Let $G$ be a torsion-free group and $C$ the ring of complex numbers. The zero divisor (idempotent, resp.) conjecture is that there is no nontrivial zerodivisor (idempotent, resp.) in $CG$.
The wiki page:
http://en.wikipedia.org/wiki/Group_ring
says "This conjecture (zero divisor conjecture) is equivalent to K[G] having no non-trivial idempotents under the same hypotheses for K and G.
"
Is this obvious true? Are there some reference for this claim?
Best Answer
Passman showed that whenever there are zero-divisors in a group ring one also has (non-zero) nilpotent elements. He shows that for any field $k$ and any torsionfree group $G$, the ring $kG$ is a prime ring, i.e. the zero-ideal is a prime ideal.
Now, if $a,b \in kG$ are non-zero and $ab=0$, then there exists $c \in kG$ such that $bca\neq 0$. (This is just another way of saying that the product of the two-sided ideal $(b)$ with the two-sided ideal $(a)$ cannot give the zero ideal, since $(0)$ is a prime ideal.)
Now, we see that $(bca)^2 = bcabca =0$. Hence, $kG$ contains nilpotent elements. The other direction is obvious.