Let be an algebraically closed field, a algebra and a simple module with We know, by Schur’s lemma, that every element of is in the form for some
Application 1. If is commutative, then
Proof. Let and be a subspace of Define the map by for all Clearly is linear and, for any and we have
That means and hence , for some Thus if , then which means that is an submodule of and so because is simple over So every non-zero subspace of is equal to Hence
Application 2. Let be the center of For every there exists (a unique) such that for all and the map defined by is a algebra homomorphism.
Proof. Define the map by for all Then because Thus for some and therefore for all The uniqueness of is trivial.
To show that is a homomorphism, let Then
and so Similarly
Definition. The homomorphism is called the central character of