Theorem 11.7.1. Let $A$ be a finite central simple algebra over $k$, and let $B$ be a simple subalgebra of $A$. Then
the centralizer $C$ of $B$ in $A$ is simple,
$[A : k] = [B : k][C : k]$, and
the centralizer of $C$ in $A$ is $B$.
11.7 The centralizer theorem
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 $B \subset C')$. $\square$
Lemma 11.7.2. Let $A$ be a finite central simple algebra over $k$, and let $B$ be a simple subalgebra of $A$. If $B$ is a central $k$-algebra, then $A = B \otimes _ k C$ where $C$ is the (central simple) centralizer of $B$ in $A$.
Proof. We have $\dim _ k(A) = \dim _ k(B \otimes _ k C)$ by Theorem 11.7.1. By Lemma 11.4.7 the tensor product is simple. Hence the natural map $B \otimes _ k C \to A$ is injective hence an isomorphism. $\square$
Lemma 11.7.3. Let $A$ be a finite central simple algebra over $k$. If $K \subset A$ is a subfield, then the following are equivalent
$[A : k] = [K : k]^2$,
$K$ is its own centralizer, and
$K$ is a maximal commutative subring.
Proof. Theorem 11.7.1 shows that (1) and (2) are equivalent. It is clear that (3) and (2) are equivalent. $\square$
Lemma 11.7.4.slogan Let $A$ be a finite central skew field over $k$. Then every maximal subfield $K \subset A$ satisfies $[A : k] = [K : k]^2$.
Proof. Special case of Lemma 11.7.3. $\square$
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