Reconstruction of Torsion-Free Abelian Groups from Rational Group Fields

For a torsion-free abelian group $G$, let $$K_{\mathbb{Q}}(G)=\mathrm{Frac}\mathbb{Q}[G]$$ be the fraction field of its rational group algebra. We prove that this field determines the group up to isomorphism: $$K_{\mathbb{Q}}(G)\cong K_{\mathbb{Q}}(H)⟺G\cong H\mathrm{.}$$ The main structural input is that, over every field $k$ of characteristic zero, the monomial defect group $${\mathrm{\Delta }}_k(G)=K_k(G{)}^{\times }\mathrm{/}(k^{\times }{X}^G)$$ is free abelian. We give a self-contained proof. It first treats the one-variable Puiseux field $F(t^{\mathbb{Q}})$: factorization from $F(t^{1\mathrm{/}n!})$ to $F(t^{1\mathrm{/}(n+1)!})$ gives split inclusions because the substituted irreducibles are square-free in characteristic zero. A transfinite decomposition of the divisible hull $\mathbb{Q}\otimes G$ then proves the general case. Given a field isomorphism, we compare the two monomial subgroups inside the common multiplicative group. Their intersection produces isomorphic subgroups $M\le G$ and $N\le H$, while the quotients $G\mathrm{/}M$ and $H\mathrm{/}N$ embed in free defect groups and are therefore free. Relative transcendence degree shows that these two free quotients have the same rank, completing the reconstruction. As consequences, rational-function stabilization of group fields exactly records free stabilization of groups, and Rickard’s bounded sequence group yields a field $F$ with $F\cong F(x,y)$ but $F\cong ̸F(x)$.