## Two applications of Schur’s lemma

Posted: July 11, 2010 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
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Let $k$ be an algebraically closed field, $A$ a $k$ algebra and $V$ a simple $A$ module with $\dim_k V < \infty.$ We know, by Schur’s lemma, that every element of $D = \text{End}_A(V)$ is in the form $\mu 1_D,$ for some $\mu \in k.$

Application 1. If $A$ is commutative, then $\dim_k V = 1.$

Proof. Let $a \in A$ and $\{0\} \neq W$ be a $k$ subspace of $V.$ Define the map $f: V \longrightarrow V$ by $f(v)=av,$ for all $v \in V.$ Clearly $f$ is $k$ linear and, for any $b \in A$ and $v \in V,$ we have

$f(bv)=a(bv)=(ab)v=(ba)v=b(av)=bf(v).$

That means $f \in D$ and hence $f = \mu 1_D$, for some $\mu \in k.$ Thus if $w \in W$, then $aw=f(w)=\mu w \in W,$ which means that $W$ is an $A$ submodule of $V$ and so $W=V,$ because $V$ is simple over $A.$ So every non-zero $k$ subspace of $V$ is equal to $V.$ Hence $\dim_k V = 1.$

Application 2. Let $Z(A)$ be the center of $A.$ For every $a \in Z(A)$ there exists (a unique) $\mu_a \in k$ such that $av=\mu_a v,$ for all $v \in V$ and the map $\chi_V : Z(A) \longrightarrow k$ defined by $\chi_V(a)=\mu_a$ is a $k$ algebra homomorphism.

Proof. Define the map $f_a : V \longrightarrow V$ by $f_a(v)=av,$ for all $v \in V.$ Then $f_a \in D$ because $a \in Z(A).$ Thus $f_a = \mu_a 1_D,$ for some $\mu_a \in k$ and therefore $av=f_a(v)=\mu_a v,$ for all $v \in V.$ The uniqueness of $\mu_a$ is trivial.

To show that $\chi_V$ is a homomorphism, let $\lambda \in k, \ a,b \in Z(A).$ Then

$\mu_{\lambda a + b} v= (\lambda a + b)v=\lambda (av) + bv = \lambda \mu_a v + \mu_b v,$

and so $\mu_{\lambda a + b} = \lambda \mu_a + \mu_b.$ Similarly

$\mu_{ab} v = (ab)v = a(bv)=a (\mu_b v) = \mu_a (\mu_b v)=(\mu_a \mu_b)v$

and so $\mu_{ab}=\mu_a \mu_b.$  $\Box$

Definition. The homomorphism $\chi_V$ is called the central character of $V.$