Proposition 5

Let A be a central simple algebra of finite degree over K. Let B be a semisimple subalgebra of A, and let $B'$ be the commutant of B.

a) The algebra B contains a maximal semisimple commutative subalgebra of A if and only if B contains $B'$.

b) Suppose that B contains $B'$, and let $g$ be a K-algebra homomorphism from B to A. There exists an inner automorphism $\theta$ of A that coincides with $g$ on B.

Let L be a maximal commutative subalgebra of A; by Lemma 3 of VIII, p. 261, L is equal to its commutant $L'$ in A. If B contains L, then its commutant $B'$ is contained in $L'$ and therefore in B.

Conversely, suppose that $B'$ is contained in B. Then $B'$ is the center of B, and it is a semisimple commutative algebra (VIII, p. 137, Proposition 2). By Corollary 1, there exists a maximal semisimple commutative subalgebra L of A containing $B'$. The commutant of L is L (VIII, p. 261, Lemma 3, a)), and that of $B'$ is equal to B (VIII, p. 259, Theorem 5). The relation $L\supset B'$ therefore implies $L\subset B$. We have proved a).

Let us prove b). Suppose that B contains $B'$, and choose a maximal semisimple commutative subalgebra L of A contained in B, which exists by a). Let $g$ be a homomorphism from B to A. By Proposition 3 of VIII, p. 262, we have the equality $[A : K] = [L : K]^2$; by the corollary of VIII, p. 263, there exists an inner automorphism $\theta_1$ of A that coincides with $g$ on L. If $f$ is the canonical injection of B into A, then the homomorphisms $g$ and $\theta_1\circ f$ have the same restriction to the center $B'$ of B because $B'$ is contained in L. By Theorem 2 of VIII, p. 256, there exists an inner automorphism $\theta$ of A such that $g=\theta \circ f$; in other words, $\theta$ extends $g$.