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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    $\begingroup$ 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. $\endgroup$ – Owen Sizemore Jul 28 '13 at 14:54
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    $\begingroup$ 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. $\endgroup$ – Julien Jul 28 '13 at 15:38
  • $\begingroup$ @OwenSizemore I believe that's an answer. You may want to post it as such. $\endgroup$ – RghtHndSd Jan 29 '14 at 5:43

By embedding $R$ into its field of fractions, we may as well assume that $R$ is a field. But then this is precisely Kaplansky's zero-divisor conjecture. This is a hard problem which is still open (2014). It has only been solved for certain classes of groups. If $R$ has characteristic $0$ then it suffices to treat $R=\mathbb{C}$ and there analytical methods are available. A reference is

Passman, Donald S. The algebraic structure of group rings. Pure and Applied Mathematics. Wiley-Interscience [John Wiley & Sons], New York-London-Sydney, 1977.

For a summary of the known results, see MO/79559.


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