If R is an integral domain (I am having 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 gn+1=e then (e−g)(e+g…+gn)=e−gn+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.

**Answer**

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=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.

**Attribution***Source : Link , Question Author : mna , Answer Author :
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