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)


\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

  1. Kim says:

    If $A$ is finite dimensional central simple k-algebra, and $B$ is finite dimensional simple k-subalgebra of $A$ and $C$ is the centralizer of $B$ in $A$, then $Z(B)=Z(C)=B\cap C$. .

    • Yaghoub Sharifi says:

      That’s right. Of course, Z(B)=B \cap C and B \cap C \subseteq Z(C) always holds for any subalgebra B of any algebra A. But for the inclusion Z(C) \subseteq B \cap C, which follows from the double centralizer theorem, we will need the conditions that you mentioned.

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