TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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Spectrum as character values

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $A$ be a nonzero commutative unital complex Banach algebra (Unital Banach algebra), let $a \in A$ , and let $σ_{A} (a)$ be the spectrum of $a$ in $A$ (Spectrum and resolvent set in a Banach algebra), with spectral radius $r (a)$ (Spectral radius). Then

$σ_{A} (a) = {χ (a) : χ \in Δ (A)},$

and consequently every character $χ \in Δ (A)$ satisfies $∣ χ (a) ∣ \leq r (a) \leq ∥ a ∥$ for every $a \in A$ .

The spectrum is taken in the ambient algebra $A$ ; the statement is not a claim about the spectrum computed in a subalgebra, and no injectivity or surjectivity of the Gelfand transform is asserted.

Facts & Assumptions

Given: An assumed Axiom of Choice, a nonzero commutative unital complex Banach algebra $A$ , and an element $a \in A$ .

[L1]

Every proper ideal of $A$ is contained in a maximal ideal, and the maximal ideals of $A$ are exactly the kernels of the characters of $A$ (Maximal ideals and characters of a commutative Banach algebra, The Axiom of Choice).

[L2]

$z \in σ_{A} (a)$ exactly when $z 1 - a$ is not invertible in $A$ ; an element of a proper ideal is never invertible (Spectrum and resolvent set in a Banach algebra).

[L3]

Every character of $A$ is unital, so $χ (1) = 1$ , and satisfies $∣ χ (x) ∣ \leq ∥ x ∥$ for all $x$ (Characters on a unital Banach algebra are continuous).

[L4]

The spectral radius is $r (a) = \max {∣ z ∣ : z \in σ_{A} (a)}$ and satisfies $0 \leq r (a) \leq ∥ a ∥$ (Spectral radius, The Axiom of Choice).

Proof

technique · direct

1.1L1L2L3

For $χ \in Δ (A)$ the element $a - χ (a) 1$ lies in $\ker χ$ , because $χ (a - χ (a) 1) = χ (a) - χ (a) χ (1) = 0$ by linearity and [L3]; as $\ker χ$ is a proper ideal, $a - χ (a) 1$ is not invertible, so $χ (a) \in σ_{A} (a)$ by [L2].

1.2L2

If $λ \in σ_{A} (a)$ then the ideal $I : = (a - λ 1) A$ generated by $a - λ 1$ is proper: were $I = A$ , there would be $b \in A$ with $b (a - λ 1) = 1$ , making $a - λ 1$ invertible, contrary to [L2].

2.1step 1.2L1L3

By [L1] the proper ideal $I$ of [step 1.2] is contained in a maximal ideal $M$ , and $M = \ker χ$ for some character $χ$ ; since $a - λ 1 \in I \subseteq M$ we get $χ (a - λ 1) = 0$ , that is, $χ (a) = λ$ by linearity and [L3].

3.1step 1.1step 2.1L4∎

Steps [step 1.1] and [step 2.1] give $σ_{A} (a) = {χ (a) : χ \in Δ (A)}$ . For each character, $χ (a) \in σ_{A} (a)$ , so $∣ χ (a) ∣ \leq \max {∣ z ∣ : z \in σ_{A} (a)} = r (a)$ by [L4], and $r (a) \leq ∥ a ∥$ by [L4]; the displayed consequence follows.

Remarks

AC is spent once, in the maximal-ideal extension. Both inclusions are otherwise algebraic: the forward inclusion only tests the character on a coset representative, and the reverse inclusion only extends an ideal.
Consequences for the Gelfand transform. Since $\hat{a} (χ) = χ (a)$ , the equality of the statement says that the range of $\hat{a}$ is exactly $σ_{A} (a)$ , which is how the norm formula $∥ \hat{a} ∥_{\infty} = r (a)$ is proved in Gelfand transform is a contractive unital homomorphism.
No isometry claim. The inequality chain $∣ χ (a) ∣ \leq r (a) \leq ∥ a ∥$ is all that the spectrum identity yields; for a general commutative Banach algebra the first inequality can be strict, as cex-gelfand-transform-of-a-banach-algebra-need-not-be-isometric records.
Reading order. The example items named by ID above are homed on later pages of the plan, so they are named rather than hyperlinked: a body link to later material must be declared as a forward reference, and Step-5b closure removes every such declaration. Rehoming those items to an earlier page (an owner-only reading-order change) would make the citations backward and restore the links.

Spectrum as character values

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources