Let G be a linearly ordered abelian group, where the group operator respects order. That is, a ≤ b and x ≤ y implies a+x ≤ b+y.
Let R be any integral domain and let R[G] be the group algebra G over R. If you're not familiar with group algebras, I can explain it prety quickly.
Let R[G] consist of finite linear combinations of elements of G, with coefficients taken from R. It's basically a ring of polynomials, where the elements of G act as indeterminants. Hence the notation R[G], which is usually reserved for polynomials. The additive identity is 0, which indicates no group elements are used. The multiplicative identity is 1e, where e is the group identity. Note that R embeds in R[G] via R*e. Nonabelian groups create interesting noncommutative rings, but in this case the group, and the ring, are commutative.
Let the valuation of a member of R[G] be the lowest group element represented, according to the linear order of G. The valuation of 0 is infinite.
When two nonzero elements are multiplied, their valuations are added together. The coefficients on the two lowest group elements are multiplied to produce the coefficient on the lowest element in the product. Since R is an integral domain, this coefficient is nonzero. Therefore R[G] is an integral domain.
The "valuation" function is actually a group homomorphism from the nonzero elements of R[G] onto G. The homomorphism v(x) extracts the least group element of x. Multiplication in R[G] corresponds to addition in G.
Let K be the fraction field of R[G], which can be represented as quotients of "polynomials" from R[G]. Note that K includes F, the fraction field of R.
Let's build a valuation homomorphism from the nonzero elements of K onto G. Again, multiplication in K corresponds to group action in G. This homomorphism is based on v(), which extracts the lowest member of G. If an element of K is represented by a/b, in lowest terms, map this fraction to v(a)-v(b). Multiply two fractions together and show this is indeed a homomorphism.
Review the four criteria for a valuation group. The first three criteria are satisfied automatically. We only need prove condition 4, the valuation of the sum.
Let's bring in an arbitrary multiplyer, a member of R[G]. If a and b are elements of K, and v(a+b) is at least as large as the lesser of v(a) and v(b), then the same relationship holds for m*a and m*b. This is because multiplication by m distributes over a+b, and multiplication by m shifts everything in G, preserving order.
If a and b are fractions, multiply through by the common denominator. This becomes our multiplier m above. Therefore we only need verify condition 4 for elements in R[G].
Let a and b be finite linear combinations of elements of G. The least element in the sum is the least element in either a or b, or perhaps something larger if the least elements of a and b cancel each other out. Therefore v(a+b) is greater than or equal to the lesser of v(a) and v(b). This proves condition 4, and makes G a valuation group.
The valuation ring associated with G is the preimage of the nonnegative elements of G. These are the fractions in K where the denominator has a valuation no higher than the numerator. The units are the fractions where the numerator and denominator have the same valuation.
In conclusion, any ordered abelian group will do. There is no limit on cardinality. Let G be any ordinal, where group action is defined by ordinal addition. Bring in the inverses, the negative ordinals, and G is complete. Let R be any ring, the integers for instance. Upgrade R to its fraction field, (you always have to do this anyways), build F[G], find its fraction field K, and G becomes the valuation group for K.