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Gelfand-Mazur

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $A$ be a unital complex Banach algebra (Unital Banach algebra) which is a division algebra: every nonzero element of $A$ is invertible (Invertible element and general linear group of a Banach algebra). Then the map

$C \to A, λ \mapsto λ 1,$

is an isomorphism of complex algebras and an isometry, and consequently $A = C 1$ and $\dim_{C} A = 1$ .

Facts & Assumptions

Given: An assumed Axiom of Choice, a unital complex Banach algebra $A$ in which every nonzero element is invertible, and the map $φ : C \to A$ , $φ (λ) = λ 1$ .

[L1]

$A$ is a complex vector space with associative bilinear multiplication, $1 a = a 1 = a$ for all $a$ , $∥ 1 ∥ = 1$ and $∥ x y ∥ \leq ∥ x ∥ ∥ y ∥$ ; in particular $1 \neq 0$ (Unital Banach algebra).

[L2]

$a \in A$ is invertible exactly when some $b$ satisfies $a b = b a = 1$ ; the only non-invertible element of a division algebra is $0$ (Invertible element and general linear group of a Banach algebra).

[L3]

$λ \in σ_{A} (a)$ exactly when $λ 1 - a$ is not invertible (Spectrum and resolvent set in a Banach algebra).

[L4]

Under the Axiom of Choice every element of a nonzero unital complex Banach algebra has nonempty spectrum (Spectrum is nonempty compact and norm bounded).

Proof

technique · direct

1.1

The map $φ$ is complex-linear and multiplicative: $φ (0) = 0$ and $φ (1) = 1$ ; $φ (λ + μ) = (λ + μ) 1 = λ 1 + μ 1$ ; $φ (λ μ) = (λ μ) 1 = (λ 1) (μ 1) = φ (λ) φ (μ)$ , using bilinearity and $1 \cdot 1 = 1$ ; and $φ$ is injective, because $λ 1 = 0$ with $λ \neq 0$ would give $1 = λ^{- 1} λ 1 = 0$ , contradicting [L1].

L1algebra

1.2

For every $a \in A$ the spectrum $σ_{A} (a)$ is nonempty by [L4].

2.1

Fix $a \in A$ and pick $λ \in σ_{A} (a)$ by [step 1.2]; then $λ 1 - a$ is not invertible by [L3], so $λ 1 - a = 0$ by the division-algebra hypothesis [L2]; hence $a = λ 1 = φ (λ)$ lies in the image of $φ$ .

step 1.2L2L3

2.2

The isomorphism is isometric: $∥ φ (λ) ∥ = ∥ λ 1 ∥ = ∣ λ ∣ ∥ 1 ∥ = ∣ λ ∣$ by [L1].

step 1.1L1algebra

3.1

Since $a$ was arbitrary, $φ$ is surjective, and by [step 1.1] it is an injective complex-algebra homomorphism; hence it is a complex-algebra isomorphism $C \to A$ and $A = C 1$ .

step 1.1step 2.1algebra

4.1

The statements of the theorem are proved: $φ$ is an algebra isomorphism by [step 3.1] and an isometry by [step 2.2], so a complex unital Banach division algebra is one-dimensional over $C$ .

step 2.2step 3.1∎

Remarks

The Axiom of Choice enters only through the nonemptiness of the spectrum. If one is willing to assume that the spectrum of every element is nonempty, the argument above is choice-free; conversely the theorem is the standard quantitative form of the fact that one-point spectra force division algebras to be scalars.
"Division algebra" cannot be weakened to "no zero divisors". The disc algebra $A (\overline{D})$ is a unital commutative complex Banach algebra without zero divisors: a product of two functions whose product vanishes on the connected disc $D$ vanishes identically by analytic continuation, so one factor is zero. It is nevertheless not a division algebra, because the coordinate function $z$ is nonzero while $σ_{A (\overline{D})} (z) = \overline{D}$ (cex-spectrum-can-shrink-in-a-larger-banach-algebra, ex-maximal-ideal-space-of-the-disc-algebra). The boundary argument for the spectrum produces only a topological zero divisor, that is, an element $a$ admitting unit vectors $b_{n}$ with $a b_{n} \to 0$ or $b_{n} a \to 0$ ; topological zero divisors need not be algebraic ones, as $1 - z$ in the disc algebra shows, so the two notions must not be conflated. The correct replacement of "no zero divisors" is "no nonzero topological zero divisors": every element of the boundary of the invertible group is a topological zero divisor, so a unital complex Banach algebra in which no nonzero element is a topological zero divisor is a division algebra.
Use in the Gelfand theory. This is the step that identifies the quotient $A / m$ of a commutative unital Banach algebra by a maximal ideal with $C$ , making characters and maximal ideals correspond; the following page of this track uses the theorem in exactly that form.
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.

Depends on

Used by

Maximal ideals and characters of a commutative Banach algebra Theorem

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Bühler and Dietmar A. Salamon, Functional Analysis — Theorem 5.20 and §5.1.1, printed pp. 209–214 and 222–223 (standard reference, not scraped)
Vahid Shirbisheh, Lectures on C-star Algebras, v2 — §2.1, printed pp. 19–24 (standard reference, not scraped)