Theorem 5

Let A be a central simple K-algebra, and let B be a semisimple subalgebra of A of finite degree over K.

a) The commutant $B'$ of B in A is a simple subalgebra, and B is the commutant of $B'$ in A. Moreover, the algebra $B\cap B'$ is a semisimple commutative algebra of finite degree over K; it is the common center of B and $B'$.

b) Suppose that B is simple. Then $B'$ is simple, and we have the equalities

(See VIII, p. 124, Definition 2 for the definition of the left degree $[A : B]_s.$)

The K-algebra $A^o$ is central simple, and the K-algebra B is semisimple and of finite degree. By Corollary 1 of VIII, p. 221, the algebra $C = B\otimes_KA^o$ is semisimple. Let M be the C-module with the same additive group as A and with external law given by the formula $(b\otimes a)a'=ba'a$ for $a, a'$ in A and $b$ in B. Let $u$ be an element of End$_{\mathbf{Z}}(A)$. Then $u$ belongs to the commutant $C'_M$ of the C-module M if and only if $u$ is right A-linear and left B-linear, in other words, if and only if $u$ belongs to the commutant of $B_M$ in the ring of homotheties of the A-module $A_s$. We consequently define an isomorphism $\gamma$ from $B'$ to $C'_M$ by the relation $\gamma (b')(x) =b'x$ for $b'$ in $B'$ and $x$ in M. Now, the ring C is semisimple, and the C-module M is generated by the element 1 of A. By Proposition 6 of VIII, p. 139, the ring $C'_M$ is semisimple, so the algebra $B'$ is semisimple.

Let $\varphi$ be the K-algebra homomorphism from $A\otimes_KA^o$ to End$_K(M)$ that sends $a\otimes a'$ to the K-linear mapping $x\mapsto axa'$ from M to M. Since the K-algebras A and $A^o$ are central simple, the only two-sided ideals of $A\otimes_KA^o$ are 0 and $A\otimes_KA^o$ (VIII, p. 218, Corollary). We have $C_M=\varphi (B\otimes A^o)$ and $C'_M=\varphi (B'\otimes K)$. The homomorphism $\varphi$ is not zero; it is therefore injective. Since the ring C is semisimple, we have $C''_M= C_M$ by Proposition 5 of VIII, p. 139. It follows that the subalgebra $B\otimes_KA^o$ of $A\otimes_KA^o$ is the commutant of the subalgebra $B'\otimes_KK$. The commutant of $B'\otimes_KK$ in $A\otimes_KK$ is therefore equal to $(B\otimes_KA^o)\cap (A\otimes_KK)$, that is, to $B\otimes K$ by Proposition 19 of II, §7, No. 9, p. 311. Hence the commutant of $B'$ in A is equal to B. The algebra $L = B\cap B'$ is the center of B. Since B is a semisimple algebra of finite degree over K, the algebra L is commutative and semisimple and has finite degree over K. Since B is the commutant of $B'$ in A, the center of $B'$ is also equal to $L = B\cap B'$ (VIII, p. 77). We have proved a).

Now suppose that the algebra B is simple. By Corollary 2 of VIII, p. 222, the ring C is simple. By Proposition 4 of VIII, p. 123 applied to the C-module M, whose commutant is isomorphic to $B'$, the ring $B'$ is simple and M is a $B'$-module of finite length. In other words, $B'$ is a simple subring of the simple ring A, and the left degree $[A : B']_s$ is an integer $m\geqslant 1$. Viewed as a left $B'$-module, A has a finite basis $(a_1, . . . , a_m)$. Moreover (loc. cit.), $\varphi$ induces by restriction an isomorphism from C to $C''_M=$ End$_{B'}$(A), and the mapping $c\mapsto (ca_1, . . . , ca_m)$ from C to $A^m$ is therefore bijective. Consequently, C is a free right A-module of dimension $m$. Now, we have $C = B\otimes A^o$, so C is a free right A-module of dimension [B : K]; it follows that $[A : B']_s=m= [B : K]$. From Proposition 6 of VIII, p. 125, we deduce

since we also have

and [B : K] is finite and nonzero, we conclude that we have the equality $[A : B]_s= [B': K] ($Set Theory, III, §6, No. 3, p. 188, Corollary 3). We have proved b).

Let A be a central simple K-algebra of finite degree. There can exist semisimple commutative subalgebras B of A satisfying $[A : K]\not=$ $[B : K] [B': K]$ (Exercise 1 of VIII, p. 269).