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 $1$?
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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).
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$\begingroup$ I think it was known that free groups are left orderable since Magnus before it was known that locally indicable implies left orderable, such was Burns. $\endgroup$ – Benjamin Steinberg yesterday
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$\begingroup$ @BenjaminSteinberg I am not an expert in history of algebra at all, but Hidman talks a lot about free groups and seems to pretend to be the first to prove Kaplansky conjecture for free groups. $\endgroup$ – Fedor Petrov yesterday
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$\begingroup$ The argument being straightforward for locally indicable groups (for convenience I included it in an answer, it's unnecessary to to go through left-orderability. Maybe the latter is useful to prove the non-existence of zero divisors at all. $\endgroup$ – YCor yesterday
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$\begingroup$ (of course, Higman's paper preceeds Kaplansky conjecture, so he does not use such words) $\endgroup$ – Fedor Petrov yesterday
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$\begingroup$ Yes. But I had thought Higman went through orderability but I dont remember $\endgroup$ – Benjamin Steinberg 23 hours ago
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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.
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$\begingroup$ Your "no zero divisor" should read as "no non-trivial idempotent", or the proof must be modified. $\endgroup$ – Fedor Petrov yesterday
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1$\begingroup$ Yes this is Higman's proof except he does the obvious modification for zero divisors. I had thought he had used ordering but I guess not $\endgroup$ – Benjamin Steinberg 23 hours ago
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2$\begingroup$ @BenjaminSteinberg what is the obvious modification for zero divisors? If $uv=0$, we may suppose that the supports of $u$ and $v$ generate $G$, then there exists a homomorphism $G\to \mathbb{Z}$ which is non-zero either on $u$ or on $v$, but why on both? $\endgroup$ – Fedor Petrov 23 hours ago
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$\begingroup$ @BenjaminSteinberg I'm also not sure what you mean by obvious modification, as the homomorphism from the group generated by $\mathrm{Supp}(u)\cup\mathrm{Supp}(v)$ onto $\mathbf{Z}$ might kill $\mathrm{Supp}(v)$. $\endgroup$ – YCor 22 hours ago
