Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For a compact space $K$, the maximal ideals in the ring $C(K)$ of continuous real-valued functions on $K$ are easily identified with the points of $K$ (a point defines the maximal ideal of functions vanishing at that point).

Now take $K=\mathbb{R}$. Is there a useful characterization of the set of maximal ideals of $C(\mathbb{R})$, the ring of continuous functions on $\mathbb{R}$? Note that I'm not imposing any boundedness conditions at infinity (if one does, I think the answer has to do with the Stone-Čech compactification of $\mathbb{R}$ - but I can't say I'm totally clear on that part either). Is this ring too large to allow a reasonable description of its maximal ideals?

share|improve this question
2  
Isn't the answer the same? A maximal ideal will just be one that if you mod out by it you get a field, so certainly the points of the real line will give maximal ideals, by evaluation. Also, any nonunit will be contained in some maximal ideal, but a nonunit must vanish somewhere, and so will be in that. Am I missing something? –  Charles Siegel Nov 3 '09 at 0:07
7  
That says that the ideals given by points of R are dense in the Zariski topology on maximal ideals, not that they are all of them. Just because every nonunit is contained in one of those maximal ideals doesn't mean that there aren't any other maximal ideals. –  Eric Wofsey Nov 3 '09 at 0:21
12  
Explicitly, consider the ideal of compactly supported functions. This is an ideal which is not contained in any of the maximal ideals corresponding to points (since for any point there is a compactly supported function not vanishing at it), but it must be contained in some maximal ideal (since Zorn's lemma is true). –  Anton Geraschenko Nov 3 '09 at 1:36

2 Answers 2

up vote 22 down vote accepted

Peter Johnstone's book Stone Spaces (p. 144) proves that for any X, maximal ideals in C(X) are the same as maximal ideals in C_b(X) (bounded functions), i.e. the Stone-Cech compactification \beta X. Indeed, if I is a maximal ideal, let Z(I) be the set of all zero sets of elements of I; this is a filter on the lattice of all closed sets that are zero sets of functions. Then I is contained in J(Z(I)), the set of functions whose zero sets are in Z(I), so by maximality they are equal. But also, by maximality, Z(I) must be a maximal filter on the lattice of zero sets, and we get a bijection between maximal filters of zero sets and maximal ideals in C(X). Now the exact same discussion applies to C_b(X) to give a bijection between maximal filters of zero sets and maximal ideals of C(X) (since the possible zero sets of bounded functions are the same as the possible zero sets of all functions). But the maximal ideals of C_b(X) are just \beta X.

The difference between C_b(X) and C(X) is that for C_b(X), the residue fields for all of these maximal ideals are just C, while for C(X) you can get more exotic things. Indeed, if a maximal ideal in C(X) has residue field C, then every function on X must automatically extend continuously to the corresponding point of \beta X. This can actually happen for noncompact X, e.g. the ordinal \omega_1.

Section IV.3 of Johnstone's book has a pretty thorough discussion of this stuff if you want more details.

share|improve this answer

This isn't really an answer to your question, but I'd like to see it here next time I come looking, so I'll post it. The following result is basically Theorem 2.1 in C-differentiable spaces by Juan A. Navarro González and Juan B. Sancho de Salas.

Theorem: For any manifold M, the maximal ideals of C(M) whose residue field is ℝ is exactly in bijection with the points of M.

Proof: It's clear that points give you distinct maximal ideals with residue field ℝ, so we just need to show that every such ideal comes from a point. Suppose m is a maximal ideal in C(M) such that C(M)/m=ℝ and ∩g∈m{g=0}=∅.

Choose a sequence of compact sets K1⊂K2⊂...⊂M such that Ki is in the interior of Ki+1 and M=∪Ki (you can do this since M is hausdorff and second countable). For each i, choose a function fi which is 0 on Ki but 1 outside of Ki+1, and define f=∑fi. Note that for any r∈ℝ, the set {x|f(x)=r} is a closed subset of some Ki, so it is compact.

Since we have a surjection C(M)→ℝ whose kernel is m, there is some r∈ℝ so that f-r∈m. Since ∩g∈m{g=0}=∅, the open sets {g≠0}g∈m is a cover of M, and in particular cover the compact set {f=r}. So there is some finite collection g1, g2, ..., gn∈m so that {g1=0}∩...∩{gn=0}∩{f=r}=∅. But then (g1)²+...+(g1)²+(f-r)²∈m is a nowhere vanishing function, so it is a unit, so m=C(M), a contradiction.

share|improve this answer
1  
So the R-valued points of Spec(C(M)) are just M ... nice! –  Andrew Critch Nov 3 '09 at 4:16
    
So why doesn't this apply to the maximal ideal containing the compactly supported functions? The residue field is not $\mathbb{R}$? What is it? The field of functions which don't vanish at infinity? –  Joe Hannon Jan 20 at 1:41
    
@JoeHannon: "What is it?" is an excellent question. I don't know of a way to actually produce a maximal ideal containing compactly supported functions, only that one must exist by the axiom of choice. The argument in this answer shows that the residue field for such a maximal ideal cannot be $\mathbb R$. I'd expect different such maximal ideals to have different residue fields. Producing one example would be pretty interesting. –  Anton Geraschenko Jan 20 at 19:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.