# Joan Lindsay Orr


Let $$A$$ be a unital algebra and $$L$$ a proper left ideal. The quotient algebra $$A/L$$ is a vector space, and each $$a\in A$$ acts on $$A/L$$ by left multiplication as $$T_a(x + L) = ax + L$$. The map $$\pi:a\mapsto T_a$$ is a multiplicative, linear map from $$A$$ to the algebra of linear maps on $$A/L$$ and so is a representation of $$A$$ on $$A/L$$, called the left regular representation.

The left regular representation is irreducible if and only if $$L$$ is a maximal left ideal.

Proof. Suppose the left regular representation is irreducible and that $$L\subseteq L' \subseteq A$$ is a left ideal. Then $$W = \pi(L')$$ is clearly a subspace of $$A/L$$ and gives a subrepresentation, so that $$W$$ is either $$0$$ or $$A/L$$. Thus $$L' = L' + L = \pi^{-1}(\pi(L'))$$ is either $$L$$ or $$A$$. Since these are the only two possibilities, $$L$$ must be a maximal left ideal.

Conversely, suppose $$L$$ is a maximal left ideal and suppose $$0 \subseteq W \subseteq A/L$$ is a subrepresentation. Clearly $$L' = \pi^{-1}(W)$$ is a left ideal conatining $$L$$ and so is either $$L$$ or $$A$$. Thus $$W = \pi(\pi^{-1}(W))$$ is either $$0$$ or $$A/L$$.

A primitive ideal is the kernel of an irreducible representation.

Given a proper left ideal $$L$$, the kernel of the left regular representation is $$\{a\in A : ax \in L \text{ for all } x\in A\}$$. For $$T_a = 0$$ iff $$\pi(ax)=0$$ for all $$x\in A$$ iff $$ax\in L$$ for all $$x\in A$$.