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).


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