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

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?

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If $I$ is the kernel of a ring homomorphism $A\to\mathbb{C}$, then $A/I$ is (isomorphic to) a subring of $\mathbb{C}$ (a subfield if $I$ is maximal). But it contains $\mathbb{C}$, so… – egreg Oct 22 '13 at 19:47
@egreg There's a subtlety there. There exists a proper subring of $\mathbb{C}$ that is isomorphic to $\mathbb{C}$. – Zhen Lin Oct 22 '13 at 19:49
I think that if the field properly contains $\mathbb C$ you can find an ideal $I \subset J$ so that $A/J = \mathbb C$. – N. S. Oct 22 '13 at 19:50
@ZhenLin Yes, but here we're talking about homomorphisms of $\mathbb{C}$-algebras. – egreg Oct 22 '13 at 19:50
@egreg how can you say it is a subring of $\mathbb{C}$? couldn't it contain $\mathbb{C}$ properly? – Henrique Tyrrell Oct 22 '13 at 19:51

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.

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