I am curious to know if the following question can be answered by my naive approach. I understand that there is a huge body of work in rings of continuous function and some really deep ideas.
Consider the ring of continuous functions on a compact Hausdorff space $X$ (taking $X=[0,1]$ is also fine as far as this question is concerned), denoted $C(X)$. Let $I$ be a proper ideal. Is $I=I(S)$ for some $S$, where $I(S):=\{f\in C(X)\mid f(s)=0\, \forall s\in S\}$??
My naive approach: We know that $I$ is contained in some maximal ideals which are of the form $I(p)$ for some point $p\in X$ according to the above notation. We take all such maximal ideals containing $I$ and intersect them. So $I\subset \bigcap I(p)=I(\bigcup \{p\})$. I was hoping to show an equality here, but I am not sure how to approach this or whether it is even true that they are equal?? I wonder if taking $f\in \bigcap I(p)$ and $f\notin I$ leads to a contradiction by breaking some maximality condition; like now we have $I\subset (I,f)\subset I(\bigcup \{p\})$ an so on…
Any help will be appreciated. Thanks in advance.
Best Answer
Here is a simple example: $X=[-1,1]$. The principal ideal of $C(X)$ generated by the function $x$ has a unique $0$ the point $0$, but it does not equal $\{f \ | f(0)=0\}$. For instance, the function $x \sin \frac{1}{x} \not \in (x)$.