Prime ideals in $C[0,1]$ and ultrafilters I'm looking for prime ideals in the ring $C[0,1]$ of continuous functions $[0,1]\longrightarrow \mathbb{R}$. I raised that question recently and got good answers but now I'd like to improve a bit the construction with ultrafilters. So, could you help me?
Consider an ultrafilter $\mathcal{F}$ of subsets of $[0,1]$ such that for each $x\in [0,1]$ there exists $U\in \mathcal{F}$ such that $x\notin U$ (there exists such ultrafilter indeed). Now we build an ideal $I\lhd C[0,1]$ which consists of all continuous $f:[0,1]\longrightarrow\mathbb{R}$ such that $f^{-1}(0)\in \mathcal{F}$. It is easy to see that $I$ is an ideal indeed:
Suppose $f,g\in I$, i.e. $f^{-1}(0),g^{-1}(0)\in \mathcal{F}$. Then $(f+g)^{-1}(0)$ contains $f^{-1}(0)\cap g^{-1}(0)$ which is in $\mathcal{F}$, so $f+g\in I$ also. Other properties can be proved similarly. 
Also it can be shown that $I$ is a prime ideal. Is it true that $I$ is never maximal?
Each maximal ideal in $C[0,1]$ is $\mathrm{Ker}(ev_x)$ ($ev_x$ is an evaluation map), so we are to show that $\forall x\in [0,1]$ there exists $f\in I$ such that $f(x)\ne 0$. So, if we take $x\in [0,1]$ we can find $U\in \mathcal{F}$ such that $x\notin U$. However we cannot finish this reasoning by 'but there exists $f:[0,1]\longrightarrow \mathbb{R}$ such that $U=f^{-1}(0)$'. Well, could you improve this?
 A: Note that the zero set of a continuous function cannot be an arbitrary set, it must be a closed set (and more, a closed $G_\delta$ set, but since we're dealing with a metric space, all closed sets are zero sets of continuous functions); so you don't deal with any filter, but a family of closed sets that has the filter properties, except that only closed supersets belong to the family, not all supersets. If you start with an arbitrary filter, the relevant family is the family of closed sets belonging to the filter.
The ideal corresponding to a filter $\mathcal{F}$ is a maximal ideal if and only if $\mathcal{F}$ contains a singleton set. That is the case for all fixed ultrafilters (the ultrafilters of the form $\{ A \subset [0,1] : x \in A\}$ for some $x \in [0,1]$).
So such ideals can be maximal, and some are.
Note that since $[0,1]$ is compact, the intersection of all closed sets belonging to a filter is never empty, and all members of the ideal vanish on that intersection, so for every proper ideal $I$, there is a nonempty closed set on which all members of $I$ vanish.
