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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: 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?

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According to T.Y.Lam, "The first course in noncommutative rings", p.95, for a torsion-free group $G$ and a domain $k$, the following properties are equivalent: (6.17) $kG$ is reduced, (6.18) $kG$ is a domain (conjectured to be always true). So I think that it should have been "nilpotents" in place of "idempotents". – Victor Protsak Jul 27 '10 at 3:03
The equivalence of zerodivisor and nilpotent is already known in D. Passmann's book. Probably, the wiki page is making a mistake. – yeshengkui Jul 27 '10 at 6:00
up vote 4 down vote accepted

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.

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You say: "The other direction is obvious". The question asked about a possible equivalence between zero divisor conjecture and idempotent conjecture. How does existence of non-trivial nilpotents imply existence of non-trivial idempotents (elements $e$ other than $0,1$ satisfying $e^2=e$)? – Victor Protsak Aug 5 '10 at 15:23
I was thinking about the other direction of my claim about zero-divisors and nilpotent elements. In fact I was just explaining some detail of Passman's argument that you mentioned already in your comment. – Andreas Thom Aug 5 '10 at 15:31

Clearly one implies the other as $x^2=x$ means $x(x-1)=0$.

I doubt they are known to be equivalent since the sources I found: the K-theory handbook and Alain Valette survey (see Conjecture 2) listed them as separate conjectures. Of course, it would be hard to prove they are not equivalent, since both might be true!

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The zero divisor conjecture implies the idempotent conjecture as $1$ and $0$ are the only idempotents in a ring with unity and no zero-divisors, nothing sure about the other way around

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