---
tag: 3D6XH41A3B
kind: statement
role: theorem
name: Gelfand–Mazur theorem
msc: [46H05, 12J05]
uses: [09FTAET5Z4, 0WZXK05ME7]
about: [ZFW9Q3WTAQ]
sources:
  - url: https://en.wikipedia.org/wiki/Gelfand%E2%80%93Mazur_theorem
same_as:
  wikidata: Q2226677
  wikipedia: Gelfand–Mazur theorem
---

## Statement

Let $A$ be a complex Banach algebra with unit $1 \ne 0$.
Suppose every nonzero element of $A$ is invertible, so that $A$ is a [[division ring|0WZXK05ME7]].
Then every $a \in A$ equals $\lambda 1$ for a unique $\lambda \in \C$, and the map $\C \to A$, $\lambda \mapsto \lambda 1$, is an isomorphism of algebras.

## Remarks

Proof idea: the spectrum $\set{\lambda \in \C : a - \lambda 1 \text{ is not invertible}}$ of each $a \in A$ is nonempty.
Suppose it were empty, and let $\varphi$ be a continuous linear functional on $A$.
Then $\lambda \mapsto \varphi((a - \lambda 1)^{-1})$ would be a bounded entire function that tends to $0$, hence zero by Liouville's theorem.
As this holds for every $\varphi$, the Hahn–Banach theorem would give $(a - \lambda 1)^{-1} = 0$, a contradiction.
So some $a - \lambda 1$ is not invertible; in a division algebra this forces $a - \lambda 1 = 0$.
The complex scalars are needed.
The quaternions $\mathbb{H}$ with the Euclidean norm on $\R^4$ form a real Banach algebra in which every nonzero element is invertible, but $\mathbb{H}$ is not $\R$ or $\C$.
Consequence: a field that contains $\C$ and has an absolute value extending the usual one on $\C$, complete for it, is $\C$ itself.

## Explanations

### L1 human

A Banach algebra is a space where one can add, multiply, scale by complex numbers and measure size, and where limits exist.
The theorem says that if such an algebra lets you divide by every nonzero element, it is nothing more than the complex numbers.
So there is no room for a bigger "complex number system" with a good notion of size.

### L2 human

The theorem is a basic step in the theory of commutative Banach algebras.
If $A$ is a commutative unital complex Banach algebra and $\mathfrak{m}$ a maximal ideal, then $\mathfrak{m}$ is closed and $A/\mathfrak{m}$ is a Banach division algebra, hence $\C$.
This gives the characters $A \to \C$ that underlie the Gelfand transform.
Some proofs of Ostrowski's theorem, that a field complete for an Archimedean absolute value is $\R$ or $\C$, use the Gelfand–Mazur theorem.