每日定理9

Isaacs, $\textit{Character Theory of Finite Groups}$, Theorem(1.16)

Let $A$ be a semisimple algebra and let $M$ be an irreducible $A$-module. Let $D=E_A(M)$. Then $E_D(M)=A_M$.

Pf: Without loss of generality, assume $M\subseteq A^{\circ}$ and let $I=M(A)$.

• $A_M\subseteq E_D(M)$ clearly
• Let $\vartheta\in E_D(M)$ then $(m\alpha)\vartheta=(m\vartheta)\alpha$ for $\alpha\in D$
• If $m\in M$, define $\alpha_m:M\rightarrow M$ by $(x)\alpha_m=mx$ and $\alpha_m\in E_A(M)=D$
• $(mn)\vartheta=(n\alpha_m)\vartheta=(n\vartheta)\alpha_m=m(n\vartheta)$

每日定理9

原文：https://www.cnblogs.com/zhengtao1992/p/10782695.html