A note on multiplicative commutators of division rings

TL;DR

This paper provides a counterexample in division ring theory, showing that the multiplicative commutator subgroup may not generate the entire division ring as a vector space over its center, challenging a previous conjecture.

Contribution

It presents the first known example disproving the conjecture that the multiplicative commutator subgroup always generates the division ring as a vector space.

Findings

Counterexample division ring where commutators do not generate the entire ring

Disproof of the conjecture on vector space generation by multiplicative commutators

Implication for understanding the structure of division rings

Abstract

We give an example of a division ring $D$ whose multiplicative commutator subgroup does not generate $D$ as a vector space over its centre, thus disproving the conjecture posed in the paper "Vector space generated by the multiplicative commutators of a division ring, J. Algebra Appl. 12 (2013), no. 8".

Equations

$$\Big{(}\sum_{i=m}^{\infty}a_{i}t^{i}\Big{)}\Big{(}\sum_{j=n}^{\infty}b_{j}t^{j}\Big{)}=\sum_{r=n+m}^{\infty}\Big{(}\sum_{i+j=r}a_{i}\sigma^{i}(b_{j})\Big{)}t^{r}.$$

$$v:D^{\ast }$$

$$\sum _{i=n}^{\infty }{a}_i{t}^i$$

$$D^{\prime}\subseteq\ker(v)=\Big{\{}L+\sum_{i>0}a_{i}t^{i}\Big{\}}.$$

Full text

The author acknowledges Australian Research Council grant DP160101481. This work was done at the University of Münster, where the author was a Humboldt Fellow.

Let $D$ be a division ring. Denote by $F:=Z(D)$, the centre of $D$ and $D^{\prime}$ its multiplicative commutator subgroup, i.e., the group generated by the set of multiplicative commutators \big{\{}xyx^{-1}y^{-1}\mid x,y\in D\backslash\{0\}\big{\}}. There are classical results due to Herstein, Kaplansky and Scott, among others, showing that the group $D^{\prime}$ is “dense” in $D$ (see for example [3, §13]). In [1] the authors study the $F$-vector space $T(D)$ generated by the set of multiplicative commutators. They prove that if $T(D)$ is radical over $F$, then $D=F$, and if $\dim_{F}T(D)<\infty$, then $\dim_{F}D<\infty$. Furthermore, they prove that $T(D)$ contains all separable elements of $D$ and thus if $D$ is an algebraic division ring over its centre with $\operatorname{char}(D)=0$, then $T(D)=D$.

They then conjecture [1, Abstract and Conjecture 1] that a division ring is generated by all multiplicative commutators as a vector space over its centre, i.e., $T(D)=D$ for any arbitrary division ring.

Here we give a counterexample to this conjecture. In fact we show that the multiplicative subgroup $D^{\prime}$ can not recover $D$ as a vector space and $[D:T(D)]=\infty$.

We recall the Hilbert classical construction of division rings (see [2, §1]). Let $L$ be a field, $\sigma\in\operatorname{Aut}(L)$ and $F$ be the fixed field of $\sigma$. Let $D=L((x,\sigma))$ be the division ring of formal Laurent series, consisting of elements $\sum_{i=n}^{\infty}a_{i}t^{i}$, where $a_{i}\in L$ and $n\in\mathbb{Z}$, addition defined component-wise and multiplication by

[TABLE]

By [2, §1, Lemma 4] if the order of $\sigma$ is infinite then $Z(D)=F$ and $[D:F]=\infty$ and if the order is a finite number $n$, then $Z(D)=F((t^{n}))$ and $[D:Z(D)]=n^{2}$. It is easy to see that $D=L((x,\sigma))$ is a valued division ring with value group $\mathbb{Z}$ as follows

[TABLE]

Therefore we have

[TABLE]

Now if we choose an automorphism $\sigma$ of infinite order, since $Z(D)=F$, Equation 2 shows that T(D)\subseteq\Big{\{}L+\sum_{i>0}a_{i}t^{i}\Big{\}}, and thus $[D:T(D)]=\infty$.