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


Leave a Reply

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

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

Google+ photo

You are commenting using your Google+ 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 )


Connecting to %s