TITLE: A problem on Milnor's book(about smooth manifolds)
QUESTION [0 upvotes]: Suppose that $M$ is a smooth manifold. Let $F$ be $C^\infty(M,\mathbb{R})$, all the smooth functions on $M$. Then each $x\in M$ defines a ring homomorphism $F\to\mathbb{R}$ $f\mapsto f(x)$. The writer said in the book that if there is a countable basis for the topology of $M$, every ring homomorphism $F\to\mathbb{R}$ is obtained in this way. But I don't know why. Could anyone give me some hints? How to use the condition the existence of a countable basis? Thanks!

REPLY [1 votes]: I suppose there is a nicer way, but I don't see it right now, so it's time to unwrap $Z$-filters.
Before everything else, we note that a (unital) ring homomorphism $F \to \mathbb{R}$ is necessarily $\mathbb{R}$-linear, and its kernel is a) a maximal ideal of $F$, and b) a linear subspace of codimension $1$. Conversely, if $\mathfrak{m}$ is a maximal ideal of $F$ which has codimension $1$, then the field $F/\mathfrak{m}$ is canonically isomorphic to $\mathbb{R}$, and thus induces a ring homomorphism $F\to \mathbb{R}$. The correspondence between these maximal ideals and the ring homomorphisms $F \to \mathbb{R}$ is bijective.
For $f\in F$, we define $Z(f) = f^{-1}(0)$, and we set $\mathfrak{Z}(F) = \{ Z(f) : f\in F\}$. A $Z$-filter on $M$ is a family $\mathscr{Z} \subset \mathfrak{Z}(F)$ with the following properties:

$M \in \mathscr{Z}$, $\varnothing \notin \mathscr{Z}$,
$Z_1, Z_2 \in \mathscr{Z} \implies Z_1 \cap Z_2 \in \mathscr{Z}$,
if $Z_1 \in \mathscr{Z}$ and $Z_2 \in \mathfrak{Z}(F)$ with $Z_1 \subset Z_2$, then $Z_2 \in \mathscr{Z}.$

$Z$-filters on $M$ are related to proper ideals of $F$:
If $\mathfrak{a} \subset F$ is a proper ideal, then
$$V(\mathfrak{a}) := \{ Z(f) : f \in \mathfrak{a}\}$$
is a $Z$-filter. Since $Z(0) = M$, and an $f\in F$ with $Z(f) = \varnothing$ is a unit, $V(\mathfrak{a})$ satisfies the first point. Since $Z(f^2+g^2) = Z(f) \cap Z(g)$, and $f,g\in\mathfrak{a} \implies f^2 + g^2 \in \mathfrak{a}$, the second point also holds for $V(\mathfrak{a})$. To see that $V(\mathfrak{a})$ also satisfies the third point, let $Z_1 = Z(f)$ for some $f\in \mathfrak{a}$, and $Z_2 = Z(g)$ for some $g\in F$. Then $f\cdot g \in \mathfrak{a}$, and $Z(f\cdot g) = Z(f) \cup Z(g) = Z(g) \in V(\mathfrak{a})$.
Conversely, for a $Z$-filter $\mathscr{Z}$, let
$$I(\mathscr{Z}) := \{ f \in F : Z(f) \in \mathscr{Z}\}.$$
Then $I(\mathscr{Z})$ is a proper ideal of $F$. Since $M = Z(0) \in \mathscr{Z}$, we have $0 \in I(\mathscr{Z})$, and since $\varnothing \notin \mathscr{Z}$, $I(\mathscr{Z})$ contains no units. If $f,g \in I(\mathscr{Z})$, then $Z(f + g) \supset Z(f) \cap Z(g)$, whence $f+g \in I(\mathscr{Z})$ by points 2 and 3. Finally, if $f\in I(\mathscr{Z})$ and $g\in F$, then $Z(f\cdot g) = Z(f) \cup Z(g) \in \mathscr{Z}$ by point 3, so $f\cdot g \in I(\mathscr{Z})$.
It's immediate that for proper ideals $\mathfrak{a} \subset \mathfrak{b}$ we have $V(\mathfrak{a}) \subset V(\mathfrak{b})$, and for $Z$-filters $\mathscr{Z} \subset \mathscr{W}$ we have $I(\mathscr{Z}) \subset I(\mathscr{W})$. Further, we have $\mathfrak{a} \subset I(V(\mathfrak{a}))$, and (this one is not immediate, but not hard to see)
$$V(I(\mathscr{Z})) = \mathscr{Z}.$$
A $Z$-ultrafilter is a $Z$-filter that is not properly contained in any other $Z$-filter.
Thus we have a bijective correspondence between the maximal ideals of $F$ and the $Z$-ultrafilters. If $\mathfrak{m}$ is a maximal ideal of $F$, and $\mathscr{Z}$ a $Z$-filter with $V(\mathfrak{m}) \subset \mathscr{Z}$, then $I(\mathscr{Z})$ is a proper ideal of $F$ with $\mathfrak{m} \subset I(V(\mathfrak{m})) \subset I(\mathscr{Z})$, so by the maximality we have equality, and $V(\mathfrak{m}) = V(I(\mathscr{Z})) = \mathscr{Z}$, so $V(\mathfrak{m})$ is a $Z$-ultrafilter. Conversely, if $\mathscr{U}$ is a $Z$-ultrafilter, and $I(\mathscr{U}) \subset \mathfrak{a}$, where $\mathfrak{a}$ is a proper ideal, then $\mathscr{U} = V(I(\mathscr{U})) \subset V(\mathfrak{a})$. Since $\mathscr{U}$ is a $Z$-ultrafilter, $\mathscr{U} = V(\mathfrak{a})$ follows, and then $\mathfrak{a} \subset I(V(\mathfrak{a})) = I(\mathscr{U})$ shows the maximality of $I(\mathscr{U})$.
Now, $Z$-ultrafilters come in two varieties. The first kind are the principal $Z$-ultrafilters $\mathscr{U}(x) := \{ Z \in \mathfrak{Z}(F) : x\in Z\}$. It is easy to see that $I(\mathscr{U}(x)) = \{ f \in F : f(x) = 0\}$ is the kernel of the evaluation homomorphism at $x$, so these are just the ones we are interested in. The second kind are the free $Z$-ultrafilters. A $Z$-filter $\mathscr{Z}$ is called free if
$$\bigcap \mathscr{Z} = \varnothing.$$
We need to show that for a free $Z$-ultrafilter $\mathscr{U}$, the maximal ideal $I(\mathscr{U})$ does not correspond to a ring homomorphism $F\to \mathbb{R}$, and that indeed every non-free $Z$-ultrafilter is principal. The latter is easy, if $\mathscr{U}$ is a non-free $Z$-ultrafilter, let $x \in \bigcap \mathscr{U}$. Then clearly $\mathscr{U} \subset \mathscr{U}(x)$, and since $\mathscr{U}$ is a $Z$-ultrafilter, we have equality. Further, for $x \neq y$ we have $\mathscr{U}(x) \neq \mathscr{U}(y)$.
Now we come to the part using the second countability of $M$. Since $M$ is a second countable locally compact space, we can write $M$ as the union of a sequence $(K_n)$ of compact subsets of $M$, which we can choose so that $K_n \subset \operatorname{int} (K_{n+1})$ for all $n$. And we can construct an $f_0 \in F$ such that
$$ 2^{-n} \leqslant f_0(x) \leqslant 2^{4-n}$$
for all $x \in K_n \setminus K_{n-1}$. Then $f_0$ is a unit in $F$, and if $\mathscr{U}$ is a free $Z$-ultrafilter on $M$, then
$$\lambda\cdot 1 - f_0 \notin I(\mathscr{U})$$
for all $\lambda \in \mathbb{R}$, so the codimension of $I(\mathscr{U})$ is at least $2$. To see this, note that for every $\lambda$ we have $Z(\lambda \cdot1 -f_0) \subset K_n$ for some $n$ (depending on $\lambda$), but no $Z \in \mathscr{U}$ can be contained in any $K_n$ (otherwise $\{ Z \cap K_n : Z \in \mathscr{U}\}$ would be a family of compact sets with the finite intersection property, so $$\bigcap_{Z \in \mathscr{U}} (Z\cap K_n) \neq \varnothing,$$ and $\mathscr{U}$ could not be free). Thus the maximal ideals correspnding to free $Z$-ultrafilters do not correspond to ring hommorphisms $F \to \mathbb{R}$.
To see how this can fail if $M$ is not second countable, consider the long line $L$.  Then every continuous function $f \colon L \to \mathbb{R}$ is eventually constant (the argument here works with a small modification), and we see that
$$f \mapsto \lim_{x \to \Omega} f(x)$$
is a (unital) ring homomorphism $C^{\infty}(L,\mathbb{R}) \to \mathbb{R}$ which is not an evaluation homomorphism.
{"set_name": "stack_exchange", "score": 0, "question_id": 2317277}