Transcendence degree of the fraction field of $k[G]$ for torsion-free abelian group $G$

Question

Let $k$ be a field of characteristic $p$ and $G$ be a torsion-free abelian group. Then the group ring $k[G]$ is an integral domain , let $k(G)$ denote its field of fractions . Then can we say anything about the transcendence degree of $k(G)$ over $F_p$ in terms of $k$ and/or $G$ ? What about the transcendence degree of $k(G)$ for field $k$ of characteristic $0$ ?

Answer 1

Let $A\subset G$ be the subgroup generated by a maximal $\mathbf{Z}$-free family (so $A$ is free and $G/A$ is torsion). Then $k(A)=k(G)$. In addition, $A\simeq\mathbf{Z}^{(I)}$, where $I\subset A$ is a basis, and then we see that $k(A)$ is the field of rational fractions in the indeterminates $(x_i)_{i\in I}$ and $k(A)\subset k(G)$ is an algebraic extension. So their common transcendency degree over $k$ is the cardinal $|I|$, that is, equal to the $\mathbf{Q}$-rank $d=\dim_{\mathbf{Q}}(A\otimes_{\mathbf{Z}}\mathbf{Q})$ of $A$.

If $t$ is the transcendency degree of $k$ over $\mathbf{F}_p$, then it follows that the transcendency degree of $k(G)$ over $\mathbf{F}_p$ is $t+d$.