## 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 $a \in K. \ \Box$

## 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

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

## The double centralizer theorem (1)

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

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).