Here you can see part (1). In the following, Z \langle x,y \rangle is the algebra generated by non-commuting variables x,y and with coefficients in Z. So an element of  Z \langle x,y \rangle is a finite sum of terms in the form \alpha x_1x_2 \ldots x_n, where n is any positive integer, \alpha \in Z and each x_i is either 1 or x or y. For example f = \alpha_0 + \alpha_1 xyx + \alpha_2 yx^3y^2 where \alpha_i \in Z.

Lemma. Let Z be the center of a domain R. If R is not left (resp. right) Ore, then R contains a copy of Z \langle x,y \rangle.

Proof. Choose 0 \neq r_1,r_2 \in R such that Rr_1 \cap Rr_2 = (0). Then r_1,r_2 are left linearly independent over R and we’re done by Jategaonkar’s lemma. \Box

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

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

A left Ore domain which is not right Ore. Let D be a division ring and \sigma : D \longrightarrow D be a homomorphism which is not onto.  Since \sigma (1_D)=1_D \neq 0, we have \ker \sigma \neq D and thus \ker \sigma = (0), because D has no non-zero proper ideal. Therefore \sigma is injective and hence the left skew polynomial ring R=D[x,\sigma] is a domain.

Claim 1. R is not right Ore.

Proof. Suppose R is right Ore. Since we’ve assumed that \sigma is not onto, there exists d \in D - \sigma (D). Now we must have xR \cap dxR \neq (0). So, there exist non-zero elements g, h \in R such that xg=dxh. Clearly g, h must have the same degree, say n. Let d_1x^n, \ d_2x^n be the leading terms of g, h respectively. So the leading terms of xg , dxh, which have to be equal, are  \sigma (d_1)x^{n+1}, \ d \sigma (d_2) x^{n+1} respectively. Equating these leading terms gives us d=\sigma (d_1 d_2^{-1}) \in \sigma (D), which is a contradiction. \Box

Claim 2. Euclidean algorithm holds in R.

Proof. That means we need to prove for every f, g \in R, there exists r, s \in R such that f=sg + r, where either r = 0 or \deg r < \deg g. The proof is by induction over \deg f: if \deg f < \deg g, then we may choose s = 0 and r = f. Let d_1 x^n, \ d_2 x^m, \ n \geq m, be the leading terms of f, g respectively. Then f and d_1x^{n-m}d_2^{-1}g have the same leading terms and thus h = f - d_1x^{n-m}d_2^{-1}g has a degree < n. Thus, by induction, there exist r_1, s_1 \in R such that h = s_1g + r_1, where either r_1 = 0 or \deg r_1 < \deg g. Finally we have h=s_1g + r_1 = f - d_1 x^{n-m}d_2^{-1}g and so f = (s_1 + d_1x^{n-m}d_2^{-1})g + r_1. \ \Box

Claim 3. R is left Ore.

Proof. Let I be a non-zero left ideal of R and choose a non-zero g \in I with the smallest degree. For every f \in I, by Claim 2, there exist r, s \in R such that f = sg + r, where either r = 0 or \deg r < \deg g. But r = f - sg \in I and thus we cannot have \deg r < \deg g. Therefore r=0 and f \in Rg, i.e. I=Rg. So every left ideal of R is principal and, as a result, R is left Noetherian and therefore left Ore. \Box


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