A theorem of Azumaya

Posted: December 22, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
Tags: , , ,

Theorem. Let D be a division algebra with the center Z and suppose that K is a maximal subfield of D. Let R=D \otimes_Z K. Then

1) \ R is a simple ring and thus left primitive. In fact:

2) \ D is a faithful simple left R module and \text{End}_R(D) \cong K.

Proof. 1) This part follows from the first part of this corollary but we will give a self-contained proof: let \{ k_j : \ j \in J \} be a Z basis for K. Suppose that R is not simple. Let (0) \neq I be an ideal of R. Choose n to be the smallest integer for which there exists 0 \neq x=\sum_{j=1}^n d_j \otimes k_{t_j} \in I. Then d_1 \neq 0 and therefore, by replacing x with (d_1^{-1} \otimes 1)x if necessary, we may assume that d_1 = 1. Now, for any d \in D we have

\sum_{j=2}^n (dd_j-d_jd) \otimes k_{t_j} = (d \otimes 1)x - x(d \otimes 1) \in I,

which gives us dd_j=d_jd, by minimality of n. So d_j \in Z, for all j. Thus x = 1 \otimes k, for some 0 \neq k \in K.  But then

1_R = 1 \otimes 1 = (1 \otimes k^{-1})x \in I

and so I=R.

2) Defining (d_1 \otimes k)d_2=d_1d_2k, for all d_1,d_2 \in D, \ k \in K, and extending it linearly will make D a left R module. Clearly D is a faithful R module because \text{ann}_R D \neq R is an ideal of R and R is a simple ring by 1).  To prove that D is a simple R module let d_1 \neq 0 and d_2 \in D. Then (d_2d_1^{-1} \otimes 1)d_1=d_2 and hence Rd_1=D. Finally, \text{End}_R D \cong K is just a special case of this theorem. Just note that, since K is commutative, K^{op}=K and since it is a maximal subfield, C_D(K)=K. \ \Box

Remark. By part 2) of the theorem and the density theorem, R is a dense subring of \text{End}_K(D). Also if \dim_K D = n < \infty, then R \cong \text{End}_K(D) \cong M_n(K).


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