A double-centralizer theorem for central simple algebras

Posted: January 30, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , ,

See also The double centralizer theorem.

Throughout this post k is a field and M_n(k), as usual, is the algebra of n \times n matrices with entries in k.

Notation. Let A be a ring and let B be a subring of A. Let X \subseteq A. Then we will denote by C_B(X) the centralizer of X in B, i.e. C_B(X)=\{b \in B : \ bx=xb, \ \forall x \in X \}. If X=\{a\}, then we will write C_B(a) instead of C_B(\{a\}).

The following lemma is a well-known result in linear algebra and so I will not prove it here.

Lemma. Let R=M_n(k) and let a,b \in R. If C_R(a) \subseteq C_R(b), then b \in k[a]. In other words, C_R(C_R(a))=k[a], for all a \in R. \ \Box

The lemma can be extended to any finite dimensional central simple algebra:

Theorem. (W. L. Werner, 1969) Let A be a finite dimensional central simple k-algebra and let a,b \in A. If C_A(a) \subseteq C_A(b), then b \in k[a]. In other words, C_A(C_A(a))=k[a], for all a \in A.

Proof. Let R = A \otimes_k A^{op}. We proved in here that R \cong M_n(k), for some integer n.

Claim . C_R(a \otimes_k 1) \subseteq C_R(b \otimes_k 1).

Proof of the claim. Let \{b_1, \cdots , b_m \} be a k-basis for A^{op} and let r \in C_R(a \otimes_k 1). So

r = \sum_{i=1}^m a_i \otimes_k b_i,

for some a_i \in A. Since r \in C_R(a \otimes_k 1), we have

\sum_{i=1}^m aa_i \otimes_k b_i=\sum_{i=1}^m a_ia \otimes_k b_i.

So \sum_{i=1}^m (aa_i-a_ia) \otimes_k b_i=0 and hence, since b_i are k-linearly independent, we have aa_i-a_ia=0, for all i. That means a_i \in C_A(a) for all i. Therefore

r=\sum_{i=1}^m a_i \otimes_k b_i \in C_A(a) \otimes_k A^{op} \subseteq C_A(b) \otimes_k A^{op}.


C_R(a \otimes_k 1) \subseteq C_A(b) \otimes_k A^{op}. \ \ \ \ \ \ \ \ \ \ \ \ (*)

But clearly if x \otimes_k y \in C_A(b) \otimes_k A^{op}, then

(x \otimes_k y)(b \otimes_k 1)=xb \otimes_k y = bx \otimes_k y = (b \otimes_k 1)(x \otimes_k y).

So C_A(b) \otimes_k A^{op} \subseteq C_R(b \otimes_k 1) and the claim now follows from (*). \ \Box

So, by the lemma and the claim, b \otimes_k 1 \in k[a \otimes_k 1]=k[a] \otimes_k 1. Thus b \in k[a] and the proof of the theorem is complete. \Box

Remark. Note that Z(C_A(a))=C_A(C_A(a)), where Z(C_A(a)) means the center of C_A(a). A consequence of the theorem is that for any a \in A, C_A(a) is commutative iff C_A(a)=k[a].


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