If a field contains the complex field, then it is $\mathbb{C}$

Question

This question is originated from a book by Gaal, (Linear Analysis and Representation Theory). Theorem 7 from section 6, chapter 1 reads as follows, and I quote:

"Theorem 7: Let $A$ be a complex, commutative Banach algebra. Then every regular maximal ideal of $A$ is the kernel of some non-trivial homomorphism $h:A \rightarrow \mathbb{C}$ and conversely, the kernel of every h is a regular maximal ideal. The relative identities of I are all mapped by $h$ into $1$."

The proof is straight forward and uses the fact that $I$ being maximal, the quotient $A/I$ is a field, which was proved just before this theorem. The problem is, he then asserts that $A/I$ is topologically isomorphic to $\mathbb{C}$. Couldn't the field properly contain $\mathbb{C}$? For instance, be the completion of the field of quotient complex funcions?

Answer 1

This is actually the Gelfand-Mazur theorem. You can see the proof of it in SUNDER, V. S., Functional Analysis: Spectral Theory, theorem 3.2.5. It is stated as below:

Theorem (Gelfand-Mazur theorem): The following conditions on a unital commutative Banach algebra $\mathcal{A}$ are equivalent:

(i) $\mathcal{A}$ is a division algebra - i.e., every non-zero element is invertible; ($\mathcal{A}$ is a field)

(ii) $\mathcal{A}$ is simple - i.e., there exist no proper ideals in $\mathcal{A}$; and

(iii) $\mathcal{A}=\mathbb{C}1_\mathcal{A}$.

You can find the book on Sunder's webpage.