Skip to main content
added 94 characters in body
Source Link
Fedor Petrov
  • 108.8k
  • 9
  • 264
  • 459

No, it can not. $R$ is a group ring of the free group with $n$ generators. This group is locally indicable (any non-trivial subgroup has a homomorphism onto $\mathbb{Z}$), thus by result of Higman (Higman G. The Units of Group Rings // Proc. London Math. Soc. 1940. Vol. 46. P. 231–248) it satisfies the Kaplansky zero divisors conjecture: its group ring over a field does not have zero divisors (therefore alsoin particular, it does not contain non-trivial idempotents conjecture).

$R$ is a group ring of the free group with $n$ generators. This group is locally indicable (any non-trivial subgroup has a homomorphism onto $\mathbb{Z}$), thus by result of Higman (Higman G. The Units of Group Rings // Proc. London Math. Soc. 1940. Vol. 46. P. 231–248) it satisfies the Kaplansky zero divisors conjecture (therefore also idempotents conjecture).

No, it can not. $R$ is a group ring of the free group with $n$ generators. This group is locally indicable (any non-trivial subgroup has a homomorphism onto $\mathbb{Z}$), thus by result of Higman (Higman G. The Units of Group Rings // Proc. London Math. Soc. 1940. Vol. 46. P. 231–248) it satisfies the Kaplansky zero divisors conjecture: its group ring over a field does not have zero divisors (in particular, it does not contain non-trivial idempotents).

Source Link
Fedor Petrov
  • 108.8k
  • 9
  • 264
  • 459

$R$ is a group ring of the free group with $n$ generators. This group is locally indicable (any non-trivial subgroup has a homomorphism onto $\mathbb{Z}$), thus by result of Higman (Higman G. The Units of Group Rings // Proc. London Math. Soc. 1940. Vol. 46. P. 231–248) it satisfies the Kaplansky zero divisors conjecture (therefore also idempotents conjecture).