[proofplan] We verify three things: well-definedness (every $\ker \varphi$ is a maximal ideal), injectivity (distinct characters have distinct kernels), and surjectivity (every maximal ideal is the kernel of a character). Well-definedness uses that $\ker \varphi$ is an ideal of codimension 1, hence maximal among proper ideals in any unital ring. Injectivity uses that for any character $\varphi$ and any $x \in A$, the element $x - \varphi(x) \cdot 1$ lies in $\ker \varphi$ — hence two characters with the same kernel agree pointwise. Surjectivity is the deepest part: given a maximal ideal $M$, the quotient $A/M$ is a unital commutative Banach algebra (after verifying $M$ is closed) and a field (since $M$ is maximal in the commutative case); the **Gelfand–Mazur Theorem** then identifies $A/M$ isometrically with $\mathbb{C}$, and the quotient map becomes a character with kernel $M$. [/proofplan]

[step:Show every proper ideal has proper closure, hence maximal ideals are closed] Let $J \subseteq A$ be a proper ideal. We show $\overline{J}$ is also a proper ideal. First, $\overline{J}$ is an ideal: closure of a subspace is a subspace; closure under multiplication by elements of $A$ uses continuity of multiplication. Specifically, for $u \in \overline{J}$ and $a \in A$, choose $u_n \in J$ with $u_n \to u$; then $a u_n \in J$ and $a u_n \to a u$ in $A$ by submultiplicativity ($\|a u_n - a u\| \le \|a\| \|u_n - u\| \to 0$), so $au \in \overline{J}$. Similarly $ua \in \overline{J}$. Second, $\overline{J}$ is proper: by [Properties of Invertible Elements](/theorems/2668) part (i), $G(A)$ is open in $A$. Since $J$ is proper, $1 \notin J$ (else $J$ would contain every element via $a = a \cdot 1 \in J$ for $a \in A$). More strongly, $J \cap G(A) = \varnothing$: if $u \in J \cap G(A)$, then $1 = u^{-1} u \in J$ (using that $J$ is an ideal), contradicting properness. Since $G(A)$ is open and $J \cap G(A) = \varnothing$, the closure satisfies $\overline{J} \cap G(A) = \varnothing$ as well (the open complement $A \setminus G(A)$ contains $J$, hence its closure). In particular $1 \notin \overline{J}$, so $\overline{J} \ne A$. Now if $M \in \mathfrak{M}_A$ is a maximal ideal, $M$ is proper, so $\overline{M}$ is a proper ideal by the above. Maximality forces $\overline{M} = M$, so $M$ is closed. [/step]

[step:Verify well-definedness: $\ker \varphi$ is a maximal ideal] Let $\varphi \in \Phi_A$. Then $\varphi : A \to \mathbb{C}$ is a non-zero algebra homomorphism, hence a non-zero linear map onto a one-dimensional vector space. Therefore $\varphi$ is surjective onto $\mathbb{C}$, and its kernel $\ker \varphi$ is a linear subspace of codimension $1$. $\ker \varphi$ is an ideal: for $u \in \ker \varphi$ and $a \in A$, \begin{align*} \varphi(au) = \varphi(a) \varphi(u) = \varphi(a) \cdot 0 = 0, \end{align*} so $au \in \ker \varphi$ (and $ua \in \ker \varphi$ by commutativity of $A$). $\ker \varphi$ is proper since $\varphi(1) = 1 \ne 0$ (a non-zero character preserves the unit, as in the proof of [Characters are Continuous](/theorems/2675), Step 2), so $1 \notin \ker \varphi$. $\ker \varphi$ is maximal: any ideal $I$ properly containing $\ker \varphi$ contains some $a$ with $\varphi(a) \ne 0$; then $1 - a/\varphi(a) \in \ker \varphi \subseteq I$ and $a/\varphi(a) \in I$ (since $a \in I$ and $I$ is closed under scalar multiplication), so $1 = (1 - a/\varphi(a)) + a/\varphi(a) \in I$, forcing $I = A$. Hence $\ker \varphi \in \mathfrak{M}_A$. [/step]

[step:Verify injectivity of the map $\ker$] Let $\varphi, \psi \in \Phi_A$ with $\ker \varphi = \ker \psi$. For any $x \in A$, set \begin{align*} u := x - \varphi(x) \cdot 1. \end{align*} Then $\varphi(u) = \varphi(x) - \varphi(x) \varphi(1) = \varphi(x) - \varphi(x) \cdot 1 = 0$ (using $\varphi(1) = 1$), so $u \in \ker \varphi = \ker \psi$, giving \begin{align*} 0 = \psi(u) = \psi(x) - \varphi(x) \psi(1) = \psi(x) - \varphi(x). \end{align*} Hence $\psi(x) = \varphi(x)$ for all $x \in A$, so $\psi = \varphi$. [/step]

[step:Verify surjectivity using Gelfand–Mazur]Let $M \in \mathfrak{M}_A$. By Step 1, $M$ is a closed proper ideal. We show $M$ is the kernel of some character. [claim:The quotient $A / M$ is a unital commutative Banach algebra] Since $M$ is a closed proper subspace of the Banach space $A$, the quotient $A/M$ is a Banach space under the quotient norm $\|x + M\| := \inf\{\|x - u\| : u \in M\}$, with quotient map $q : A \to A/M$, $x \mapsto x + M$, continuous of operator norm $\|q\|_{\mathcal{L}(A, A/M)} \le 1$. The multiplication on $A/M$ is well-defined: for $x, y \in A$ and $u, v \in M$, \begin{align*} (x + u)(y + v) - xy = xv + uy + uv \in M \end{align*} since $M$ is an ideal. Hence $(x + M)(y + M) := xy + M$ is independent of representatives. Submultiplicativity of the quotient norm: for any $\varepsilon > 0$ choose $u, v \in M$ with $\|x + u\| \le \|x + M\| + \varepsilon$, $\|y + v\| \le \|y + M\| + \varepsilon$. Then \begin{align*} \|xy + M\| \le \|(x + u)(y + v)\| \le \|x + u\| \|y + v\| \le (\|x + M\| + \varepsilon)(\|y + M\| + \varepsilon), \end{align*} and letting $\varepsilon \downarrow 0$ gives $\|xy + M\| \le \|x + M\| \|y + M\|$. The unit $1 + M$ is non-zero (since $1 \notin M$ as $M$ is proper) and satisfies $(1 + M)(x + M) = x + M$. Commutativity of $A/M$ follows from commutativity of $A$. [/claim] [proof] Self-contained above; uses only completeness of the quotient of a Banach space by a closed subspace (a standard Banach-space fact: the quotient norm makes $A/M$ a Banach space when $M$ is closed) and the ideal property to define multiplication. [/proof] [claim:$A / M$ is a field] We show every non-zero element of $A/M$ is invertible. Let $\bar{x} = x + M \ne 0$ in $A/M$, i.e.\ $x \notin M$. Consider the ideal $J := M + Ax = \{u + ax : u \in M, a \in A\}$ of $A$. This is a sum of an ideal $M$ and a left-ideal (which equals a two-sided ideal in the commutative case) $Ax$, hence is itself an ideal. Since $x = 0 + 1 \cdot x \in J$ and $x \notin M$, $J$ properly contains $M$. By maximality of $M$, $J = A$. Hence there exist $u \in M$ and $a \in A$ with $1 = u + ax$. Passing to the quotient, \begin{align*} 1 + M = (u + ax) + M = ax + M = (a + M)(x + M), \end{align*} so $\bar{a} = a + M$ is the inverse of $\bar{x}$ in $A/M$. By commutativity, $\bar{a}$ is also a left inverse, so $\bar{x} \in G(A/M)$. Since every non-zero element of $A/M$ is invertible and $A/M$ is a unital commutative ring, $A/M$ is a field. [/claim] [proof] The argument is the standard ring-theoretic fact: in a commutative unital ring, the quotient by a maximal ideal is a field. We have made it explicit using only the fact that $M + Ax \supsetneq M$ (since $x \notin M$) is an ideal, hence equals $A$ by maximality, producing the inverse. [/proof] By the two claims, $A/M$ is a unital commutative complex Banach algebra in which every non-zero element is invertible, i.e.\ a complex unital normed division algebra. Apply the [Gelfand–Mazur Theorem](/theorems/2670) (whose hypothesis is exactly: complex unital normed division algebra), to obtain an isometric isomorphism \begin{align*} \theta : \mathbb{C} \to A/M, \qquad \theta(\lambda) = \lambda \cdot 1_{A/M}, \qquad \|\theta(\lambda)\|_{A/M} = |\lambda|. \end{align*} Its inverse $\theta^{-1} : A/M \to \mathbb{C}$ is also an isometric isomorphism of unital algebras. Define \begin{align*} \varphi : A &\to \mathbb{C} \\ x &\mapsto \theta^{-1}(x + M) = \theta^{-1}(q(x)). \end{align*} That is, $\varphi = \theta^{-1} \circ q$. Since $q : A \to A/M$ is a unital algebra homomorphism (linear and multiplicative, $q(1) = 1 + M$) and $\theta^{-1}$ is a unital algebra isomorphism (linear, multiplicative, $\theta^{-1}(1 + M) = 1$), the composition $\varphi$ is a unital algebra homomorphism $A \to \mathbb{C}$. Moreover $\varphi(1) = 1 \ne 0$, so $\varphi$ is non-zero and hence $\varphi \in \Phi_A$. Finally, \begin{align*} \ker(\varphi) = \ker(\theta^{-1} \circ q) = q^{-1}(\ker(\theta^{-1})) = q^{-1}(\{0_{A/M}\}) = M, \end{align*} where the second equality uses that $\theta^{-1}$ is a bijection (so $\theta^{-1}(z) = 0$ iff $z = 0$), and $q^{-1}(\{0\}) = M$ by definition of the quotient map. Hence $\ker(\varphi) = M$, proving surjectivity of the map $\ker : \Phi_A \to \mathfrak{M}_A$.[/step]

[guided]The structure of this step is **quotient by a maximal ideal → field → $\mathbb{C}$ via Gelfand–Mazur → character**. Let us trace each link. **Why is $A/M$ a Banach algebra?** Three pieces have to fit together: (a) the quotient norm $\|x + M\|$ is well-defined and complete because $M$ is *closed* (Step 1), (b) multiplication is well-defined on cosets because $M$ is an *ideal*, (c) submultiplicativity $\|\bar{x}\bar{y}\| \le \|\bar{x}\|\|\bar{y}\|$ comes from picking near-optimal representatives. **Why is $A/M$ a field?** Because $M$ is *maximal*. Concretely, if $\bar{x} \ne 0$ in $A/M$, the ideal $M + Ax$ in $A$ properly contains $M$ (it contains $x \notin M$), so by maximality $M + Ax = A$. Writing $1 = u + ax$ with $u \in M$ shows $\bar{a}\bar{x} = 1$ in $A/M$. So every non-zero element is invertible. **Why does Gelfand–Mazur apply?** Theorem 2670 says: every complex unital normed division algebra is isometrically isomorphic to $\mathbb{C}$. We have just shown $A/M$ is such a thing — a complex (because $A$ is) commutative (because $A$ is) unital Banach algebra (verified) which is a field (verified, in particular every non-zero element is invertible). The norm structure comes from completeness; "complex" because $A$ is over $\mathbb{C}$. So $A/M \cong \mathbb{C}$ isometrically. **Why is the composition a character?** Composing the surjective unital algebra homomorphism $q : A \to A/M$ with the isomorphism $\theta^{-1} : A/M \to \mathbb{C}$ produces a unital algebra homomorphism $\varphi : A \to \mathbb{C}$. It is non-zero since $\varphi(1) = 1$. **Why does $\ker \varphi = M$?** A general fact: $\ker(g \circ f) = f^{-1}(\ker g)$ for any composition; here $\ker(\theta^{-1}) = \{0\}$ (since $\theta^{-1}$ is injective), and $q^{-1}(\{0_{A/M}\}) = M$ by definition of quotient. A subtle point worth emphasising: the role of *closedness of $M$* (Step 1). Without closedness, $A/M$ would not be a Banach space — completeness would fail. And without completeness, Gelfand–Mazur would not apply (its proof uses Liouville for $A$-valued functions, which needs $A$ complete). The fact that **maximal ideals are automatically closed in a unital BA** is exactly what makes the bijection $\Phi_A \leftrightarrow \mathfrak{M}_A$ possible.[/guided]

[step:Combine the three properties] By Step 2, $\ker$ maps $\Phi_A$ into $\mathfrak{M}_A$. By Step 3, $\ker$ is injective. By Step 4, $\ker$ is surjective. Hence \begin{align*} \ker : \Phi_A \to \mathfrak{M}_A \end{align*} is a bijection. [/step]