Two applications of Schur’s lemma

Posted: July 11, 2010 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
Tags: , ,

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.

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