The answer is Yes, and this is the [ultraproduct](http://en.wikipedia.org/wiki/Ultraproduct) construction. Let U be any nonprincipal ultrafilter on the set of primes. This is simply the dual filter to a maximal ideal on the set of primes, containing all finite sets of primes. (In other words, U contains the Frechet filter.) 

The quotient R/U is a field of characteristic 0. 

The ultraproduct construction is completely general, and has nothing to do with rings or fields. If M<sub>i</sub> for i in I is any collection of first order structures, and U is an ultrafilter on the subsets of I, then we may form the ultraproduct &Pi; M<sub>i</sub>/U, which is the set of equivalence classes by the relation f equiv g iff { i in I | f(i) = g(i) } in U. Similarly, the structure is imposed on the ultraproduct coordinate-wise, and this is well-defined. The most important theorem is Los's theorem, which says that &Pi; M<sub>i</sub>/U satisfies a first order statement phi([f]) if and only if { i in I | M<sub>i</sub> satisfies phi(f(i)) } in U. 

In your case, since every F<sub>p</sub> is a field, the ultraproduct is also a field. And since the set of p bigger than any fixed n is in U, then the ultraproduct will have 1+...+1 (n times) not equal to 0, for any fixed n &gt; 0. That is, the ultraproduct will have characteristic 0.