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.

resultant$R(F_1,F_2)$ vanishes in $k$. – Joe Silverman Apr 22 '12 at 15:38