We will assume throughout that is a ring with unity and a multiplicatively closed subset of with and

**Definition 1**. is called a **left (resp. right) denominator** set if:

1) is left (resp. right) Ore;

2) for and with there exists such that (resp. for and with there exists such that ).

**Theorem**. The left (resp. right) quotient ring of with respect to exists if and only if is a left (resp. right) denominator set.

*Proof.* (*Sketch*) We’ve already proved that if the left (resp. right) quotient ring of with respect to exists, then is a left (resp. right) denominator set. For the converse, assuming that is a left denominator (the “right” version is the same), we’ll construct the left quotient ring of by first defining a relation on : we say if and only if there exist such that and

*Claim*. The relation is an equivalence relation.

*Proof*. The reflexive property is proved by choosing Obviously is symmetric. In order to prove that is transitive suppose that So there exist such that and There also exist such that and On the other hand, since is left Ore, we have and thus there exist and such that So and hence there exists such that i.e. Finally if we put then i.e.

Next, we will show the equivalence class of the element by and we let be the set of all We are going to put a ring structure on Let be two elements of By left Ore condition, and thus there exist and such that Now define Also, since there exist and such that Let and define It is proved that the addition and multiplication that we’ve defined are well-defined and satisfy all the conditions needed to make a ring. Let It’s easy to see that and Finally define by See that is a ring homomorphism. Now if and only if if and only if there exist such that and Thus Therefore is the left quotient ring of with respect to

**Remark**. Every finite subset of can be written as This is easy to see: let By Remark 3 in this post, there exist such that Let for every , and see that So for all .