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 K_{1}⊂K_{2}⊂...⊂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 g_{1}, g_{2}, ..., g_{n}∈m so that {g_{1}=0}∩...∩{g_{n}=0}∩{f=r}=∅. But then (g_{1})²+...+(g_{1})²+(f-r)²∈m is a nowhere vanishing function, so it is a unit, so m=C(M), a contradiction.

densein 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 0:21somemaximal ideal (since Zorn's lemma is true). – Anton Geraschenko Nov 3 '09 at 1:36