Theorem 11.6.1. Let A be a finite central simple k-algebra. Let B be a simple k-algebra. Let f, g : B \to A be two k-algebra homomorphisms. Then there exists an invertible element x \in A such that f(b) = xg(b)x^{-1} for all b \in B.
Proof. 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, see Lemmas 11.3.2 and 11.4.6. Then M has two B \otimes _ k L^{op}-module structures defined by m \cdot _1 (b \otimes l) = lmf(b) and m \cdot _2 (b \otimes l) = lmg(b). The k-algebra B \otimes _ k L^{op} is simple by Lemma 11.4.7. Since B is simple, the existence of a k-algebra homomorphism B \to A implies that B is finite. Thus B \otimes _ k L^{op} is finite simple and we conclude the two B \otimes _ k L^{op}-module structures on M are isomorphic by Lemma 11.4.6. Hence we find \varphi : M \to M intertwining these operations. In particular \varphi is in the commutant of L which implies that \varphi is multiplication by some x \in A, see Lemma 11.4.6. Working out the definitions we see that x is a solution to our problem. \square
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