Past exam of the non-commutative rings course of the University of Oxford / 2023 / 2 / c

View $\hat{R}$ as a noncommutative formal power series ring, and let $\epsilon :\hat{R}\to \mathbb{Q}$ take the coefficient of the empty word. This is a ring homomorphism, since the empty word has only the factorization $1\cdot 1$. In particular the two-sided ideal $I$ generated by $x_1,x_2$ lies in $\ker \epsilon$.

In fact $I=\ker \epsilon$. Every noncommutative formal power series with zero constant term can be partitioned according to the first letter of its words and writtenBoth summands belong to the two-sided ideal $I$.

For $a=1+h\in 1+I$, define the formal geometric seriesThis defines an element of $\hat{R}$: since $h$ has zero constant term, every word in $h^k$ has length at least $k$. Thus the coefficient of a fixed word of length $d$ receives contributions only from $0\le k\le d$, each a finite sum.

For each finite partial sum $g_N={\sum }_{k=0}^N(-h{)}^k$, associativity gives both identitiesIn every fixed word degree the last term vanishes once $N$ is large enough. It follows coefficient by coefficient that $ag=ga=1$. HenceThe argument is formal and uses no analytic convergence. More generally, every noncommutative formal power series with a nonzero rational constant term is a unit, after factoring out that scalar. This is the unit criterion for noncommutative formal power series.