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

and so

**Definition**. The homomorphism is called the **central character** of