Theorem 6

Let A be a central simple K-algebra, B a subalgebra of A of finite degree, and $B'$ its commutant in A.

a) Suppose that B is central simple. Then $B'$ is central simple, and the K-algebra homomorphism $\theta : B\otimes_KB'\rightarrow A$ that sends $b\otimes b'$ to $bb'$ is an isomorphism.

b) Suppose that B is semisimple, and let $L = B\cap B'$. Then $B'$ is a semisimple algebra. The commutant $L'$ of L in A is a semisimple ring with center L, and the ring homomorphism $\psi : B\otimes_LB'\rightarrow L'$ that sends $b\otimes b'$ to $bb'$ is an isomorphism.

Let us prove a). If B is central simple, then $B'$ is central simple by Theorem 5 of VIII, p. 259. Hence the K-algebra $B\otimes_KB'$ is simple (VIII, p. 222, Corollary 2), and the homomorphism $\theta : B\otimes_KB'\rightarrow A$ is injective. Now, by the equality of $[A : B']_s$ and [B : K] (Theorem 5), the left $B'$-modules $B\otimes_KB'$ and A are free of the same finite dimension; they are therefore $B'$-modules of the same finite length. By Corollary 2 of II, §1, No. 10, p. $213,\theta$ is bijective.

Let us prove b). By Theorem 5, the algebra L is commutative, of finite degree over K, and semisimple. By loc. cit. applied to L, its commutant $L'$ in A is a semisimple algebra, and L is the commutant of $L'$ in A, so L is the center of $L'$. Since L is the center of the semisimple rings $L', B$, and $B'$, we can identify $L'$ with a finite product of simple rings $L'_i$ (for $i\in I$), so that we have

where $L_i$ is the center of $L'_i$ and where $B_i$ and $B'_i$ are subalgebras of $L'_i$ with center $L_i$ that are each other’s commutants in $L'_i$. We view $L'_i$ as a central simple algebra over the commutative field $L_i$ and $B_i$ as a central simple $L_i$-algebra of finite degree. By assertion a), the canonical mapping $\psi_i: B_i\otimes_{L_i}B'_i\rightarrow L'_i$ that sends $b_i\otimes b'_i$ to $b_ib'_i$ is a ring isomorphism. Now, we can identify $B\otimes_LB'$ with $\prod_{i\in I}(B_i\otimes_{L_i}B'_i)$, so that $\psi$ is the product of the family of mappings $(\psi_i)_{i\in I}$. Therefore, $\psi$ is a ring isomorphism.