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.



Emptyness of a projective variety
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". 


