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The answer is Yes, and this is the 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_i 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 Π M_i/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 Π M_i/U satisfies a first order statement phi([f]) if and only if { i in I | M_i satisfies phi(f(i)) } in U.

In your case, since every F_p 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 > 0. That is, the ultraproduct will have characteristic 0.

Edit: I confess I missed the part initially about containing the algebraic numbers, and so there is more to be done, as Kevin points out. What Los's theorem gives you is that something will be true in R/U just in case the set of p for which F_p has the property is in the ultrafilter U.

So you need to do some work still. What you need to know is that for any finite list of equations, that there is an infinite set of primes p for which the equations have a solution in F_p. Is this true? I'm not sure (and this may be outstripping my algebra knowledge). This property is equivalent to asking whether every finite list of equations over Z is true in at least one F_p, since one can always add one more equation so as to exclude any particular F_p. If you have this for all finite collections of equations, then you can form a filter containing those sets, which would form a descending sequence of subsets of primes, and then extend this to an ultrafilter.

  In this case, any particular equation would have a solution in F_p for a set of p in U, and os the ultrapower R/U would have a solution. In this case, the ultrapower will contain the algebraic numbers. So I will let the algebraists answer whether every finite list of polynomials over Z has a solution in at least one F_p. If so, the answer to your question is yes, as I have explained. If not, then the answer to your question is no, since every quotient of this ring will be an ultraproduct.

Let me note that if one uses the product of all finite fields, instead of just the F_p, then you can certainly find the desired solutions, and so the ultraproduct will be as desired.

The answer is Yes, and this is the 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_i 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 Π M_i/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 Π M_i/U satisfies a first order statement phi([f]) if and only if { i in I | M_i satisfies phi(f(i)) } in U.

In your case, since every F_p 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 > 0. That is, the ultraproduct will have characteristic 0.

Edit: I confess I missed the part initially about containing the algebraic numbers, and so there is more to be done, as Kevin points out. What Los's theorem gives you is that something will be true in R/U just in case the set of p for which F_p has the property is in the ultrafilter U.

What you need to know is that for any finite list of equations, that there is an infinite set of primes p for which the equations have a solution in F_p. This property is equivalent to asking whether every finite list of equations over Z is true in at least one F_p, since one can always add one more equation so as to exclude any particular F_p. Is this true? (I wasn't sure.)

But according to what Kevin says in the comments below, it is true, and this is precisely what you need for the construction to go through. You can form a filter containing those sets, which would form a descending sequence of subsets of primes, and then extend this to an ultrafilter. In this case, any particular equation would have a solution in F_p for a set of p in U, and so the ultrapower R/U would have a solution. In this case, the ultrapower will contain the algebraic numbers.

The answer is Yes, and this is the 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_i 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 Π M_i/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 Π M_i/U satisfies a first order statement phi([f]) if and only if { i in I | M_i satisfies phi(f(i)) } in U.

In your case, since every F_p 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 > 0. That is, the ultraproduct will have characteristic 0.

The answer is Yes, and this is the 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_i 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 Π M_i/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 Π M_i/U satisfies a first order statement phi([f]) if and only if { i in I | M_i satisfies phi(f(i)) } in U.

In your case, since every F_p 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 > 0. That is, the ultraproduct will have characteristic 0.

Edit: I confess I missed the part initially about containing the algebraic numbers, and so there is more to be done, as Kevin points out. What Los's theorem gives you is that something will be true in R/U just in case the set of p for which F_p has the property is in the ultrafilter U.

So you need to do some work still. What you need to know is that for any finite list of equations, that there is an infinite set of primes p for which the equations have a solution in F_p. Is this true? I'm not sure (and this may be outstripping my algebra knowledge). This property is equivalent to asking whether every finite list of equations over Z is true in at least one F_p, since one can always add one more equation so as to exclude any particular F_p. If you have this for all finite collections of equations, then you can form a filter containing those sets, which would form a descending sequence of subsets of primes, and then extend this to an ultrafilter.

In this case, any particular equation would have a solution in F_p for a set of p in U, and os the ultrapower R/U would have a solution. In this case, the ultrapower will contain the algebraic numbers. So I will let the algebraists answer whether every finite list of polynomials over Z has a solution in at least one F_p. If so, the answer to your question is yes, as I have explained. If not, then the answer to your question is no, since every quotient of this ring will be an ultraproduct.

Let me note that if one uses the product of all finite fields, instead of just the F_p, then you can certainly find the desired solutions, and so the ultraproduct will be as desired.

The answer is Yes, and this is the 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 most important theorem is Los's theorem, which says that a first order statement about [f]_U is true in R/U just in case the set of p for which it is true in F_p about f(p) is in U. Since for most F_p, the number 1+...+1 (n times) is not 0, for n > 0, we get that this statement is true for all n in R/U.

Similarly, every F_p is a field, so this is also true in R/U.

The answer is Yes, and this is the 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_i 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 Π M_i/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 Π M_i/U satisfies a first order statement phi([f]) if and only if { i in I | M_i satisfies phi(f(i)) } in U.

In your case, since every F_p 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 > 0. That is, the ultraproduct will have characteristic 0.