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Let $G,H$ be torsion-free abelian groups such that $k[G] \cong k[H]$ for any field $k$. Then is it true that $G \cong H$ ?

If this is not true, then what if I change the hypothesis to $R[G]\cong R[H]$ for any non-zero commutative unital ring; is the conclusion true then ?

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  • $\begingroup$ I wonder what happens if one asks the same question but $k$ is fixed. $\endgroup$ Jun 27 '17 at 17:47
  • $\begingroup$ @Max : true , I wonder that too . $\endgroup$
    – user
    Jun 27 '17 at 18:06
  • $\begingroup$ @Max : by Eric's answer , $k[G]^\times \cong k^\times \times G$ , so for a fixed $k$ , we atleast have $k^\times\times G \cong k^\times \times H$ ; where $k^\times , G,H$ are all abelian groups , so we can infer that $G \cong H$ whenever $k^\times $ has the cancellation property in the category of abelian groups ; in particular when $k^\times$ is finitely generated , so at least when $k$ is finite $\endgroup$
    – user
    Jun 27 '17 at 18:12
  • $\begingroup$ @Max : so in particular , having $k[G]^\times \cong k^\times \times G$ is a desirable property for our conclusion to hold ; and it is a conjecture of Kaplansky (the Units Conjecture) that whenever $k$ is a field and $G$ is torsion free , then $k[G]^\times \cong k^\times \times G$ $\endgroup$
    – user
    Jun 27 '17 at 18:24
  • $\begingroup$ But this won't be enough : there is no cancellability property for arbitrary groups (although there is one for finite groups, but here $G$ is torsion free) $\endgroup$ Jun 27 '17 at 18:35
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The group of units of $k[G]$ is exactly $k^\times\times G$. In particular, if $k=\mathbb{F}_2$, then the group of units is $G$, so $G$ can be recovered from the ring $\mathbb{F}_2[G]$.

To prove this, note that any torsion-free abelian group $G$ admits a total ordering compatible with the group structure (choose a totally ordered basis for $G\otimes\mathbb{Q}$ and use the lexicographic order). Considering $k[G]$ to be $G$-graded, suppose $u\in k[G]$ is a unit with inverse $v$. Let $a$ be the minimum degree of a homogeneous part of $u$ and $b$ be the maximum degree, and let $c$ be the minimum degree of a homogeneous part of $v$ and $d$ be the maximum degree. Then the product $uv$ has homogeneous parts of degree $a+c$ and also $b+d$. Since $uv=1$, we must have $a+c=b+d$. Since $a\leq b$ and $c\leq d$, this is only possible if $a=b$ and $c=d$, which means $u$ and $v$ are homogeneous. That is, $u$ is a unit of $k$ times an element of $G$.

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  • $\begingroup$ what is the grading that you give to $k[G]$ ? $\endgroup$
    – user
    Jun 27 '17 at 17:15
  • $\begingroup$ A homogeneous element of degree $g$ is just anything of the form $ag$ for $a\in k$. $\endgroup$ Jun 27 '17 at 17:52
  • $\begingroup$ also I have made some comments following this question of mine , in reply to a comment of Max . Could you please enlighten more ? Thanks again $\endgroup$
    – user
    Jun 27 '17 at 18:41
  • $\begingroup$ Related to this MO question : do you know what happens if we assume that $k(G) \cong k(H)$ for any field $k$? (Here $k(G)$ denotes the field of fractions of $k[G]$, whenever $G$ is a torsion-free abelian group). $\endgroup$
    – Watson
    Jun 27 '17 at 22:02

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