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 same question for field $k$ of characteristic $0$ ? An answer to this might help in solving <https://math.stackexchange.com/questions/2338911/fraction-field-of-group-ring-of-field-over-torsion-free-abelian-group>


Also asked on <https://math.stackexchange.com/questions/2440013/transcendence-degree-of-the-fraction-field-of-kg-for-torsion-free-abelian-gr>



Transcendence degree of the fraction field of $k[G]$ for torsion free abelian group $G$
It seems that the transcendence degree of $k(G)$ over $k$ should be the dimension of the $\mathbb Q$-vector space $G \otimes\_\mathbb Z \mathbb Q$. Indeed, passing from $G$ to $G \otimes \mathbb Q$ corresponds to adding roots of existing elements, so it does not alter the transcendence degree. Thus we are reduced to $\mathbb Q$-vector spaces. Choosing a basis $\{e\_i\ |\ i \in I\}$ gives algebraically independent elements $\{e\_i\ |\ i \in I\}$ of $k(G)$ over $k$ such that $k(G)$ is algebraic over $k(\{e\_i\ |\ i \in I\})$; therefore the $e\_i$ form a transcendence basis.


