Smooth manifold $M$ is completely determined by the ring $F$. 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.
 A: 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.
A: Suppose $e:F\to\mathbb{R}$ is any ring-homomorphism.  Note that if $f\in F$ is such that $f>0$ everywhere, then $e(f)>0$:  $f$ is a unit in $F$, so $e(f)\neq 0$, and $\sqrt{f}$ is smooth, so $e(f)=e(\sqrt{f}^2)=e(\sqrt{f})^2\geq 0$.  I claim furthermore that $e$ is an $\mathbb{R}$-algebra homomorphism, so $e(c)=c$ if $c$ is a constant function.  Indeed, we know this must be true if $c\in\mathbb{Q}$; for arbitrary $c\in\mathbb{R}$, now use the fact that $e(c-q)>0$ if $q<c$ and $q\in\mathbb{Q}$ and $e(q-c)>0$ if $q>c$ and $q\in\mathbb{Q}$.
Now suppose that $e$ is not given by evaluation at any point.  Then for each $x\in M$, there is a function $f_x\in F$ such that $e(f_x)\neq f_x(x)$.  Letting $c=e(f_x)$ and replacing $f_x$ with $f_x-c$, we may assume $f_x(x)\neq 0$ and $e(f_x)=0$.  Replacing $f_x$ with its square, we may further assume that $f_x\geq 0$ everywhere.  For any compact $K\subseteq M$, finitely many of the sets $\{y:f_x(y)>0\}$ cover $K$, and so adding together the corresponding $f_x$'s, we get an element $f_K\in F$ such that $f_K\geq 0$ everywhere, $f_K>0$ on $K$, and $e(f_K)=0$.  Multiplying by an appropriate constant, we can further assume that $f_K>1$ on $K$.
Now, as you suggest, take $g\in F$ such that $g>0$ and $g^{-1}([0,c])$ is compact for each $c$.  Since $g>0$, $e(g)>0$.  Let $c=e(g)$ and $K=g^{-1}([0,2c])$.  Then $e(g+cf_K)=e(g)=c$ since $e(f_K)=0$.  But $g+cf_K>c$ everywhere (if $x\not\in K$, $g(x)\geq 2c>c$, and if $x\in K$, $cf_K(x)>c$).  So $h=g+cf_k-c$ is strictly positive, and so $e(h)>0$.  But $e(h)=e(g)-e(c)=c-c=0$.  This is a contradiction.
