# Prime ideals in the ring of germs of continuous functions

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?

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On a loosely related note - the space of codimension $d$ ideals of $\mathcal{C}^\infty(M,\mathbb{R})$ can be topologized in such a way that it is a compactification of configuration space of $d$ distinct unordered points on $M$. (arxiv.org/abs/1006.4352) –  Vít Tuček Nov 16 '11 at 0:12
The discussion of this answer: mathoverflow.net/questions/40736/… is somewhat related to this question. –  Emerton Nov 16 '11 at 13:08
related question mathoverflow.net/questions/35793/prime-ideals-in-c0-1 –  Nikita Kalinin Nov 17 '11 at 18:37
I believe you will find the free reference by Lambek, Gillman and Fine useful: at.yorku.ca/i/a/a/l/94.htm –  Jose Capco Nov 17 '11 at 19:34

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 $g\in I$.

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Obvious secondary question: Can one prove the existence of any other prime ideals without invoking choice? Obvious tertiary question: Why are so many questions on mathoverflow about what you can prove without invoking choice? –  Will Sawin Nov 15 '11 at 23:33
@Tom: You could do the same with any ultrafilter that converges to 0; it need not concentrate on a countable sequence of points $a_n$. –  Andreas Blass Nov 15 '11 at 23:57
Very pretty! Thank you! –  Dylan Wilson Nov 16 '11 at 1:27
@Will: Concerning the tertiary question, I notice that many such questions around here are titled in a provocative style that Dylan did not go for. He could have called his question "Does this ring have any non-maximal prime ideals?" if his concern had been logical rather than about wanting to see an example. –  Tom Goodwillie Nov 17 '11 at 17:15

Let me give a self-contained proof of the existence of prime ideals of the type you require, for $\mathbb R$ and many other spaces .
Given a topological space $X$, let ${\mathfrak m }_x\subset C(X)$ denote the ideal of functions vanishing at $x$ and ${\mathfrak m }_x^{nb}\subset {\mathfrak m }_x$ the ideal of functions vanishing on a ( variable!) neighbourhood of $x$.

Lemma: In a completely regular space $X$, the intersection of ${\mathfrak m }_a \cap {\mathfrak m }_b \; (a\neq b)$ contains no prime ideal
Indeed, consider disjoint neighbourhoods $U_a,U_b$ of $a,b$ , a function $f_a\in C(X)$ with $f_a(a)=1$ and support in $U_a$, and an analogous function $f_b$.
If for some prime ${\mathfrak p}$ we had ${\mathfrak p}\subset {\mathfrak m }_a \cap {\mathfrak m }_b$, then since $f_af_b=0\in {\mathfrak p}$ we would have, for example, $f_a\in {\mathfrak p}$ and a fortiori $f_a\in {\mathfrak m }_a$. Contradiction since $f_a(a)=1$

General result: If in a completely regular space $X$ there is a point $a\in X$ such that ${\mathfrak m }_a^{nb}\subsetneq {\mathfrak m }_a$ then there exists a prime ideal ${\mathfrak p}\subset C(X)$ not of the form ${\mathfrak m }_x$ .
Indeed, the ideal ${\mathfrak m }_a^{nb}$ is radical ( ${\mathfrak m }_a^{nb} =\sqrt {\mathfrak m }_a^{nb}$) hence is the intersection of the primes containing it. By the lemma, the only prime of the form ${\mathfrak m }_x$ containing ${\mathfrak m }_a^{nb}$ is ${\mathfrak m}_a$, hence there must exist a prime not of the form ${\mathfrak m }_x$ containing ${\mathfrak m }_a^{nb}$.

Remarks
1) Of course $\mathbb R$ satisfies the hypothesis of the theorem but so do also all usual completely regular spaces at non isolated points $a$. The hypothesis ${\mathfrak m }_x^{nb}\subsetneq {\mathfrak m }_a$ is very weak: why should a function zero at $a$ vanish in a whole neighbourhhod of $a$ ?!
2) For the aficionados of the axiom of choice, Zorn's lemma is hidden in the fact that a radical ideal in a ring is the intersection of the prime ideals which contain it.

Edit (November 30th, 2011)
I have just come across a remarkable paper proving that for any decent compact space $X$ ( for example a compact cube $[a,b]^n \subset \mathbb R^n$ ), the Krull dimension of $C(X)$ is infinity. The author modestly claims that the result is hidden in Gillman-Jerison's well known book Rings of Continuous Functions.

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Dear Georges, Very nice! In the statement of the General result, shoudl $\mathfrak m_x^{nb}$ read $\mathfrak m_a^{nb}$? Regards, Matthew –  Emerton Nov 16 '11 at 12:44
Dear @Matt, you are right of course: I have corrected the typo. Thank you very much for your attentive reading and the kind words. –  Georges Elencwajg Nov 16 '11 at 13:58
I'm confused, why can't you prove the existence of other primes in this ring more simply as follows: The intersection of all the prime ideals is the nilradical. The nilradical of this particular ring is zero, but it is not an integral domain, therefore there must be other primes. The proof of it not being an integral domain should work for any regular space... we just need Urysohn, I think. –  Dylan Wilson Nov 17 '11 at 1:46
(I guess I should say that "the ring" I'm talking about is the germ of continuous functions at a point.) –  Dylan Wilson Nov 17 '11 at 1:46
Dear @Dylan, you are right. You don't even need "integral domain": any reduced local ring which is not a field has a non-prime maximal ideal. This applies for example to the integral domain of germs of analytic functions at a point of a (real or complex) manifold. Since Tom had given an answer in terms of germs, I gave examples also for global functions. And then Tom edited his answer to take my answer into account... Actually, I like that sort of back and forth movement, including comments like yours: thanks! –  Georges Elencwajg Nov 17 '11 at 9:43

You can take a look of the following:- http://www2.imperial.ac.uk/~naddingt/notes/cont_ag.pdf . It contains an example of a prime ideal for smooth functions.

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It looks like it only contains an example of a radical ideal... –  Dylan Wilson Nov 16 '11 at 18:38
If you restrict the radical ideal to the smooth functions then it'll be prime (that's an example of prime ideal in the ring of smooth functions) but not in the ring of continuous functions. –  Krish Nov 17 '11 at 8:54