The present state of knowledge is the following.

1. Nobody knows a single example of two non-isomorphic torsion–free abelian groups whose rational group–algebra fields of fractions are isomorphic.
2. No proof is known that this can never happen.

In other words, the question is open.

Below is a summary of what is known and of the main ideas that have been tried.

A. Easy cases in which the answer is known to be “no”.

A1. Finite rank, free groups.
If G ≅ ℤ^r and H ≅ ℤ^s , then
ℚ(G) = ℚ(t_1^{±1}, … , t_r^{±1}) and
ℚ(H) = ℚ(u_1^{±1}, … , u_s^{±1}).
These fields are isomorphic iff r = s, so in the free case the field already determines the rank, hence the group.

A2. One group free, the other not.
Bastos–Viswanathan (Canad. Math. Bull. 31 (1988) 139-146) show that ℚ(G) is purely transcendental over ℚ iff G is free. Therefore ℚ(G) cannot be isomorphic to ℚ(H) when one of the groups is free and the other is not.

A3. Rank-one groups that are contained in one another.
If G ⊂ H are rank-1 and the inclusion is proper, then ℚ(G) ⊊ ℚ(H). (One sees this by looking at valuations: the canonical Krull valuation on ℚ(H) has value group H; its restriction to ℚ(G) has value group G, hence cannot be surjective.)

B. What makes the general problem hard.

For any ordered torsion-free abelian group G one can embed ℚ[G] in the Hahn series field ℚ((G)) (Mal’cev–Neumann). The canonical valuation on ℚ((G)) restricts to a Krull valuation v_G on ℚ(G) with value group G and residue field ℚ. All of our structural information about ℚ(G) comes from studying valuations that extend v_G. Unfortunately a given field can possess many essentially different Krull valuations, and (so far) no condition has been found that singles out v_G in a purely field- theoretic way.

C. Why the obvious attacks have not yet worked.

• If ℚ(G) ≅ ℚ(H) via some isomorphism σ, the valuation v_G need not be transported to v_H; σ might well send v_G to a valuation whose value group is only a quotient or a subgroup of H.

• Trying to recover G from the set of all value groups that occur for Krull valuations of ℚ(G) also fails: one can show that many different torsion-free abelian groups appear as value groups for suitable valuations of the same field.

D. Connections with other open problems.

A negative answer to the present question (i.e. “ℚ(G) determines G”) would settle the still-open cancellation problem for pure transcendental extensions quoted in the motivation: “Is there a field F with F ≅ F(x, y) but F ≄ F(x) ?” Indeed, Göbel–Rodríguez have constructed torsion-free abelian groups G such that G ≅ G ⊕ ℤ ⊕ ℤ but G ≄ G ⊕ ℤ. If ℚ(G) determined G, then ℚ(G)(x, y) = ℚ(G ⊕ ℤ ⊕ ℤ) would be isomorphic to ℚ(G), whereas ℚ(G)(x) = ℚ(G ⊕ ℤ) would not—giving the desired field.

E. References for partial results.

1. G. G. Bastos & T. M. Viswanathan, “Torsion-free abelian groups, valuations and twisted group rings,” Canad. Math. Bull. 31 (1988), 139–146.

2. L. Fuchs, Infinite Abelian Groups, vol. I, secs. 27–28 (background on torsion-free rank-1 groups and their embeddings).

3. D. Passman, The Algebraic Structure of Group Rings, chap. 11 (Krull valuations on group algebras).

At present neither an affirmative nor a negative answer is known; a new idea seems to be required to make further progress.