Theorem 11.7.1. Let $A$ be a finite central simple algebra over $k$, and let $B$ be a simple subalgebra of $A$. Then

1. the centralizer $C$ of $B$ in $A$ is simple,

2. $[A : k] = [B : k][C : k]$, and

3. the centralizer of $C$ in $A$ is $B$.

Proof. Throughout this proof we use the results of Lemma 11.4.6 freely. Choose a simple $A$-module $M$. Set $L = \text{End}_ A(M)$. Then $L$ is a skew field with center $k$ which acts on the left on $M$ and $A = \text{End}_ L(M)$. Then $M$ is a right $B \otimes _ k L^{op}$-module and $C = \text{End}_{B \otimes _ k L^{op}}(M)$. Since the algebra $B \otimes _ k L^{op}$ is simple by Lemma 11.4.7 we see that $C$ is simple (by Lemma 11.4.6 again).

Write $B \otimes _ k L^{op} = \text{Mat}(m \times m, K)$ for some skew field $K$ finite over $k$. Then $C = \text{Mat}(n \times n, K^{op})$ if $M$ is isomorphic to a direct sum of $n$ copies of the simple $B \otimes _ k L^{op}$-module $K^{\oplus m}$ (the lemma again). Thus we have $\dim _ k(M) = nm [K : k]$, $[B : k] [L : k] = m^2 [K : k]$, $[C : k] = n^2 [K : k]$, and $[A : k] [L : k] = \dim _ k(M)^2$ (by the lemma again). We conclude that (2) holds.

Part (3) follows because of (2) applied to $C \subset A$ shows that $[B : k] = [C' : k]$ where $C'$ is the centralizer of $C$ in $A$ (and the obvious fact that $B \subset C')$. $\square$

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