TITLE: Smooth manifold $M$ is completely determined by the ring $F$.
QUESTION [7 upvotes]: For any smooth manifold $M$, the collection $F = C^\infty(M, \mathbb{R})$ of smooth real valued functions on $M$ can be made into a ring, and every point $x \in M$ determines a ring homomorphism $F \to \mathbb{R}$ in $F$. If $M$ is compact, every maximal ideal in $F$ arises in this way from a point of $M$.
My question is, if there is a countable basis for the topology of $M$, how do I see that every ring homomorphism $F \to \mathbb{R}$ is obtained in this way?
Progress. We probably want to make use of an element $f \ge 0$ in $F$ such that each $f^{-1}[0, c]$ is compact? But it is not clear to me what do from there.

REPLY [3 votes]: If $M$ is second-countable, let $\varphi: F \to \mathbb{R}$ be a linear homomorphism. Construct an $f$ as follows: take a countable partition of unity (with compact supports, by second countability) $p_i$. Then$$f = p_1 + 2p_2 + 3p_3 + \dots$$is well-defined since at any given point all but finitely many are zero. However, it is clear that the preimage of $[0, c]$ is always compact, since when $k > c$< points outside the union of the compact support of $p_1$ through $p_k$ will necessarily evaluate to at least $k$. Let $\varphi(f) = t$.
Notice that $\varphi$ is actually an $\mathbb{R}$-algebra homomorphism. Obviously, it is a $\mathbb{Q}$-algebra homomorphism, and the fact that squares must be nonnegative preserves ordering of scalars so continuity does the rest.
Suppose for the sake of contradiction that for each $x$, we had some $f_x$ so that $\varphi(f_x) \neq f_x(x)$. By translating, squaring, and scaling, we can find nonnegative $f_x$ in the kernel of $\varphi$ with $f_x(x) > t$. In fact, this open condition is true on an open neighborhood of $x$. The compact set $f^{-1}([0, t])$ can be covered by finitely many of these open neighborhoods, so we can build a summed function $f'$ which is everywhere nonnegative and at least $t$ on that compact set. Then $\varphi(f + f' - t) = 0$, but $f + f' - t$ is positive everyyhere and hence a unit, contradiction.