Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2009 / B / 4

A character of an algebra is a nonzero multiplicative complex-linear map $\chi :A\to \mathbb{C}$. A maximal proper ideal is a proper ideal contained in no larger proper ideal. For a character, $\chi (1)=1$ and $\chi (a)\in \sigma (a)$: otherwise $a-\chi (a)1$ would be invertible although its character value is zero. The spectral bound gives $|\chi (a)|\le \Vert a\Vert$, so the character is continuous. It is onto, and its kernel is a closed maximal ideal because $A/\ker \chi \cong \mathbb{C}$.

Conversely let $M$ be maximal. Its closure is an ideal and remains proper: otherwise an element of $M$ would lie within distance one of the unit, hence would be invertible by the Neumann series, forcing $1\in M$. Maximality then gives $\overline{M}=M$. The Banach algebra $A/M$ is a field. For any $b$ in this quotient choose $\lambda \in \sigma (b)$, using nonemptiness of the complex spectrum. In a field only zero is noninvertible, so $b-\lambda 1=0$. Hence every quotient element is a scalar multiple of the unit. The quotient map therefore determines a character ${\chi }_M$, uniquely normalized by ${\chi }_M(1)=1$. These constructions are inverse:For the convolution algebra, write $e_j$ for the sequence supported at $j$ with value one. Then $e_j{e}_k=e_{j+k}$, $e_0=1$, and $e_1{e}_{-1}=1$. If $z=\chi (e_1)$, continuity and $\Vert e_{\pm 1}{\Vert }_1=1$ imply $|z|\le 1$ and $|z^{-1}|\le 1$. Thus $|z|=1$, so uniquely $z=e^{it}$ for $0\le t<2\pi$, and $\chi (e_j)=e^{ijt}$ for every integer $j$. Absolute convergence in ${\ell }^1$ now givesConversely each displayed map is bounded, nonzero, and multiplicative, since the absolutely convergent double sum can be rearranged along $k+l=j$. Thus these are all characters, and all maximal proper ideals areThey are distinct because the character is uniquely recovered from its kernel and its value on the unit.