## A theorem of Azumaya

Posted: December 22, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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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 $R \cong \text{End}_K(D) \cong M_n(K).$