[proofplan]
The three statements are proved in cascade. Statement (i) is the heart: invertible elements have non-zero image under every character (multiplicativity), and conversely, a non-invertible element generates a proper principal ideal which by Zorn extends to a maximal ideal — and by the bijection between maximal ideals and characters (theorem 2676), this maximal ideal is the kernel of some character that vanishes on $x$. Statement (ii) follows by applying (i) to $\lambda \cdot 1 - x$ for each $\lambda \in \mathbb{C}$. Statement (iii) follows from (ii) and the definition of spectral radius as the supremum of $|\lambda|$ over $\lambda \in \sigma_A(x)$.
[/proofplan]
[step:Prove $(\Rightarrow)$ in (i): invertible elements have non-zero character images]
Suppose $x \in G(A)$ with inverse $y \in A$, i.e.\ $xy = yx = 1$. Let $\varphi \in \Phi_A$. Since $\varphi$ is a unital algebra homomorphism, $\varphi(1) = 1$ and $\varphi$ is multiplicative, so
\begin{align*}
\varphi(x)\varphi(y) = \varphi(xy) = \varphi(1) = 1.
\end{align*}
Hence $\varphi(x) \neq 0$ (and $\varphi(y) = \varphi(x)^{-1}$). This holds for every $\varphi \in \Phi_A$.
[/step]
[step:Prove $(\Leftarrow)$ in (i) via maximal ideals]
Suppose $x \notin G(A)$. We construct $\varphi \in \Phi_A$ with $\varphi(x) = 0$.
Consider the principal ideal
\begin{align*}
J := Ax = \{ax : a \in A\} \subseteq A.
\end{align*}
This is an ideal of $A$ (it is a subspace, and for any $b \in A$, $b \cdot (ax) = (ba)x \in J$, with right-multiplication symmetric by commutativity).
We claim $J$ is proper. If $J = A$, then $1 \in J$, so there exists $a \in A$ with $ax = 1$. By commutativity, $xa = ax = 1$, so $a$ is a two-sided inverse of $x$, contradicting $x \notin G(A)$. Hence $J$ is proper, i.e.\ $J \subsetneq A$.
Apply Zorn's Lemma to the partially ordered set $\mathcal{F}$ of proper ideals of $A$ containing $J$, ordered by inclusion. This is non-empty (contains $J$). Every chain $\{I_\alpha\}$ in $\mathcal{F}$ has the upper bound $\bigcup_\alpha I_\alpha$, which is again an ideal (the union of a chain of ideals); it is proper because $1 \notin I_\alpha$ for any $\alpha$ (else $I_\alpha = A$), so $1 \notin \bigcup_\alpha I_\alpha$. By Zorn, $\mathcal{F}$ contains a maximal element $M$, which is a maximal ideal of $A$ containing $J$.
By the [Maximal Ideals are Kernels of Characters](/theorems/2676) bijection, $M = \ker(\varphi)$ for some $\varphi \in \Phi_A$. Since $x = 1 \cdot x \in Ax = J \subseteq M = \ker(\varphi)$, we conclude $\varphi(x) = 0$.
Combining with the previous step, $x \in G(A) \iff \varphi(x) \neq 0$ for all $\varphi \in \Phi_A$, proving (i).
[guided]
We want to manufacture a character vanishing on $x$ when $x$ is not invertible. The idea is to find a maximal ideal containing $x$ and use the bijection with characters.
**Why does $x$ generate a proper ideal?** The principal ideal $Ax = \{ax : a \in A\}$ contains $x$ (take $a = 1$). It is the smallest ideal of $A$ containing $x$. Could it be all of $A$? Only if $1 \in Ax$, i.e.\ $1 = ax$ for some $a$. But then $a$ would be an inverse of $x$ (using commutativity), contradicting $x \notin G(A)$. So $Ax$ is proper.
**Why is there a maximal ideal containing $Ax$?** This is the standard Zorn's Lemma argument. Consider proper ideals containing $Ax$, ordered by inclusion. The union of any chain of proper ideals is a proper ideal (the key point: $1$ is in none of them, so $1$ is not in the union, so the union is proper). Zorn gives a maximal element $M$, which is by construction a maximal ideal containing $Ax$, hence containing $x$.
**Why does this give us a character vanishing on $x$?** By [theorem 2676](/theorems/2676), the map $\varphi \mapsto \ker \varphi$ is a bijection between characters and maximal ideals. So $M = \ker \varphi$ for a unique $\varphi \in \Phi_A$. Since $x \in M = \ker \varphi$, $\varphi(x) = 0$.
The contrapositive form of (i) is what we have proved here: $x \notin G(A) \iff \exists \varphi \in \Phi_A$ with $\varphi(x) = 0$. Combined with the forward direction (Step 1), this gives (i).
[/guided]
[/step]
[step:Deduce (ii) by applying (i) to $\lambda \cdot 1 - x$]
By definition,
\begin{align*}
\sigma_A(x) = \{\lambda \in \mathbb{C} : \lambda \cdot 1 - x \notin G(A)\}.
\end{align*}
Applying (i) to $\lambda \cdot 1 - x \in A$ gives
\begin{align*}
\lambda \cdot 1 - x \notin G(A) \iff \exists\, \varphi \in \Phi_A : \varphi(\lambda \cdot 1 - x) = 0.
\end{align*}
For any $\varphi \in \Phi_A$, by linearity and the unital property $\varphi(1) = 1$,
\begin{align*}
\varphi(\lambda \cdot 1 - x) = \lambda \varphi(1) - \varphi(x) = \lambda - \varphi(x).
\end{align*}
Hence $\varphi(\lambda \cdot 1 - x) = 0 \iff \lambda = \varphi(x)$. Combining,
\begin{align*}
\lambda \in \sigma_A(x) \iff \exists\, \varphi \in \Phi_A : \lambda = \varphi(x) \iff \lambda \in \{\varphi(x) : \varphi \in \Phi_A\}.
\end{align*}
This proves (ii): $\sigma_A(x) = \{\varphi(x) : \varphi \in \Phi_A\}$.
[/step]
[step:Deduce (iii) from (ii) by taking suprema of moduli]
The spectral radius is defined as
\begin{align*}
r_A(x) = \sup\{|\lambda| : \lambda \in \sigma_A(x)\}.
\end{align*}
Substituting the description of $\sigma_A(x)$ from (ii):
\begin{align*}
r_A(x) = \sup\{|\lambda| : \lambda \in \{\varphi(x) : \varphi \in \Phi_A\}\} = \sup\{|\varphi(x)| : \varphi \in \Phi_A\}.
\end{align*}
This proves (iii). The supremum is attained when $\Phi_A \ne \varnothing$ and $\sigma_A(x)$ is compact (theorem 2669), but we do not need attainment for the formula.
[/step]
