2 typo

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 1 0 on Ki but 0 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.

1

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 1 on Ki but 0 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.