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 $\alpha_m\in E_A(M)=D$. $(mn)\vartheta=(n\alpha_m)\vartheta=(n\vartheta)\alpha_m=m(n\vartheta)$
• Let $e\in I=AnA$ with $0\neq n\in M$. $m=me=m\sum(a_inb_i)=\sum((ma_i)(nb_i))$
• $m\vartheta=\sum((ma_i)(nb_i))\vartheta=\sum(ma_i)((nb_i)\vartheta)=m\sum a_i((nb_i)\vartheta$