Proposition 6.3.8 (Maximal Regular Ideals and Characters).label Let $A$ an abelian Banach algebra. Let $\cali{M}(A)\define\curl{\frak{m}\tleq A:\frak{m}\text{ is maximal and regular}}$, $\OM(A)\define\curl{\om:A\to \com|\om\text{ is a nonzero homomorphism}}$ then there is a bijection
\begin{align*}\cali{M}(A)\to \OM(A)&&\frak{m}\mto \om_{\frak{m}}\end{align*}
with inverse
\begin{align*}\OM(A)\to \cali{M}(A)&&\om\mto \om\inv(0)\end{align*}
Proof. By Lemma 6.3.7 each $\frak{m}\in\cali{M}(A)$ give rise to an $\om_{\frak{m}}\in\OM(A)$ such that $\frak{m}=\om_{\frak{\om}}\inv(0)$. If $\om\in\OM(A)$ then $\frak{m}_{\om}\define \om\inv(0)$ is a maximal ideal of $A$ since $A/\om\inv(0)\iso \com$ which has no nontrivial ideal and regular since $1\in\com\suf \om(A)$. Furthermore, isomorphism $A/\om\inv(0)\iso\com$ is unique; hence $\om=\om_{\frak{m}_\om}$$\square$