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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2 Answers
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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).
answered Jan 16, 2021 at 17:47
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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. $\endgroup$ Commented Jan 16, 2021 at 17:56
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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. $\endgroup$ Commented Jan 16, 2021 at 18:13
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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. $\endgroup$– YCorCommented Jan 16, 2021 at 18:28
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$\begingroup$ (of course, Higman's paper preceeds Kaplansky conjecture, so he does not use such words) $\endgroup$ Commented Jan 16, 2021 at 18:50
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$\begingroup$ Yes. But I had thought Higman went through orderability but I dont remember $\endgroup$ Commented Jan 16, 2021 at 19:39
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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. $\endgroup$ Commented Jan 16, 2021 at 18:47
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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 $\endgroup$ Commented Jan 16, 2021 at 19:43
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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? $\endgroup$ Commented Jan 16, 2021 at 19:50
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1$\begingroup$ Look at Higman's paper. It is on page 242. He does an extremely similar argument. He doesn't quite push into laurent polynomials. He uses the indexing function to imitate the proof for Laurent polynomials $\endgroup$ Commented Jan 16, 2021 at 20:34