## The double centralizer theorem (1)

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