Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

We denote the ring of all continuous real-valued functions on $X$ by $C(X)$. The ring of all bounded continuous real valued functions on $X$ is denoted by $C_b(X)$. One of the goals of the study of $C(X)$ is to connect algebraic properties of the ring $C(X)$ with the topological properties of the space $X$.

For a simple and well-known connection there is the following theorem:

Theorem: The topological space $X$ is connected iff the ring $C(X)$ has no idempotent element other than $0$ and $1$.

My question comes from the relation between ring homomorphisms of the rings $C(X)$, $C(Y)$ and topological spaces $X$, $Y$.

Q1: Let $X$ and $Y$ be two $T_{3\frac{1}{2}}$topological spaces. If $X$ is a zero dimensional space and $C(X)$ is ring isomorphic to $C(Y)$ (i.e. $C(X)\cong C(Y)$) can we deduce that $Y$ is also a zero dimensional space?

(we recall that a topological space $X$ is zero dimensional if it has a base of clopen subsets.)

The same question can be asked by exchanging the ring $C(X)$ with the Banach algebra $C_b(X)$ as follows:

Q2: Let $X$ and $Y$ be topological spaces with the previous assumption. If $X$ is a zero dimensional space and $C_b(X)$ is ring isomorphic to $C_b(Y)$ (i.e. $C_b(X)\cong C_b(Y)$) can we deduce that Y is also a zero dimensional space?

share|improve this question
    
@Qiaochu: The theorem is trivial and holds for every ringed space. –  Martin Brandenburg Jun 15 '12 at 14:15
    
@Martin, could you expand on your comment? I don't think about T3.5 spaces that are not locally compact Hff very often, so the question is not so obvious to me. –  Yemon Choi Jun 15 '12 at 18:17
    
A locally ringed space $X$ is connected if and only if $\mathrm{\Spec}(\mathscr{O}_X(X))$ is connected, and the latter holds if and only if $\mathscr{O}_X(X)$ is connected, i.e., has no idempotents other than $0$ or $1$. @Martin Does this hold for every ringed space? –  Keenan Kidwell Jun 15 '12 at 19:34
    
@Martin: whoops. I got a little confused there. –  Qiaochu Yuan Jun 15 '12 at 20:43
1  
I was refering to what AliReza has named a "Theorem" in his question. Qiaochu has deleted his comment that this might not be true for geneal $X$. [just to clarify my comment above] @Keenan: Perhaps ringed spaces with nontrivial stalks. –  Martin Brandenburg Jun 16 '12 at 14:49
add comment

2 Answers

If $X$ and $Y$ are realcompact spaces (for example, if they are compact, or $\sigma$-compact, or are separable metric spaces), then the isomorphism of $C(X)$ and $C(Y)$ implies the homeomorphism of $X$ and $Y$. See L. Gillman and M. Jerison, Rings of Continuous Functions, Van Nostrand, 1960 (Theorem 8.3). This implies the requested zero-dimensionality result. For compact sets, there is also the Banach-Stone theorem stating that $X$ and $Y$ are homeomorphic if $C(X)$ and $C(Y)$ are linearly isometric.

share|improve this answer
    
Thank you dear Anatoly Kochubei. I think you Put an Extra condition in the Question. You suppose that $X$ and $Y$ are two real-compact spaces. But the space $X$ in my Question has only the zero-dimensionality condition.As you Know there is a non real-compact zero dimensional space.(fore example [$0, ω_1$))which is not real compact. For the second answer You suppose the compactness, But you Know that if $C_b(X)$ is isomorphic with $C_b(Y)$ we could not deduce that $C(X)$ is isomorphic to $C(Y)$. Please defend your answer in general case, not with the special case. –  Ali Reza Jun 15 '12 at 14:39
    
I just gave the conditions, under which your questions get positive answers. Why are you sure that no additional conditions are needed? –  Anatoly Kochubei Jun 15 '12 at 19:13
add comment

In general the answer to both questions is no: there are pseudocompact zero-dimensional spaces (see, e.g., this answer) whose Cech-Stone compactifications are not zero-dimensional. For an $X$ like that one has $C(X)=C_b(X)$ and this ring is isomorphic to $C(\beta X)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.