We all know that the ring of germs of continuous functions at a point on, say $\mathbb{R}$, has a unique maximal ideal- namely, those functions that vanish at that point.


**Can anyone think of a single other example of a prime ideal?**


We know that there have to be lots and lots of them, since there are no non-zero nilpotent elements in the ring. You can't try and construct such a thing by looking at the germ of functions that vanish at some bigger set of points, because then you can use bump functions to produce two germs whose product is in the ideal but neither of whom belong to the ideal themselves.


The only other possible things we could think of would only have a hope of working if we were dealing with smooth functions (though we still haven't come up with an example there!).


Is it even possible to write down an example of such a thing for continuous functions, or are they just too ugly?



Prime ideals in the ring of germs of continuous functions
Choose a sequence $a\_n$ of distinct points converging to $0$. "Choose" an ultrafilter on this countable set. Consider the germs of functions $f$ such that the set of all $a\_n$ where $f(a\_n)=0$ belongs to the filter. (This condition on the function depends only on its germ.) 


EDIT The following refers to and complements George Elencwajg's answer. In particular the ideal $\frak m\_a$ and $\frak m\_a^{nb}$ are defined there.


Recall: An ultrafilter on the set $X$ gives you a maximal ideal in the ring of all real-valued functions, and these are the only prime ideals. If $X$ has a topology, then intersecting with $C(X)$ these yield prime ideals which may or may not be maximal. If $X$ is compact Hausdorff then every ultrafilter converges to (i.e. contains all neighborhoods of) a unique point.


For the rest of this discussion I assume $X$ is compact Hausdorff.


(1) Every proper ideal in $C(X)$ is contained in $\frak m\_a$ for a (unique) point $a\in X$. In particular these are the only maximal ideals.


Proof: Otherwise $I$ has elements $f\_1,\dots ,f\_n$ such that the open sets $X-f^{-1}(0)$ cover $X$. But then $f\_1^2+\dots +f\_n^2$ is a a unit in $C(X)$.


(2) Every prime ideal $P$ of $C(X)$ must contain the ideal $\frak m\_a^{nb}$ for some point $a\in X$.


Proof: Otherwise there exist $f\_1,\dots ,f\_n$ all outside $P$, vanishing respectively on open sets $U\_1,\dots ,U\_n$ that cover $X$. The product of the $f\_i$ is zero, contradiction.


Thus 


(3) All prime ideals of $C(X)$ correspond to prime ideals of the various local rings (rings of germs) $C(X)/\frak m\_a^{nb}$.


Of course when the prime is determined by an ultrafilter then the point in question is the point that it converges to.


(4) The prime determined by an ultrafilter cannot be maximal unless the ultrafilter is principal.


Proof: If the ideal is $\frak m\_a$, then a continuous function vanishing only at $a$ will belong to the ideal, implying that the singleton $a$ belongs to the filter.


I suppose it can happen that different ultrafilters sometimes give the same prime ideal, but I'm not sure.


I suppose it can happen that there are prime ideals not determined by any ultrafilter, but I'm not sure. Such an ideal would **not** have the following property : If $f\in I$ and if $f^{-1}(0)\subset g^{-1}(0)$ then $g\in I$. 


