---
tag: 4Z6EED4616
kind: definition
name: Ore localization
msc: [16U20, 16S85]
defines: SG7AQ9T8HD
uses: [0B1261B5ED, AN9SKEZ3VY, TJP4ARC2X7, 0VTH8KPCG2, 0XVEZDE9J8, 4C1AE84N5N]
sources:
  - url: https://en.wikipedia.org/wiki/Ore_condition
---

## Definition

Let $R$ be a [[ring|0B1261B5ED]] with identity and let $S \subseteq R$ be a [[multiplicative set|AN9SKEZ3VY]].
A *right ring of fractions* of $R$ with respect to $S$ is a ring $Q$ together with a [[ring homomorphism|0VTH8KPCG2]] $\varphi \colon R \to Q$ such that:

1. $\varphi(s)$ is a [[unit|0XVEZDE9J8]] of $Q$ for every $s \in S$;
2. every element of $Q$ has the form $\varphi(a) \varphi(s)^{-1}$ with $a \in R$ and $s \in S$;
3. the kernel of $\varphi$ is $\set{a \in R : a t = 0 \text{ for some } t \in S}$.

When it exists, it is unique up to a unique isomorphism compatible with $\varphi$, and it is written $R S^{-1}$.
It is called the *Ore localization* of $R$ at $S$.

## 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|TJP4ARC2X7]] 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|4C1AE84N5N]] $S^{-1} R$.
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 $R$.