Theorem 7.3.8 (Gelfand-Mazur).label Let $A$ be a Banach algebra and a division ring then $A\iso \com$ isometrically.
Proof. Let $x\in A$ then $A$ is a division ring so $xx\inv$ is the multiplicative unit. By Theorem 7.3.7, pick $\lam\in \spec{A}{x}$. Observe $x-\lam$ is not invertible so $x=\lam 1$ by assumption that $A$ is a division ring. The scalar $\lam$ is unique so the map
\begin{align*}\Phi:\com\to A&&z\mto z1\end{align*}
is bijective. Furthermore $\Phi$ is linear, multiplicative by construction and by convention Definition 7.1.5 it preserves the norm. Thus $\Phi$ is an isometric isomorphism.$\square$
Post a Comment