**Theorem**. Let be a division algebra with the center and suppose that is a maximal subfield of Let Then

is a simple ring and thus left primitive. In fact:

is a faithful simple left module and

*Proof. *1) This part follows from the first part of this corollary but we will give a self-contained proof: let be a basis for Suppose that is not simple. Let be an ideal of Choose to be the smallest integer for which there exists Then and therefore, by replacing with if necessary, we may assume that Now, for any we have

which gives us by minimality of So for all Thus , for some But then

and so

2) Defining for all and extending it linearly will make a left module. Clearly is a faithful module because is an ideal of and is a simple ring by 1). To prove that is a simple module let and Then and hence Finally, is just a special case of this theorem. Just note that, since is commutative, and since it is a maximal subfield,

**Remark.** By part 2) of the theorem and the density theorem, is a dense subring of Also if then