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

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.