Is there a nontrivial (what I mean is below) example of a compact Hausdorff space $X$ and a family $\mathscr{F}$ of subsets of $X$ with the following pair of properties?

  • $\mathscr{F}$ is closed under finite intersection.

  • $\mathscr{F}$ is "saturated with respect to finite covers," in the sense that if $A\in\mathscr{F}$ and $A=B\cup C$, then at least one of $B,C$ contains a set that is in $\mathscr{F}$.

For any topological space $X$, for fixed $p\in X$, $\mathscr{F}=\{A\mid A\ni p\}$ is such a family, so by "nontrivial" first of all I mean "not of this form." Secondly, if $\emptyset\in\mathscr{F}$, then the second requirement is automatically fulfilled, so by "nontrivial" I also mean $\emptyset\notin\mathscr{F}$.

Motivation (you can ignore this part but comments on it are welcome): Here's where the commutative algebra and algebraic geometry come in. An awesome theorem (which I learned from Atiyah-MacDonald's commutative algebra book) says that if $X$ is a compact Hausdorff space and $C(X)$ the ring of continuous real-valued functions on $X$, $\operatorname{Spm}C(X)$, equipped with the Zariski topology, is homeomorphic to $X$. I am trying to understand the relationship between $\operatorname{Spm}C(X)$ and $\operatorname{Spec}C(X)$. First of all, is $\operatorname{Spec}C(X)$ actually bigger than $\operatorname{Spm}C(X)$? So, I am trying to explicitly construct a prime ideal of $C(X)$ that's not maximal. If $\mathscr{F}$ is any family of subsets of $X$ that is closed under intersection, then $\{f\in C(X)\mid f|_A = 0\text{ for some }A\in\mathscr{F}\}$ is an ideal. If $\mathscr{F}$ is also "saturated with respect to finite covers" in the sense above, it seems to me that this ideal is also prime. My definition of "trivial" is designed to eliminate constructions that would lead to either the unit ideal (this happens if $\emptyset\in\mathscr{F}$) or a maximal ideal (this is what happens if $\mathscr{F}=\{A\mid A\ni p\}$ for some $p\in X$.

Thanks in advance for your thoughts.

  • $\begingroup$ you ask that the members of $\mathscr{F}$ be closed in the title but not in the question...so how about all subsets? or, in fact any collection closed under finite intersection + the emptyset $\endgroup$ – uncookedfalcon Apr 16 '13 at 2:36
  • $\begingroup$ See en.wikipedia.org/wiki/Ultrafilter, but I don't think this construction works. $\endgroup$ – Qiaochu Yuan Apr 16 '13 at 3:59
  • $\begingroup$ @uncookedfalcon - Thanks, I removed "closed" from the title; also, I'd better kick out the possibility that $\emptyset\in\mathscr{F}$, because (see motivation) if it is, then the associated ideal is the unit ideal. $\endgroup$ – Ben Blum-Smith Apr 16 '13 at 4:10
  • $\begingroup$ mathoverflow.net/questions/35793/prime-ideals-in-c0-1 might be what you're looking for. $\endgroup$ – Henno Brandsma Apr 16 '13 at 7:39

This question seems to have been answered in the comments.

A propos of Qiaochu's comment, it seems that if I add the requirement that $\mathscr{F}$ be upward closed, then what I'm asking for is equivalent to a free ultrafilter on $X$. These exist assuming AC or (weaker) the ultrafilter lemma.

Henno linked to a MO question that addresses my motivation. Free ultrafilters do allow one to construct prime ideals of $C(X)$ that are not maximal as in the OP, but there are easier constructions (albeit also involving AC). For example, for any multiplicative subset $S\subset C(X)$ not containing zero but containing at least one function vanishing at each $x\in X$, any ideal maximal in the lattice of ideals disjoint from $S$ will be prime but not maximal. (The example at MO is $S=$ the set of nonzero polynomials.)

I've learned a lot, thanks guys. I'll leave the question open for a few days in the hopes of more persectives, but I'm pretty satisfied now.


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