## Kaplansky-Amitsur theorem (2)

Posted: December 24, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
Tags: , , , ,

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

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