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

  • $\begingroup$ 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 $\mathrm{tr.deg}(k(G)/\mathbb{F}_p) = \mathrm{rank}(G)$. $\endgroup$ – Peter Heinig Oct 17 '17 at 16:08
  • $\begingroup$ Thank you for this question. How might it help in solving math.stackexchange.com/questions/2338911 ? $\endgroup$ – Watson Oct 17 '17 at 17:53
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    $\begingroup$ @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 $G \cong H$ ) $\endgroup$ – user111524 Oct 17 '17 at 17:59
  • $\begingroup$ @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.) $\endgroup$ – Watson Oct 17 '17 at 18:32
  • $\begingroup$ @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 ... $\endgroup$ – 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.

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  • $\begingroup$ why $\{e_i : i\in I\}$ gives algebraically independent (over $k$ ?) elements ? $\endgroup$ – user111524 Oct 17 '17 at 16:17
  • $\begingroup$ 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 ? $\endgroup$ – user111524 Oct 17 '17 at 16:21
  • $\begingroup$ @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 $k(G)$. $\endgroup$ – Wojowu Oct 17 '17 at 16:59
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    $\begingroup$ 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). $\endgroup$ – YCor Oct 17 '17 at 17:06
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    $\begingroup$ @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. $\endgroup$ – R. van Dobben de Bruyn Oct 18 '17 at 0:00

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