MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a Tychonoff topological space. Consider the ring $C(X)$ of all continuous real valued functions on $X$. For what conditions on an ideal $I$ of $C(X)$, we could deduce that the quotient ring is isomorphic to a ring of the form $C(Y)$.i.e.

Question : An ideal $I$ has the algebraic property $\mathcal{P}$ if and only if The exists a Tychonoff topological space $Y$ so that: $$\frac{C(X)}{I} \cong C(Y)$$

share|cite|improve this question
If $X$ is compact I think it is necessary and sufficient that $I$ be closed in the sup norm; the quotient will be a C*-algebra and then one can apply Gelfand-Naimark (probably this is not necessary). – Qiaochu Yuan Aug 15 '12 at 17:00
Have you tried looking in Gilman and Jerison? – Yemon Choi Aug 15 '12 at 18:11

To expand Qiaochu Yuan's comment. If $X$ is a Hausdorff compact space, any closed ideal $I$ of the Banach algebra $C(X)$ is the kernel of the restriction map $C(X)\rightarrow C(Y)$ for some closed subset $Y$ of $X$, namely $Y:=\cap_{f\in I}f^{-1}(0)$ (that is, all functions that vanish on $Y$. This follows from the Stone-Weierstrass theorem applied to the quotient space $X/Y$, obtained collapsing $Y$ to a point). The above restriction map is surjective by the Tietze extension theorem. On the quotient, we have the ordered and isometric algebra isomorphism $C(X)/I \rightarrow C(Y)$.

share|cite|improve this answer
Hello Dear Majer. Thank you very much for your answer. But as you Know Qiaochu gave the same answer as a comment. I think this answer is partial for my Question. I didn't restrict the topological space to a compact one. I think its important to consider the general form of tychonoff spaces such as all metric spaces and etc without assuming compactness of these spaces. – Ali Reza Aug 15 '12 at 17:40
Yes, this was meant to show explicitly the form of $Y$ in the compact case, and also, that Gelfand-Naimark is not necessary as Qiaochu Yuan says. If we assume $X$ a normal space, Tietze theorem still ensures that the restriction map to any closed set $Y$is surjective, and the conclusion holds true for the corresponding ideal $I(Y)$. But for normal spaces I don't think any closed (in the compact-open sense) ideal is of this form, though I don't have examples; nor I think that the space $Y$ in the isomorphism needs to be a subset of $X$) – Pietro Majer Aug 15 '12 at 17:47
If $X$ is non-compact then one has the same result (with basically the same proof) for $C_0(X)$, the space of continuous functions vanishing at infinity. One can easily imagine a counterpart for the space of bounded continuous functions as well using the Stone-Cech compactification. But $C(X)$ itself is a pretty pathological object when $X$ is non-compact, and I'm wondering if it is really the object that you want to consider. – Paul Siegel Aug 15 '12 at 17:57
Dear Paul. there was one thing in your note that I didn't understand. You believe that we have the same result if we consider the case $\C_{0}(X)$. but as you Know if $X$ is not compact then $\C_{0}(X)$ is not unitary.i.e. $1\notin C_{0}(X)$. Maybe you are interested in considering this case but as you Know these two forms of rings does not have the same behavior. Also For the Rings of the form $C^*(X)$ as you mentioned we could change it to the ring $C(\beta X)$ and consider the notes of Qiaochu and Pietro – Ali Reza Aug 15 '12 at 18:30
The unitalization of $C_0(X)$ is $C(\tilde{X})$ where $\tilde{X}$ is the one-point compactification of $X$, and the closed ideals of $C_0(X)$ are in one-to-one correspondence with the nontrivial closed ideals of $C(\tilde{X})$ since an ideal which contains $1$ is the whole ring. So you once again get that any closed ideal in $C_0(X)$ is the set of all functions in $C_0(X)$ which vanish on a fixed closed set (depending on $I$). – Paul Siegel Aug 16 '12 at 23:19

Since you ask for a duality between quotients of $C(X)$ and the closed subsets of $X$, the following does not answer your question directly. Nevertheless, I hope it might be of interest. I am a tireless proponent of the thesis that if you want to extend the duality between compact spaces and algebra of continuous functions thereon, then the appropriate context is the space $C^b(X)$ of bounded, continuous functions, not with the norm but with the strict topoogy. This was introduced by Buck for functions on locally compact spaces and then extended, using different methods, to completely regular ones. Many of the results for Gelfand-Naimark theory for compact spaces can be carried over in the natural way, in particular, precisely the duality between closed subsets of $X$ and quotient algebras with respect to ideals---one need only demand that the ideal be closed in the above topology. This theory is presented in the book "Saks spaces and applications to functional analysis".

share|cite|improve this answer

Repeating Yemon Choi's cue, while being a bit more specific, problem 10D from Gillman and Jerison should get you started. It gives some necessary conditions (like $I$ must be a $z$-ideal) and relates the potential space $Y$ to a closed subset of the Hewitt realcompactification $\upsilon X$ of $X$. After a quick glance, I'm not sure how far you have to go to get sufficient conditions. Let me know if you have trouble getting the relevant information from the book.

share|cite|improve this answer

Hi Alireza, An ideal $I$ of $C(X)$ is an annihilator ideal if $int\bigcap Z[I]\subset intZ(f)$, $f\in C(X)$, then $f\in I$. Now let $I$ be an annihilator ideal of $C(X)$, $X$ a completely regular space and $Y=\bigcap Z[I]$. Then define $\phi$ of $C(X)$ on to $C(Y)$, which $\phi(f)=f|_{Y}$. Then it is easy to see that $\phi$ is a ring homomorphism. By above definition we have, $ker\phi=\{f: Y\subset Z(f)\}=I$. So $\frac{C(X)}{I} ≅C(Y)$.

share|cite|improve this answer

Your Answer


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.