Proposition 6.2.8 (Division Banach Algebra).label Let $A$ be a Banach algebra and a division ring then $A\iso \com$
Proof. Let $x\in A$ then by Theorem 6.2.7, let $\lam\in \spec{A}{x}$. By definition $x-\lam$ is not invertible so $x=\lam$ by assumption.$\square$