## A double-centralizer theorem for central simple algebras

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

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

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

Thus

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

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 $C_A(a)=k[a].$