Isaacs, $\textit{Character Theory of Finite Groups}$, Theorem(1.15)

Let $A$ be a semisimple algebra and let $M$ be an irreducible $A$-module. Then

SRE实战 互联网时代守护先锋，助力企业售后服务体系运筹帷幄！一键直达领取阿里云限量特价优惠。
  1. $M(A)$ is a minimal ideal of $A$
  2. if $W$ is irreducible, then it is annihilated by $M(A)$ unless $W\cong M$
  3. the map $x\mapsto x_M$ is one-to-one from $M(A)$ onto $A_M\subseteq End(M)$
  4. $\mathcal{M}(A)$ is a finite set

Pf: 1.

  • The map $\vartheta_x: y\mapsto xy$ satisfies $\vartheta_x\in E_A(A^\circ)$
  • $M(A)$ is an $E_A(A^\circ)$-module.

     2.

  • If $W\not\cong M$, then $W(A)\cap M(A)=0$
  • $W(A)M(A)\subseteq W(A)\cap M(A)=0$
  • $A^\circ$ has a submodule $W_0\cong W$ and $W_0M(A)=0$
  • Since $W\cong W_0$, they have the same annihilator in $A$

     3.

  • $A=\sum\cdot_{M\in\mathcal{M}(A)}M(A)$ for onto
  • If $x\in M(A)$ and $x_M=0$, then $x=1x\in A^\circ x=0$
  • To prove 1, let $I<M(A)$ and there exists $M_0\subseteq M(A),~M_0\cong M,~M_0\not\subseteq I$
  • $M_0I\subseteq M_0\cap I=0$, which means the elements in $I$ can annihilate $M_0$ and hence also $M$

     4.

  • $A=\sum\cdot_{M\in\mathcal{M}(A)}M(A)$ is finite dimensional.
扫码关注我们
微信号：SRE实战
拒绝背锅 运筹帷幄