## Ore domains (2)

Posted: January 4, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
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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. $\Box$