\(
\newcommand{\CC}{\mathbb{C}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\TT}{\mathbb{T}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\H}{\mathcal{H}}
\newcommand{\e}{\epsilon}
\newcommand{\x}{\mathbf{x}}
\newcommand{\y}{\mathbf{y}}
\)

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 \(x\in A\).