Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\mathbb{Q}[G]$ is an integral domain. Let $\mathbb{Q}(G)$ be its field of fractions.

Are there non-isomorphic torsion free abelian groups $G$ and $H$ such that $\mathbb{Q}(G)\cong\mathbb{Q}(H)$? If so, are there easy examples?

This is motivated by this question. There are examples of torsion free abelian groups $G$ where $G\cong G\oplus\mathbb{Z}\oplus\mathbb{Z}$, but $G\not\cong G\oplus\mathbb{Z}$. Since $\mathbb{Q}(G\oplus\mathbb{Z}\oplus\mathbb{Z})\cong\mathbb{Q}(G)(X,Y)$ and $\mathbb{Q}(G\oplus\mathbb{Z})\cong\mathbb{Q}(G)(X)$, a negative answer to my question would give a positive answer to the motivating question.

I've found a little about this construction in the literature, but not much. For example, in

*Gervasio G. Bastos and T. M. Viswanathan*, MR 942063 **Torsion-free abelian groups, valuations and twisted group rings**, *Canad. Math. Bull.* **31** (1988), no. 2, 139--146.

it is proved that $\mathbb{Q}(G)$ is a purely transcendental extension of $\mathbb{Q}$ if and only if $G$ is free abelian.

Any other relevant references would be welcome. Also, there are obvious variants: e.g., using a different field, such as $\mathbb{F}_2$ instead of $\mathbb{Q}$.