Spectrum of a Banach algebra homeomorphic to the spectrum of one of its elements

Question

In G. B. Folland - A Course in Abstract Harmonic Analysis we can read the following

" (1.15) Proposition. Let be $A$ a (complex) commutative unital Banach algebra with unit $e$, let $x_0 \in A$ and suppose that one of the following holds:

(i) $A$ is generated by $x_0$ and $e$.

(ii) $x_0$ is invertible and $A$ is generated by $x_0$ and $x_0^{-1}$.

(iii) $A$ is a simmetric $*$-algebra and $A$ is generated by $x_0$, $x_0^*$ and $e$.

Then $\hat{x_0}$ (the Gelfand trasform of $x_0$) is a homeomorphism from the spectrum $\sigma(A)$ of $A$ to the spectrum $\sigma(x_0)$ of $x_0$. "

The proof is elementary and the main motivation to state it is to explain the relationship between the spectrum of the algebra and the spectrum of an element.

My question is: there exists a converse of Prop. 1.15? That is, if $A$ is a commutative unital Banach algebra with $x_0 \in A$ such that $\hat x_0 : \sigma(A) \to \sigma(x_0)$ is a homeomorphism, then is it true that one of (i),(ii),(iii) (and maybe some other similar cases) holds?

So I am asking about a classification of commutative unital Banach algebras whose spectrum is homeomorphism to the spectrum of an element.

Thank you in advance for any suggestion.

Answer 1

The answer is yes for $C^*$ algebra and this is essentially the Stone Weierstrass theorem:

The assumption that the spectrum of $x$ is isomorphic the spectrum of $A$ (under the canonical map) essentially mean that as a function from the spectrum of $A$ to $\mathbb{C}$, $x$ is an injection, hence $x$ separate the points of the spectrum of $A$ and hence the closed involutive sub-algebra generated by $x$ is $A$ by the stone Weierstrass theorem, or, equivalently, $(iii)$ holds.

Of course in general $(i)$ and $(ii)$ have no reason to hold.

For Banach algebra one can not say anything: take $\mathbb{C}[e]$ with $e²=0$, you can put a structure of involutive Banach algebra on it, and the unit element will satisfy the isomorphism hypothesis but won't satisfy any of $(i)$, $(ii)$ or $(iii)$.