Here is the canonical example of the ultraproduct (about which you may want to read first, or this will make little sense). Say we have models for the field axioms. That is, our constants are "0" and "1", our functions are "+" and "⋅", and we have the usual field axioms. The cartesian product of fields fails to be a field. For instance, if we look at R×R, then
(1,0) × (0,1) = (0,0) = 0
shows that R×R has
zero divisors, so (1,0) and (0,1), both different from 0=(0,0), have no
inverses.
But, if we take an ultraproduct of fields, Los' theorem will guarantee that it is a field. For instance, any representative of a nonzero element of the ultraproduct is nonzero at almost every coordinate, so it has an inverse at almost every coordinate, and this inverse is unique in the ultraproduct. A bit more work shows that this putative inverse element is indeed a (and "the") multiplicative inverse of the original element.
So let's look at an interesting ultraproduct of fields. Our index set shall be the set of prime numbers P. We'll take as our models the finite fields GF(p), and some ultrafilter U on P. We'll take U to be a nonprincipal ultrafilter (see ultraproduct for why we want to do this). What is F, the ultraproduct of GF(p) over U? Los' theorem tells us it's a field, but we can find out a bit more (and we even get to choose some of its properties, by making appropriate choices for the ultraproduct U!).
The first thing we can prove is that the field F has characteristic 0. For suppose instead that ch(F)=q > 0 for some prime q. We can express this as a proposition of first-order logic: "1+1+...+1=0" where q 1's appear in the sum. Now, only GF(q) satisfies this proposition. As U is nonprincipal, Los' theorem implies that F cannot satisfy ch(F)=q. And as F must have some characteristic that is either prime or 0, its characteristic must be 0.
Note also that F is therefore infinite, despite being an ultraproduct of finite fields. First-order logic cannot express the proposition "the model is finite".
F cannot be algebraically closed. An easy way to see this is to note that in every field GF(p) except GF(2), the proposition
P: ∃y.∀x.(x⋅x)≠y
("there exists an element which is not the square of any element") is satisfied. This is because every element x of a field satisfies (-x)⋅(-x)=x⋅x, and in GF(p) when p>2, (-x) and x are distinct. Thus the mapping x→x⋅x is two-to-one except at 0, and as GF(p) is finite it cannot be
onto.
So (unless U is the principal ultrafilter generated by 2, in which case F is GF(2)) almost every index p satisfies P. So F must also satisfy it:
F has elements which are not squares of any other element.
In particular, F is not algebraically closed.
Is √-1 in F?
Does F include an element "i", which is the square root of -1?
It depends. We get to choose!
The question is whether or not F satisfies the proposition
Q: ∃x.(x⋅x)+1=0
To answer, we have to check the set of p's for which GF(p) satisfies Q.
It turns out that GF(p) satisfies Q iff p=2 or p=4k+1 for some k. (To see this, note that the multiplicative group of a finite field is cyclic, and check the cases p=2, p=4k+1 and p=4k+3.) There are infinitely many primes of the form 4k+1, and infinitely many of the form 4k+3.
So F will satisfy Q iff the set {2, 5, 13, 17, ...} = {p: p=2 or p=4k+1 and p is prime} ∈ U. This is an infinite set with an infinite complement, so there exist ultrafilters U which include it, and ultrafilters which include its complement.
We can specify in advance this property of U. So we can require F to include an element "√-1", and we can require it not to include such an element.
There's nothing very special about "-1" above. For any integer a that has a square root in infinitely many GF(p)'s and has no square root in infinitely many other GF(p)'s, we can choose which one should hold. If, for instance, a=4, then all GF(p)'s have a square root (because "(1+1)⋅(1+1) = 1+1+1+1" in every field). So we cannot have an F with no square root of 4. But if we can show that 3 has square roots in GF(p) for infinitely many p's, but has no square roots for infinitely many other p's, then we get to choose whether or not 3 will have a square root in F.
We even get to make simultaneous choices. We cannot make contradictory simultaneous choices (if 3 and 5 have square roots, then 15 must also have a square root, as it's the product of the two), but we can make any other choice. The astute reader will note that I leave a portion of this claim unproved. An attempt at a contradictory choice gets blocked by the fact that we cannot define an ultrafilter of the required sets of indices, as a contradiction implies two nonintersecting sets in the ultrafilter, which is impossible. In fact, we are precisely blocked by not being allowed to choose a set of properties which only finitely many of the GF(p)'s enjoy.
What is this field?
I'm unsure what this field really is. For all I know, it could be some algebraic extension of Q. It certainly has to include a copy of Q (as must any field of characteristic 0). But it can be made to include other elements (e.g. a square root of -1, as above). And it's possible that it has to include other elements, no matter which choices we make for U.
If you know more about the "identity" of this field, please /msg me.