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Gelfand-Mazur theorem (Theorem)

Theorem - Let $ \mathcal{A}$ be a unital Banach algebra over $ \mathbb{C}$ that is also a division algebra (i.e. every non-zero element is invertible). Then $ \mathcal{A}$ is isometrically isomorphic to $ \mathbb{C}$.

Proof : Let $ e$ denote the unit of $ \mathcal{A}$.

Let $ x \in \mathcal{A}$ and $ \sigma(x)$ be its spectrum. It is known that the spectrum is a non-empty set in $ \mathbb{C}$.

Let $ \lambda \in \sigma(x)$. Since $ x-\lambda e$ is not invertible and $ \mathcal{A}$ is a division algebra, we must have $ x-\lambda e = 0$ and so $ x=\lambda e$

Let $ \phi : \mathbb{C} \longrightarrow \mathcal{A}$ be defined by $ \phi(\lambda)=\lambda e$.

It is clear that $ \phi$ is an injective algebra homomorphism.

By the above discussion, $ \phi$ is also surjective.

It is isometric because $ \Vert\lambda e\Vert = \vert\lambda\vert \Vert e\Vert = \vert\lambda\vert$

Therefore, $ \mathcal{A}$ is isometrically isomorphic to $ \mathbb{C}$. $ \square$



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Cross-references: isometric, surjective, homomorphism, algebra, injective, clear, spectrum, unit, proof, isometrically isomorphic, invertible, division algebra, Banach algebra, unital
There is 1 reference to this entry.

This is version 4 of Gelfand-Mazur theorem, born on 2007-08-17, modified 2008-05-01.
Object id is 9871, canonical name is GelfandMazurTheorem.
Accessed 696 times total.

Classification:
AMS MSC: 46H05 (Functional analysis :: Topological algebras, normed rings and algebras, Banach algebras :: General theory of topological algebras)
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