$B$ is a Banach algebra which is a division algebra, then there is a unique isometric isomorphism of $B$ on to $\mathbb{C}$

Question

if $B$ is a Banach algeba which is a division algebra, then there is a unique isometric isomorphism of $B$ on to $\mathbb{C}$

I know that each element$\ x$ of $B$ is type $\lambda_x.e$ for unique scalar $\lambda_x \in \mathbb{C}$

then map $$\psi:B \to \mathbb{C}$$ such that $\psi(x)=\lambda_x$ is isometric isomorphism.

also $\sigma(x)=\{\lambda_x\}$. But how to prove uniqueness.

Hint is given that if $\psi '$ is other isometric isomorphism , then $\psi '(x)\in \sigma(x)$.

But I have problem in proving this hint. Any suggestion?

Answer 1

Consider $a:=x-\psi'(x)e$. As an algebra isomorphism, $\psi'$ is $\Bbb C$-linear, injective, and maps $e$ to $1$.
Then $\psi'(a) =\psi'(x) -\psi'(x)\cdot1=0$, thus $a= 0$ by injectivity, which is not invertible, so indeed $\psi'(x) \in\sigma(x) $.

Answer 2

In fact a stronger uniqueness statement is totally trivial: If $\phi_1$ and $\phi_2$ are isomorphisms (isometric or not) from a Banach algebra onto $\Bbb C$ then $\phi_1=\phi_2$:

Let $\psi=\phi_1\circ\phi_2^{-1}$. Then $\psi$ is a Banach-algebra automorphism of $\Bbb C$. Since $\psi$ is linear, $\psi(z)=az$. Since $\psi$ is injective, $a\ne0$; now $\psi(zw)=\psi(z)\psi(w)$ shows that $a=1$.