Posts Tagged ‘double centralizer theorem’

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

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As in part (1), k is a field, A is a finite dimensional central simple k-algebra and B is a simple k-subalgebra of A. We will also be using notations and the result in the lemma in part (1), i.e. R = A \otimes_k B^{op}, M is the unique simple R-module,  A \cong M^n, D = \text{End}_R(M) and C_A(B) \cong M_n(D). Finally, as usual, we will denote the center of any algebra S by Z(S). We now prove a nice relationship between dimensions.

Lemma. \dim_k C_A(B) \cdot \dim_k B = \dim_k A.

Proof. We have R \cong M_m(D) and M \cong D^m, for some integer m. Thus A \cong D^{mn}, as k-modules, and hence

\dim_k A = mn \dim_k D. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

We also have

\dim_k A \cdot \dim_k B = \dim_k R = \dim_k M_m(D)=m^2 \dim_k D \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

and

\dim_k C_A(B)=\dim_k M_n(D)=n^2 \dim_k D. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

Eliminating \dim_k D in these three identities will give us the result. \Box

Theorem. If B is a central simple k-algebra, then C_A(B) is a central simple k-algebra too and we have A =BC_A(B) \cong B \otimes_k C_A(B).

Proof. By the lemma in part (1), C_A(B) is simple and thus, since B is central simple, B \otimes_k C_A(B) is a simple algebra. Thus the map \phi : B \otimes_k C_A(B) \longrightarrow BC_A(B) defined by \phi(b \otimes_k c)=bc is a k-algebra isomorphism. Hence B \otimes_k C_A(B) \cong BC_A(B) and so

                                  \dim_k BC_A(B)=\dim_k B \otimes_k C_A(B)=\dim_k A,

by the above lemma. Therefore BC_A(B)=A. Clearly if c is in the center of C_A(B), then c commutes with every element of both B and C_A(B). Hence  c commutes with every element of A and thus c \in k. So Z(C_A(B))=k, i.e. C_A(B) is a central simple k-algebra. \Box

The Double Centralizer Theorem. C_A(C_A(B))=B.

Proof. Obviously B \subseteq C_A(C_A(B)). So, to prove that B=C_A(C_A(B)), we only need to show that \dim_k B = \dim_k C_A(C_A(B)).  Applying the above lemma to C_A(B) gives us

\dim_k C_A(C_A(B)) \cdot \dim_k C_A(B) = \dim_k A. \ \ \ \ \ \ \ \ \ \ \ \ \ (*)

Plugging \dim_k C_A(B)=\frac{\dim_k A}{\dim_k B}, which is true by the above lemma, into (*) finishes the proof. \Box

Corollary. If B is a subfield of A, then \deg A = (\deg C_A(B))(\dim_k B).

Proof. We first need to notice a couple of things. First, since B is commutative, B \subseteq C_A(B). Since B is commutative, C_A(C_A(B))=Z(C_A(B)). Thus, by the above theorem, Z(C_A(B))=B and so C_A(B) is a central simple B-algebra. Now, by the lemma

(\deg A)^2=\dim_k A = (\dim_B C_A(B))(\dim_k B)^2=(\deg C_A(B))^2 (\dim_k B)^2. \ \Box

Throughout k is a field, A is a finite dimensional central simple k-algebra and B is a simple k-subalgebra of A. We will use the notation for centralizers given in this post. The goal is to prove that C_A(C_A(B))=B. This is called the double centralizer theorem for an obvious reason. We proved another double centralizer theorem in here. In there B=k[a], for some a \in A and B did not have to be simple. So that double centralizer theorem has nothing to do with this one. We first show that the centralizer of a simple subalgebra of a finite dimensional central simple k-algebra is simple.

Lemma. The k-subalgebra C_A(B) is simple.

Proof. Since A is central simple and B, and hence B^{op}, is simple, the algebra R=A \otimes_k B^{op} is also simple by the first part of the corollary in this post. Clearly R is finite dimensional over k because A is so. So, as we mentioned before in Remark 1, R has a unique simple R-module M and any R-module is isomorphic to the direct sum of a finite number of copies of M. Thus, since A has a structure of an R-module, we must have

A \cong M^n, \ \ \ \ \ \ \ \ \ \ (1)

for some integer n. Let

D = \text{End}_R(M).

Since M is a simple R-module, D is a division ring by Schur’s lemma.  On the other hand, as we proved in this post,

C_A(B) \cong \text{End}_R(A). \ \ \ \ \ \ \ \ \ \ (2)

Now, (1), (2) and the remark in this post gives us

C_A(B) \cong \text{End}_R(A) \cong \text{End}_R(M^n) \cong M_n(\text{End}_R(M)) \cong M_n(D). \ \Box

To be continued in part (2).