## 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
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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. $\Box$

## Kaplansky-Amitsur theorem (2)

Posted: December 24, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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Theorem. (Kaplansky-Amitsur) Let $A$ be a left primitive algebra, $M$ a faithful simple $A$ module and $D=\text{End}_A(M).$ If  $A$ satisfies a polynomial of degree $d,$ then

i) $A \cong M_n(D),$ where $n = \dim_D M \leq [d/2].$ So $Z:=Z(A) \cong Z(D)$ is a field.

ii) $\dim_Z A \leq [d/2]^2.$

We proved i) in the previous section. So we just need to prove ii). First two easy remarks.

Remark 1. Let $F$ be a field. Then $M_m(A) \otimes_F M_n(B) \cong M_{mn}(A \otimes_F B),$ for any $F$ algebras $A,B.$

Remark 2. Let $C$ be a commutative ring, $A$ a PI $C$-algebra and $K$ a commutative $C$-algebra. If $A$ satisfies a multi-linear polynomial $f,$ then $A \otimes_C K$ will also satisfy $f.$

Proof of ii). So $A$ satisfies some multi-linear polynomial $f$ of degree at most $d.$ Clearly $D$ satisfies $f$ too because it’s a subring of $M_n(D) \cong A.$ Let $K$ be a maximal subfield of $D$ and $R=D \otimes_Z K.$ By Remark 2, $R$ also satisfies $f.$ But, by Azumaya theorem, $R$ is left primitive, $D$ is a faithful simple left $R$-module and $\text{End}_R(D) \cong K.$ Thus, by part i) of the theorem, $R \cong M_m(K),$ for some positive integer $m.$  Therefore by Remark 1

$A \otimes_Z K \cong M_n(D) \otimes_Z K \cong M_n(R) \cong M_{mn}(K).$

Hence $\dim_Z A = \dim_K A \otimes_Z K = (mn)^2.$ On the other hand, by Remark 2, $A \otimes_Z K$ satisfies $f$ and so $d \geq \deg f \geq 2mn.$ Therefore $mn \leq d/2$ and so $mn \leq [d/2].$ Finally we have

$\dim_Z A =(mn)^2 \leq [d/2]^2. \ \Box$

## Kaplansky-Amitsur theorem (1)

Posted: December 23, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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We will assume that $C$ is a commutative ring with 1. Also, given an integer $n \geq 1,$  we will denote by $C \langle x_1, \cdots , x_n \rangle$ the ring of polynomials in non-commuting variables $x_1, \cdots , x_n$ and with coefficients in $C.$

Definition. A $C$-algebra $A$ is called a polynomial identity algebra (or a PI-algebra) if there exists an integer $n \geq 1$ and a non-zero polynomial

$f \in C \langle x_1, \cdots , x_n \rangle$

such that $f(a_1, \cdots , a_n)=0$ for all $a_j \in A.$ The smallest possible degree of such $f$ is called the PI-degree of $A.$

Remark 1. If an algebra $A$ satisfies a polynomial $f,$ then obviously every subring of $A$ and homomorphic image of $A$ will satisfy $f$ too.

Remark 2. It is known and easy to prove that if an algebra $A$ satisfies a non-zero polynomial of degree $n,$ then $A$ will also satisfy a non-zero multi-linear polynomial of degree at most $n.$

Example 1. Every commutative algebra is PI because it satisfies the polynomial $x_1x_2 - x_2x_1.$

Example 2. Let $C$ be a commutative ring and $A=M_2(C).$ We show that $A$ is a PI-algebra. Let $a_1,a_2 \in A.$ Then $\text{tr}(a_1a_2 - a_2a_1)=0$ and so, by Cayley-Hamilton theorem, $(a_1a_2 - a_2a_1)^2 = cI_2,$ for some $c \in C.$ Therefore $(a_1a_2-a_2a_1)^2$ commutes with every element of $A$ and so $A$ satisfies the polynomial

$f=(x_1x_2-x_2x_1)^2x_3 - x_3(x_1x_2 - x_2x_1)^2.$

Remark 2. Let $C$ be a commutative ring and suppose that $A=M_n(C)$ satisfies a non-zero polynomial of degree $\leq 2n-1.$ Then, by Remark 1, $A$ will also satisfy a non-zero multi-linear polynomial

$f(x_1, \cdots , x_k) = \sum_{\alpha \in S_k} c_{\alpha} x_{\alpha(1)} \cdots x_{\alpha (k)},$

for some $k \leq 2n-1$ and $c_{\alpha} \in C.$ Renaming the variables, if necessary, we may assume that $c=c_{\text{id}} \neq 0.$ Then $f(e_{11},e_{12},e_{22},e_{23}, \cdots )=c e_{1 \ell}=0,$ for some $\ell.$ Therefore $c e_{ij}=e_{i1}(c e_{1 \ell}) e_{\ell j} = 0,$ for all $i,j.$ Thus $c = 0,$ which is a contradiction. So $A$ does not satisfy any non-zero polynomial of degree $\leq 2n-1.$

Theorem. Let $A$  be an algebra. Suppose $A$ is left primitive, $M$ a faithful simple left $A$ module and $D=\text{End}_A(M).$ If $A$ satisfies a polynomial $f$ of degree $d,$ then $\dim_D M=n \leq [d/2]$ and $A \cong M_n(D).$

Proof. Suppose $\dim_D M > [d/2].$ Then there exists some $k > [d/2]$ such that either $A \cong M_k (D)$ or $M_k(D)$ is a homomorphic image of some subring of $A.$ In either case, by remark 1, $M_k (D)$ and hence $M_k(Z(D))$ satisfies $f$. Thus by remark 2: $d \geq 2k \geq 2([d/2] + 1) > d$, which is nonsense. $\Box$