Embedding domains in division rings

Javier Sánchez Serdà1

Università degli Studi dell'Insubria,
Universitat Autònoma de Barcelona

Suppose we have a domain $R$ embedded in a division ring $E$. We define inductively:
$Q_0(R,E)=R$, and for $n\geq 0,$
$Q_{n+1}(R,E)=\begin{array}{c}\textrm{\small subring of $E$}
\\ \textrm{\small generated by}\end{array} \left\{r,
s^{-1}\mid r,s\in Q_n(R,E),\ s\neq0\right\}$.
Then $D=\operatornamewithlimits{\cup}\limits_{n=0}^\infty Q_n(R,E)$ is the smallest division ring that contains $R$ inside $E$. We define $h_E(R)$, the inversion height of $R$ inside $E,$ as $\infty$ if there is no $n\in\mathbb{N}$ such that $Q_n(R,E)$ is a division ring. Otherwise,

\begin{displaymath}h_E(R)=\min\{n\mid Q_n(R,E) \textrm{ is a division

Let $K$ be a commutative field. Suppose $\alpha\colon K\rightarrow K$ is a ring homomorphism which is not onto. Let $t\in K\setminus\alpha(K)$ and $k=\{a\in K\mid \alpha(a)=a\}.$ Consider the skew polynomial ring $K[x;\alpha].$ It was proved by Jategaonkar that the $k$-algebra generated by $x$ and $y=tx$ is a free $k$-algebra $k\langle x,y\rangle.$ We call these embeddings Jategaonkar embeddings.

We prove:

Jategaonkar embeddings have at most inversion height $2.$ And there are examples of Jategaonkar embeddings of height one and two.

If there is an embedding of the free algebra on two generators of height $1\leq n\leq\infty,$ then there exists an embedding of the free algebra on an infinite number of generators of inversion height $n.$

Let $R$ be the free algebra or the free group algebra on $2\leq m\leq\infty$ generators. We use examples in $(i)$ to obtain embeddings of $R$ of inversion height $1$ or $2.$

In a Jategaonkar embedding $D$ is never the universal field of fractions of $R.$


joint work with D. Herbera