Characterisation of a Commutative C* Algebra which is an Integral Domain 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.
 A: 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.
A: 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 
