Maximal ideals in the ring of continuous real-valued functions on ℝ

Question

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?

Answer 1

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.

Answer 2

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 K_i is in the interior of K_i+1 and M=∪K_i (you can do this since M is hausdorff and second countable). For each i, choose a function f_i which is 0 on K_i but 1 outside of K_i+1, and define f=∑f_i. Note that for any r∈ℝ, the set {x|f(x)=r} is a closed subset of some K_i, 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, ..., g_n∈m so that {g₁=0}∩...∩{g_n=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.