Introduction. We first need to recall a few facts from basic field theory. For the definition of separability see the introduction section in this post. We have the following facts.

Fact 1. Let E/F be an algebraic field extension and a \in E. Then F(a)/F is separable if and only if a is separable over F.

Fact 2. Let F \subseteq E \subseteq L be a chain of fields and suppose that L/F is algebraic. If both E/F and L/E are separable, then L/F is separable.

Fact 3. If E/F is a separable field extension and [E:F] < \infty, then E=F(a) for some a \in E.

Theorem. Let D be a division algebra with the center k and suppose that \dim_k D < \infty. There exists a \in D such that K=k(a) is a maximal subfield of D and K/k is separable.

Proof. Let A be the set of all subfields of D which are separable extensions of k. This set is non-empty because k \in A. Since \dim_k D < \infty, the set A with \subseteq has a maximal element, say K. Let C(K) be the centralizer of K in D. Suppose that C(K) \neq K. Then C(K) is a noncommutative division ring and K \subset C(K). Let Z(C(K)) be the center of K. Then, since K is commutative, we have Z(C(K))=C(C(K))=K, by the double centralizer theorem. Clearly C(K) is algebraic over K because [C(K):K] < \infty. Hence, by the Jacobson-Noether theorem, there exists a \in C(K) \setminus K such that a is separable over K. So we have the chain of fields k \subset K \subset K(a) where both K/k and K(a)/K are separable (see Fact 1). Thus, by Fact 2, K(a)/k is separable and so K(a) \in A. But this contradicts the maximality of K in A. This contradiction implies C(K)=K and so, by Corollary 3, K is a maximal subfield of D. Finally, by Fact 3, K=k(a) for some a \in K. \ \Box

Advertisements

Leave a Reply

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

WordPress.com Logo

You are commenting using your WordPress.com 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 )

Google+ photo

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

Connecting to %s