# John Lindsay Orr

$\newcommand{\CC}{\mathbb{C}} \newcommand{\NN}{\mathbb{N}} \newcommand{\RR}{\mathbb{R}} \newcommand{\H}{\mathcal{H}} \newcommand{\e}{\epsilon} \newcommand{\x}{\mathbf{x}} \newcommand{\y}{\mathbf{y}}$

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.

Lemma 1. 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 $A/L$.

Definition 2. A primitive ideal is the kernel of an irreducible representation.

Remark 3. 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 $x\in A$.