Past exam of the non-commutative rings course of the University of Oxford / 2023 / 2 (Ore sets and formal series)

A multiplicative subset $S$ is a left Ore set if for every $s\in S$ and $r\in R$ there are $s^{\prime }\in S$ and $r^{\prime }\in R$ withEquivalently, the subsets $Sr$ and $Rs$ intersect for each such pair. This is the common-left-multiple condition needed for Ore localization with denominators on the left. In a noncommutative domain, no extra condition concerning annihilation by denominators is needed, since all members of $S$ are regular elements of a ring.

We first prove that any two nonzero principal left ideals $Ra$ and $Rb$ intersect nontrivially. Suppose instead that $Ra\cap Rb=0$. The nonzero left idealsthen form a direct sum. Indeed, take a finite relationIt impliesAs $R$ is a noncommutative domain and $a\ne 0$, this gives $u_0=0$. Right cancellation of the nonzero element $b$ reduces the remaining relation to one of the same form with one fewer term. Repetition gives $u_1=\cdots =u_m=0$, proving the asserted direct sum.

Consequently its finite partial sums give a strictly ascending chain of left ideals, contradicting the ascending chain condition in a left Noetherian ring. Thus $Ra\cap Rb\ne 0$ whenever $a,b\ne 0$. This is the left Noetherian domain Ore lemma.

Now take $s\in S=R\setminus \{0\}$ and $r\in R$. If $r=0$, choose $s^{\prime }=1$ and $r^{\prime }=0$. Otherwise chooseSince $z\ne 0$, the coefficient $s^{\prime }$ is nonzero and hence belongs to $S$. This proves the left Ore set condition in all cases.

Use $s=x_1$ and $r=x_2$. A putative left Ore set equality would beThe left side is a single word ending in $x_2$, with coefficient $1$. The right side is a finite linear combination of words ending in $x_1$. Distinct words are linearly independent in the free associative algebra, so equality is impossible. ThereforeThe last letters matter because the denominators in the left Ore set equation occur on the right of the common left multiple.

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.

Take $D=\hat{R}$. We first check that this noncommutative formal power series ring is a noncommutative domain. For nonzero $F,G\in D$, let $p,q$ be the smallest lengths of words with nonzero coefficients in $F,G$, respectively. Their degree-$p$ and degree-$q$ parts $F_p,G_q$ are nonzero homogeneous polynomials in the free associative algebra.

Choose words $u,v$ with nonzero coefficients in $F_p,G_q$. In $F_p{G}_q$, the coefficient of $uv$ is the product of those two coefficients: a word of length $p+q$ has a unique splitting into a prefix of length $p$ and a suffix of length $q$. This product is nonzero in $\mathbb{Q}$. The degree-$p+q$ part of $FG$ is exactly $F_p{G}_q$, since all lower parts of $F$ and $G$ vanish. Thus $FG\ne 0$, proving that $D$ is a noncommutative domain.

By the defining property of the free associative algebra, there is a ring homomorphismThe same substitution defines a ring automorphism of $R$, with inverse $x_1\mapsto x_1-1$, $x_2\mapsto x_2-2$. Composing this ring automorphism with the natural inclusion $R\hookrightarrow D$ proves that $f$ is injective.

Finally,Their constant terms are $3,-1,1,2$, respectively, all nonzero. By the unit criterion for noncommutative formal power series proved in c, all four elements are units. Therefore $D$ and $f$ have every required property. The embedding changes the images of the generators, so the failure of the original monomial left Ore set in b presents no obstruction.