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Let $X$ be a compact hausdorff topological space with more than one element.Then prove that the ring $C(X)$ of complex valued continuous functions on $X$ is not an integral domain. Thanks for any help. Actually this question arose when I was trying to prove that any commutative C*-algebra which is also an integral domain must be isomorphic to C and I think this statement is correct. The problem I am having is with the case when X is connected.

Ester
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  • A single point is a compact Hausdorff space such that $C(X) = \mathbb C$, and Slade has an example below where $C(X)$ is not an integral domain. – Jim Mar 11 '15 at 07:03
  • Ok,let us take |X|>1. – Ester Mar 11 '15 at 07:09
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    Ester: Your question is still a bit unclear, but based on your edit I think you mean to be asking whether it is possible for it to be an integral domain. The way you worded it can be interpreted naturally as asking whether it is always an integral domain (although the trivial answer led me to doubt that). I suggest that you clear up the wording, and edit you title so that it indicates what the question is about. (Also look up Urysohn's lemma.) If you do clarify your question, you might want to comment to those who answered the unclear version in case they are interested. – Jonas Meyer Mar 16 '15 at 02:06
  • Yes, I was trying to apply the Urysohn's Lemma,but cannot manage a proof when X is connected. – Ester Mar 16 '15 at 03:48
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    @Ester: This looks related and interesting: http://math.stackexchange.com/q/1176279/ – Jonas Meyer Mar 17 '15 at 04:02

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Let $X$ be a compact Hausdorff space with more than one point. Then $C(X)$ is a natural, normal uniform algebra on $X$. This means that the character space of $C(X)$ is exactly $X$ and for each closed set $E\subseteq X$ and each closed set $F\subseteq X\setminus E$, there exists a function $f\in C(X)$ such that $f(y)=0$ for all $y\in E$ and $f(y)=1$ for each $y\in F$. Thus $C(X)$ necessarily contains non-zero divisors. This follows from the Tietze extension theorem.

Note that when $X$ is locally compact, we need one of the sets $K$ to be compact and the other closed, but I cannot remember which one needs to be compact and which needs to be closed. I refer you Uniform algebras by Gamelin, Introduction to function algebras by Browder or Banach algebras and automatic continuity (Chapter 4) by Dales for details on these facts.

SamM
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Let $A$ be a non-commutative C*-algebra that is also an integral domain. If $a$ be a self-adjoint element of $A$, then it is well-known that C*(a), C*-algebra generated by $a$, is a commutative C*-algebra. Since, $A$ is an integral domain, C*(a) is an integral domain, too. If we accept the above conjecture that any commutatve C*-algebra which is also an integral domain must be isomorphic to C, then C*(a) as a commutative C*-algebra which is also an integral domain must be C and it implies that $a$ is a complex number. Since $a$ is an arbitrary self-adjoint element of $A$, $A$ must be isomorphic to C and it is a contradiction beacause of non-commutativity of $A$. Hence, I think the above conjecture in incorrect.

Best regards A. Hosseini

Amin Hosseini
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  • The conjecture doesn't imply that $a\in\mathbb{C}$. It implies that $a$ corresponds to an element of $\mathbb{C}$ under an isomorphism between $\langle a\rangle_{C^*}$ and $\mathbb{C}$. – Jacob Manaker Oct 14 '20 at 17:14