ! This is exactly what I sought. This also provides an alternative proof of the converse statement. $\endgroup$ – js21 Apr 22 '12 at 13:17

Here is an argument which avoids elimination theory at the cost of being highly nonconstructive. We prove the contrapositive: if $X(F)$ is non-empty for fields $F$ of infinitely many different positive characteristics, then $X(F)$ is non-empty for a field $F$ of characteristic zero.

Let $F_i$ be an infinite collection of fields of distinct characteristics such that $X(F_i)$ is non-empty and consider the ultraproduct $F = \prod F_i/U$. By Łoś's theorem $F$ has characteristic $0$, and by choosing a point in each $X(F_i)$ we obtain a point in $X(F)$ by taking coordinatewise ultraproducts.

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    $\begingroup$ Instead of ultraproducts, you could use the compactness theorem of first-order logic directly. Let $T$ be the theory, in the language of fields plus $n$ new constant symbols $c_i$, consisting of the axioms for fields, the equations saying that the $c_i$ are roots of all the polynomials in $S$, and the formulas $1+\dots+1\neq0$ for each prime number of summands. If $X(F)$ is nonempty for fields of infinitely many prime characteristics, then each finite subset of $T$ is satisfiable. By compactness, there is a model of all of $T$, and that's a field $F$ of characteristic 0 with nonempty $X(F)$. $\endgroup$ – Andreas Blass Apr 22 '12 at 20:32
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    $\begingroup$ In the comment I just posted, $T$ should also include the formulas $c_i\neq 0$ for the $n$ new constants, so that you're talking about solutions to $S$ in projective space. $\endgroup$ – Andreas Blass Apr 22 '12 at 20:34
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    $\begingroup$ Correction to the correction: $T$ should not say that all the $c_i$ are non-zero but that at least one of them is. In other words, it should contain the disjunction $c_1\neq0\lor\dots\lor c_n\neq0$. $\endgroup$ – Andreas Blass Apr 28 '12 at 16:40

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