Throughout is a regular submonoid of the ring i.e. is multiplicatively closed, and every element of is regular.

**Remarks:**

**1)** (resp. ) is defined if and only if is left (resp. right) Ore. In this case, since every element of is regular, the corresponding map (resp. ) defined by (resp. ) is one-to-one and hence (resp. ) contains a copy of

**2)** If every element of is a unit, then for all and thus for all So, is left (and right) Ore and therefore (and ) both exist and are equal to The reason that they are equal to is that for all and thus for all

**3)** An example of **2)**: let be a left Artinian ring and a regular element of Then the chain must terminate, i.e. there exists such that Thus for some Therefore which gives us Obviously the same result holds for right Artinian rings.

**4)** If contains all the regular elements of and it is left (resp. right) Ore, then the left (resp. right) quotient ring of is denoted by

**5)** If is central, i.e. it is contained in the center of , then it’s clearly both left and right Ore and thus

**Lemma**. Suppose that is central and Let Then for all

*Proof.* Since is central, we have for all Therefore for all

We will use the above lemma to prove that the GK-dimension of an algebra is invariant under localization with respect to a central submonoid:

**Theorem**. If is an algebra over a field and is central, then

*Proof.* Let where As we’ve seen before, there exit and such that for all Let Then, by the lemma, for all integers (note that for both sides are equal to ). Therefore and hence Thus The other direction of the inequality is trivial because