Gelfand-Mazur theorem

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

Proof : Let $e$ denote the unit of $𝒜$.

Let $x\in 𝒜$ and $\sigma (x)$ be its spectrum. It is known that the spectrum is a non-empty set (http://planetmath.org/SpectrumIsANonEmptyCompactSet) in $ℂ$.

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

Let $\varphi :ℂ⟶𝒜$ be defined by $\varphi (\lambda )=\lambda e$.

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

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

It is isometric because $\parallel \lambda e\parallel =|\lambda |\parallel e\parallel =|\lambda |$

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

Title	Gelfand-Mazur theorem
Canonical name	GelfandMazurTheorem
Date of creation	2013-03-22 17:29:03
Last modified on	2013-03-22 17:29:03
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	7
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 46H05