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 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. $\Box$