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If $R$ is an integral domain (I am having $\mathbb{Z}$ or a field in mind) and $G$ a (not necessarily finite) group then we can form the group ring $R(G)$.

Note that if $g^{n+1} = e$ then $(e-g)(e+g\ldots + g^n) = e - g^{n+1} = 0$. This means if $G$ has torsion then $R(G)$ always has zero-divisors.

What about the inverse? So if $G$ is torsion-free does that imply $R(G)$ having no zero-divisors.

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Note that if $R=\mathbb{C}$ then there is the very famous Kaplansky conjecture that if $G$ is torsion-free then in $\mathbb{C}G$ there are no nontrivial idempotents (ie elements $p$ such that $p^2=p$ and $p=\neq 1, 0$. This is related because a nontrivial idempotent give you a zero divisor by looking at $p(1-p)$. The Kaplansky conjecture is true for many torsion-free groups, but remains open in general. –  Owen Sizemore Jul 28 '13 at 14:54
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Over a field, this is called Kaplansky zero divisor conjecture. See this MO thread and that other MO thread for references. I guess it is still open, since google does not lead to any complete solution. –  1015 Jul 28 '13 at 15:38
    
@OwenSizemore I believe that's an answer. You may want to post it as such. –  rghthndsd Jan 29 at 5:43
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