Here you can see part (1). In the following, is the algebra generated by non-commuting variables and with coefficients in So an element of is a finite sum of terms in the form where is any positive integer, and each is either or or For example where

**Lemma**. Let be the center of a domain If is not left (resp. right) Ore, then contains a copy of

*Proof.* Choose such that Then are left linearly independent over and we’re done by Jategaonkar’s lemma.

**Corollary**. Let be an algebra over a field. If is PI or has a finite GK-dimension, then is both left and right Ore.

*Proof.* Otherwise, by the lemma, would contain a free algebra and we know that such algebras are neither PI nor have a finite GK dimension.

**A left Ore domain which is not right Ore. **Let be a division ring and be a homomorphism which is not onto. Since we have and thus because has no non-zero proper ideal. Therefore is injective and hence the left skew polynomial ring is a domain.

*Claim 1*. is not right Ore.

*Proof*. Suppose is right Ore. Since we’ve assumed that is not onto, there exists Now we must have So, there exist non-zero elements such that Clearly must have the same degree, say Let be the leading terms of respectively. So the leading terms of which have to be equal, are respectively. Equating these leading terms gives us which is a contradiction.

*Claim 2*. Euclidean algorithm holds in

*Proof*. That means we need to prove for every there exists such that where either or The proof is by induction over : if then we may choose and Let be the leading terms of respectively. Then and have the same leading terms and thus has a degree Thus, by induction, there exist such that where either or Finally we have and so

*Claim 3*. is left Ore.

*Proof*. Let be a non-zero left ideal of and choose a non-zero with the smallest degree. For every by Claim 2, there exist such that where either or But and thus we cannot have Therefore and i.e. So every left ideal of is principal and, as a result, is left Noetherian and therefore left Ore.