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


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


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

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

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