# Gelfand-Mazur 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 (http://planetmath.org/SpectrumIsANonEmptyCompactSet) 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 $\|\lambda e\|=|\lambda|\|e\|=|\lambda|$

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

Title Gelfand-Mazur theorem GelfandMazurTheorem 2013-03-22 17:29:03 2013-03-22 17:29:03 asteroid (17536) asteroid (17536) 7 asteroid (17536) Theorem msc 46H05