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 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 the coefficient of at least one of the monomials of the highest degree in f is 1 and f(a_1, \cdots , a_n)=0, for all a_j \in A. Then 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). Then for any a_1,a_2 \in A we have \text{tr}(a_1a_2 - a_2a_1)=0. Thus 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 any element of A. So we’ve proved that A is PI because it 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

Advertisements
Comments
  1. Jose Brox says:

    Great!

    Thank you

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s