Idempotent Laurent polynomials (in noncommuting variables) Let $K$ be a field and $R=K\langle X_1,\dots,X_n,X_1^{-1},\dots,X_n^{-1}\rangle$ the Laurent polynomial ring in $n$ noncommuting variables. Can $R$ have idempotents distinct from $0$ and $1$?
 A: 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).
A: Here's a self-contained proof (which is certainly Higman's proof), following Fedor Petrov's answer.
Let $G$ be a locally indicable group (= every nontrivial f.g. subgroup has $\mathbf{Z}$ as quotient). The $KG$ has no nontrivial idempotent.
Indeed, suppose $u^2=u$ in $KG$ with $u\neq 0,1$. Then passing to the subgroup generated by $\mathrm{Supp}(u)$, we can suppose that the support of $u$ generates $G$, and that $G$ is finitely generated. Clearly $G\neq 1$. Then fix a surjective homomorphism $G\to\mathbf{Z}$. Push forward $u$ to $K[\mathbf{Z}]$ to get an idempotent, whose support generates $\mathbf{Z}$. But $K[\mathbf{Z}]=K[t^{\pm 1}]$ has no nontrivial idempotent, contradiction.
