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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