Throughout is a ring with unity, is a regular submonoid which is left Ore. So the left quotient ring of with respect to exists and is a subring of The following facts also holds for the “right” version:

** 1) **If , then and if then Also

*Proof*. The first part is trivial. Now suppose that Then and so Hence Similarly For the last part, if then and so

**2)** If then and It is also true that

*Proof*. Again, the first part is trivial. To prove the non-trivial side of the second part, let where and As we’ve showed before, there exist some and such that and thus where For the last part, suppose that where by the second part. So we have and thus

**3)** For any we have

*Proof*. Let where We have and thus for some Then, since we get which gives us because is a unit in (or because is regular in ). Thus

**4)** If is left Noetherian, then so is If and is left Noetherian, then

*Proof*. For the first part, first note that, by **1)** and **2)**, left ideals of are exactly in the form where is a left ideal of Let and Then For the second part, let be a two-sided ideal of and Then and thus So, since is left Noetherian, there exists some such that and thus Therefore for all Hence and so is a two-sided ideal of (In fact )

**5)** If is prime, then so is

*Proof*. Suppose that for some Then for all Thus which gives us i.e. Therefore either or which means either or This also proves that if is semiprime, then so is