Gelfand–Mazur theorem

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.

Going deeper

For a student

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.