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.
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.
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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