Ore localization

Remarks

Ore's theorem: $R S^{-1}$ exists if and only if $S$ is a right denominator set. This means that $S$ satisfies the right Ore condition and is right reversible: if $s a = 0$ with $s \in S$, then $a t = 0$ for some $t \in S$. In that case one may build $R S^{-1}$ as the set of classes of pairs $(a, s) \in R \times S$, where $(a, s) \sim (b, t)$ when there are $c, d \in R$ with $s c = t d \in S$ and $a c = b d$. It then has the universal property: each ring homomorphism $f \colon R \to B$ with $f(S)$ consisting of units factors uniquely through $\varphi$.

Example: if $R$ is commutative, every multiplicative set is a right denominator set, and $R S^{-1}$ is the usual localization $S^{-1} R$. Clause 3 matches the kernel description there.

Non-example: for the free algebra $R = k\langle x, y \rangle$ over a field $k$ and $S = R \setminus \set{0}$, no right ring of fractions exists. Suppose $\varphi$ satisfied clauses 1 to 3; then $\varphi$ would be injective by clause 3. By clause 2, $\varphi(y)^{-1} \varphi(x)$ would equal $\varphi(b) \varphi(t)^{-1}$, giving $x t = y b$ with $t \ne 0$, which is impossible.

The left ring of fractions $S^{-1} R$ is defined in the same way with elements $\varphi(s)^{-1} \varphi(a)$. When both exist, they are isomorphic as rings over $R$.