## Posts Tagged ‘PI algebra’

Definition 1. A ring $R$ is called Dedekind-finite if $\forall a,b \in R: \ ab=1 \Longrightarrow ba=1.$

Remark 1. Some trivial examples of Dedekind-finite rings: commutative rings, any direct product of Dedekind-finite rings, any subring of a Dedekind-finite ring.

Definition 2. A ring $R$ is called reversible if $\forall a,b \in R : \ ab = 0 \Longrightarrow ba = 0.$

Example 1. Every reversible ring $R$ is Dedekind-finite. In particular, reduced rings are Dedekind-finite.

Proof. Suppose that $ab=1$ for some $a,b \in R.$ Then $(ba-1)b=b(ab)-b=0$ and thus $b(ba-1)=0.$ So $b^2a=b$ and hence $ab^2a=ab=1.$ It follows that $ba=(ab^2a)ba=(ab^2)(ab)a=ab^2a=1.$ So $R$ is Dedekind-finite. Finally, note that every reduced ring is reversible because if $ab=0,$ for some $a,b \in R,$ then $(ba)^2=b(ab)a=0$ and thus $ba=0. \Box$

Example 2. Every (left or right) Noetherian ring $R$ is Dedekind-finite.

Proof. We will assume that $R$ is left Noetherian. Suppose that $ab=1$ for some $a,b \in R.$ Define the map $f: R \longrightarrow R$ by $f(r)=rb.$ Clearly $f$ is an $R$-module homomorphism and $f$ is onto because $f(ra)=(ra)b=r(ab)=r,$ for all $r \in R.$ Now we have an ascending chain of left ideals of $R$

$\ker f \subseteq \ker f^2 \subseteq \cdots.$

Since $R$ is left Noetherian, this chain stabilizes at some point, i.e. there exists some $n$ such that $\ker f^n = \ker f^{n+1}.$ Clearly $f^n$ is onto because $f$ is onto. Thus $f^n(c)=ba-1$ for some $c \in R.$ Then

$f^{n+1}(c)=f(ba-1)=(ba-1)b=b(ab)-b=0.$

Hence $c \in \ker f^{n+1}=\ker f^n$ and therefore $ba-1=f^n (c) = 0. \Box$

Example 3. Finite rings are obviously Noetherian and so Dedekind-finite by Example 2. More generally:

Example 4. If the number of nilpotent elements of a ring is finite, then the ring is Dedekind-finite. See here.

Note that Example 4 implies that every reduced ring is Dedekind-finite; a fact that we proved in Example 1.

Example 5. Let $k$ be a field and let $R$ be a finite dimensional $k$-algebra. Then $R$ is Dedekind-finite.

Proof. Every left ideal of $R$ is clearly a $k$-vector subspace of $R$ and thus, since $\dim_k R < \infty,$ any ascending chain of left ideals of $R$ will stop at some point. So $R$ is left Noetherian and thus, by Example 2, $R$ is Dedekind-finite. $\Box$

Remark 2. Two important cases of Example 5 are $M_n(R),$ the ring of $n \times n$ matrices over a field, and, in general, semisimple rings. As a trivial result, $M_n(R)$ is Dedekind-finite for any commutative domain $R$ because $M_n(R)$ is a subring of $M_n(Q(R))$, where $Q(R)$ is the quotient field of $R$.
So the ring of $n \times n$ matrices, where $n \geq 2,$ over a field is an example of a Dedekind-finite ring which is not reversible, i.e. the converse of Example 1 is not true. Now let $R_i = \mathbb{Z}, \ i \geq 1.$ Then $R= \prod_{i=1}^{\infty} R_i$ is obviously Dedekind-finite but not Noetherian. So the converse of Example 2 is not true.

Example 6 and Example 7 are two generalizations of Example 5.

Example 6. Every algebraic algebra $R$ over a field $k$ is Dedekind-finite.

Proof. Suppose that $ab=1$ for some $a,b \in R.$ Since $R$ is algebraic over $k,$ there exist integers $n \geq m \geq 0$ and some $\alpha_i \in k$ with $\alpha_n \alpha_m \neq 0$ such that $\sum_{i=m}^n \alpha_i b^i = 0.$ We will assume that $n$ is as small as possible. Suppose that $m \geq 1.$ Then, since $ab=1,$ we have

$\sum_{i=m}^n \alpha_i b^{i-1}=a \sum_{i=m}^n \alpha_i b^i = 0,$

which contradicts the minimality of $n.$ So $m = 0.$ Let $c = -\alpha_0^{-1}\sum_{i=1}^n \alpha_i b^{i-1}$ and see that $bc=cb=1.$ But then $a=a(bc)=(ab)c=c$ and therefore $ba=bc=1. \ \Box$

Remark 3. Regarding Examples 5 and 6, note that although any finite dimensional $k$-algebra $R$ is algebraic over $k,$ but $R$ being algebraic over $k$ does not necessarily imply that $R$ is finite dimensional over $k.$ For example, if $\overline{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$ in $\mathbb{C},$ then it is easily seen that $\dim_{\mathbb{Q}} \overline{\mathbb{Q}}=\infty.$ Thus the matrix ring $R = M_n(\overline{\mathbb{Q}})$ is an algebraic $\mathbb{Q}$-algebra which is not finite dimensional over $\mathbb{Q}.$ So, as a $\mathbb{Q}$-algebra, $R$ is Dedekind-finite by  Example 6 not Example 5.

Example 7. Every PI-algebra $R$ is Dedekind-finite.

Proof. Let $J(R)$ be the Jacobson radical of $R.$ If $J(R)=\{0\},$ then $R$ is a subdirect product of primitive algebras $R/P_i,$ where $P_i$ are the primitive ideals of $R.$ Since $R$ is PI, each $R/P_i$ is PI too and thus, by Kaplansky’s theorem, $R/P_i$ is a matrix ring over some division algebra and thus Dedekind-finite by Example 2. Thus $\prod R/P_i$ is Dedekind-finite and so $R,$ which is a subalgebra of $\prod R/P_i,$ is also Dedekind-finite. For the general case, let $S=R/J(R).$ Now, $S$ is PI, because $R$ is PI, and $J(S)=\{0\}.$ Therefore, by what we just proved, $S$ is Dedekind-finite. Suppose that $ab = 1$ for some $a,b \in R$ and let $c,d$ be the image of $a,b$ in $S$ respectively. Clearly $cd=1_S$ and so $dc=1_S.$ Thus $1-ba \in J(R)$ and so $ba=1-(1-ba)$ is invertible. Hence there exists $e \in R$ such that $e(ba)=1.$ But then $eb=(eb)ab=e(ba)b=b$ and hence $ba=(eb)a=e(ba)=1. \Box$

## Ore domains (2)

Posted: January 4, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: , , , , ,

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. $\Box$