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

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

• Worth pointing out: over at the thread cited in this OP, there's a conjecture of 'Mohan' at (math.stackexchange.com/questions/2440013/…) , to the effect that in this situation $\mathrm{tr.deg}(k(G)/\mathbb{F}_p) = \mathrm{rank}(G)$. – Peter Heinig Oct 17 '17 at 16:08
• Thank you for this question. How might it help in solving math.stackexchange.com/questions/2338911 ? – Watson Oct 17 '17 at 17:53
• @Watson : Well, the way the transcendence degree come out , $k(G) \cong k(H)$ implies $G \otimes_{\mathbb Z} \mathbb Q \cong H \otimes_{\mathbb Z} \mathbb Q$ ; so one should ask the question : when can we cancel $\mathbb Q$ in isomorphism between tensor products with torsion free abelian groups ? ( Of course this is just one approach towards possibly showing $G \cong H$ ) – user111524 Oct 17 '17 at 17:59
• @users : your thought about tensor products doesn't help, since $\mathbb Q \otimes_{\mathbb Z} \mathbb Q \cong \mathbb Q \cong \mathbb Z \otimes_{\mathbb Z} \mathbb Q$ as groups, but the torsion free abelian groups $\mathbb Z$ and $\mathbb Q$ are not isomorphic (NB : in your comment, I guess that $k(G) \cong k(H)$ is a $k$-algebra isomorphism (so that transcendance degree over $k$ is preserved). If $k = \mathbb F_p$, then any field isomorphism $k(G) \to k(H)$ is a $k$-algebra isomorphism.) – Watson Oct 17 '17 at 18:32
• @Watson : Yeah I mean to take $k=\mathbb Q$ or $\mathbb Z_p$ ... I know that $\mathbb Q$ cannot always be cancelled ... but may be in some cases it can be cancelled ... – user111524 Oct 17 '17 at 18:37

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.
• why $\{e_i : i\in I\}$ gives algebraically independent (over $k$ ?) elements ? – user111524 Oct 17 '17 at 16:17
• Are you talking about transcendence degree of $k(G)$ over $k$ ? And why is the transcendence degree of $k(G)$ and $k(G \otimes_{\mathbb Z} \mathbb Q)$ the same ? – user111524 Oct 17 '17 at 16:21
• @users Every element of $G\otimes_{\mathbb Z}\mathbb Q$ is a sum of roots of elements of the form $g\otimes 1$, so $k(G\otimes_{\mathbb Z}\mathbb Q)$ can be viewed as an algebraic extension of $k(G)$. – Wojowu Oct 17 '17 at 16:59
• More precisely $A$ is generated by a maximal free family (which doesn't imply that it's maximal among free subgroups: the latter does not exist unless $G$ itself is free). – YCor Oct 17 '17 at 17:06
• @users: the $e_i$ are algebraically independent because they generate a subalgebra isomorphic to $k[\{e_i\}]$ (without relations). You should really check these things yourself, as they are not hard. – R. van Dobben de Bruyn Oct 18 '17 at 0:00