## The double centralizer theorem (2)

Posted: January 31, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , ,

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

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

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 $Z(B)=Z(C)=B\cap C$. .
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.