The Kaplansky conjecture says that: for any field $F$ and any torsion free group $G$, the group ring $F[G]$ does not have nontrivial idempotent elements.

Do we assume that $F$ has any characteristic ?

Does the conjecture imply that any finitely generated projective $F[G]$-module is free ? or eventually stably free?

For which field $F$ and which class of groups $G$, the conjecture is known to be true?

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    $\begingroup$ For starters: mathoverflow.net/questions/79559/… $\endgroup$ – T. Amdeberhan May 23 '17 at 18:36
  • $\begingroup$ @T.Amdeberhan I read that link before asking my question. $\endgroup$ – Nguyen lan Lee May 23 '17 at 18:44
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    $\begingroup$ The answer to your first question is already in your question, since you're stating the conjecture correctly: there's no characteristic assumption, otherwise it would be explicit. $\endgroup$ – YCor May 24 '17 at 20:36

Kaplansky's zero divisor conjecture: Let $\mathbb{F}$ be a field and $G$ be a torsion-free group. Then $\mathbb{F}[G]$ does not contain a zero divisor.

The existence of a nontrivial idempotent $a$ in a ring $R$ implies the existence of a zero divisor in $R$, because $a(a-1)=0$. So, the zero divisor conjecture implies the idempotent conjecture.

The idempotent conjecture has been confirmed in special cases. For example, Formanek (1973) showed that if $G$ is a torsion-free group satisfying the ascending chain condition on cyclic subgroups and $\mathbb{F}$ is a field of characteristic $0$, then $\mathbb{F}[G]$ has no nontrivial idempotents. Also, Bass (1976) proved that if $G$ is a torsion-free linear group, then $\mathbb{C}[G]$ has no nontrivial idempotents.

These conjectures have not been confirmed for any fixed field and it seems that confirming the conjecture even for the finite field $\mathbb{F}_2$ is still out of reach.

For the zero divisor conjecture case, our work Zero divisors and units with small supports in group algebras of torsion-free groups may be helpful. Also for more details about these two conjectures you can see Zero-divisors and idempotents in group rings.

  • $\begingroup$ Thank you very much for your answer and references! Is there any result about the classification of finitely generated projective $F[G]$-modules under the assumption that the Kaplansky conjecture is true ? $\endgroup$ – Nguyen lan Lee May 25 '17 at 2:21
  • $\begingroup$ I am not aware about results related to this question. $\endgroup$ – Zahra Taheri May 25 '17 at 9:49

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