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

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

2009-11-02T23:52:29Z

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 by Eric Wofsey for Maximal ideals in the ring of continuous real-valued functions on R

2009-11-03T00:18:29Z

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 by Anton Geraschenko for Maximal ideals in the ring of continuous real-valued functions on R

2009-11-03T02:32:00Z

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.