Maximal ideals and characters of a commutative Banach algebra

If $J$ is a proper ideal and $x\in J$ is invertible with inverse $x^{-1}$, then $1=x^{-1}x\in J$ and hence $a=a\cdot 1\in J$ for every $a$, so $J=A$; contrapositively a proper ideal contains no invertible element, and in particular $1\mathrm{\notin }J$.

For every ideal $J$ the norm closure $\bar{J}$ is an ideal: it is a closed linear subspace, and for $a\in A$ the maps $x\mapsto ax$ and $x\mapsto xa$ are continuous, so $a\bar{J}\subseteq \overline{aJ}\subseteq \bar{J}$ and $\bar{J}a\subseteq \bar{J}$.

If $\{J_k\}$ is a nonempty chain of ideals of $A$, then $J:={\bigcup }_k{J}_k$ is an ideal: it is a union of a nested family of linear subspaces and is closed under multiplication by $A$. The family $\mathcal{P}$ of proper ideals of $A$ containing $I$ is nonempty because $I\in \mathcal{P}$, and it is a poset under inclusion.

Character kernels are maximal ideals. If $\chi$ is a character then $\ker \chi$ is a proper ideal: it is a linear subspace closed under multiplication, and it is proper because $\chi \mathrm{\ne }0$ gives some $a$ with $\chi (a)\mathrm{\ne }0$. It is maximal: if $M^{\mathrm{\prime }}⊋\ker \chi$ is an ideal and $a\in M^{\mathrm{\prime }}\setminus \ker \chi$, then for every $b\in A$ the element $b-(\chi (b)\mathrm{/}\chi (a))a$ lies in $\ker \chi \subseteq M^{\mathrm{\prime }}$, hence $b\in M^{\mathrm{\prime }}$ and $M^{\mathrm{\prime }}=A$.

The assignment $\chi \mapsto \ker \chi$ is injective on $\mathrm{\Delta }(A)$: if $\ker \chi =\ker \psi$ and $a\in A$, then $a-\chi (a)1\in \ker \chi =\ker \psi$, so $0=\psi (a-\chi (a)1)=\psi (a)-\chi (a)\psi (1)=\psi (a)-\chi (a)$ by [L8], whence $\chi =\psi$.

The union $J$ of a nonempty chain $\{J_k\}\subseteq \mathcal{P}$ is proper: if $1\in J$ then $1\in J_k$ for some $k$, and then $J_k=A$ by [step 1.1], contradicting $J_k\in \mathcal{P}$. So by [step 1.3] every chain in $\mathcal{P}$ has an upper bound in $\mathcal{P}$, and [L6], available under the standing Axiom of Choice [L7], supplies a maximal element $M\in \mathcal{P}$ with $I\subseteq M⊊A$.

The maximal ideal $M$ of [step 2.1] is closed. By [step 1.2] $\bar{M}$ is an ideal containing $M$; if $\bar{M}=A$ then $1\in \bar{M}$, so some $x\in M$ satisfies $\mathrm{\parallel }1-x\mathrm{\parallel }<1$, and then $x=1-(1-x)$ is invertible by [L3], contradicting [step 1.1] and the properness of $M$. Hence $\bar{M}$ is a proper ideal containing $M$, and maximality forces $\bar{M}=M$.

The quotient $A\mathrm{/}M$ is a nonzero commutative unital complex Banach algebra by [L4], using ${\mathrm{A}\mathrm{C}}_{\omega }$ from [L7], and it is a field: if $a\mathrm{\notin }M$, the ideal $M+Aa$ strictly contains $M$ and hence equals $A$ by the maximality of $M$ in [step 2.1], so there are $m\in M$ and $b\in A$ with $m+ba=1$, and then $(b+M)(a+M)=1+M$ exhibits $a+M$ as invertible.

Consequently $A\mathrm{/}M$ is a unital complex Banach division algebra, and [L5] provides an isometric algebra isomorphism $A\mathrm{/}M\cong \mathbb{C}$; write $\pi :A\to A\mathrm{/}M$ for the quotient map, a unital algebra homomorphism of norm one.

The composite of $\pi$ with the isomorphism of [step 4.1] is a nonzero complex-linear multiplicative map $\chi :A\to \mathbb{C}$ with $\chi (1)=1$, that is, a character, and its kernel is exactly $M$; thus every proper ideal is contained in a closed maximal ideal which is the kernel of a character.

By [step 5.1] every proper ideal lies in a closed maximal ideal, proving claim 1; by [step 5.1] every maximal ideal containing a proper ideal is a character kernel, by [step 1.4] every character kernel is a maximal ideal, and by [step 1.5] the assignment is injective, so the maximal ideals are exactly the character kernels and $\chi \mapsto \ker \chi$ is a bijection, proving claim 2.