Let $X$ be a (smooth) manifold. [It's well known](https://math.stackexchange.com/questions/451223/the-stone-%C4%8Cech-compactification-of-a-space-by-the-maximal-ideals-of-the-ring-of/451318#451318) that its Stone-Cech compactification $\beta X$ is homeomorphic to $\operatorname{Specm}(C(X))$, with its Zariski topology.


**Is $\beta X$ also homeomorphic to $\operatorname{Specm}(C^\infty(X))$?**



Let $X$ be a manifold. Is it true that $\beta X\cong \operatorname{Specm}(C^\infty(X))$?
Indeed, $\operatorname{Specm}(C^\infty(X))$ is homeomorphic to $\beta X$ (through an explicit homeomorphism that will be described below). Essentially all the theory described in Gillman & Jerison's classic book *Rings of Continuous Functions* (1960) applies: for completeness of MathOverflow, let me recall how this works.


If $f$ is a continuous real-valued function on $X$, its **zero-set** is the set $Z(f) := \{x\in X : f(x)=0\}$. The zero-set of a continuous function is closed, and conversely, as $X$ is a metric space, every closed set is the zero-set of a continuous function. But in fact, even more is true: *every closed set of $X$ is the zero-set of a smooth function* (as follows, e.g., from [here](https://math.stackexchange.com/questions/2279872/find-a-smooth-function-f-in-c-infty-such-that-a-f-10-and-b-f-11)). Because of this, the notions of “z-filter” and “z-ultrafilter” below do not depend on whether we are talking about zero-sets of continuous functions or of smooth functions (or, indeed, closed sets).


Now let us make the following definitions:


* A set $\mathscr{F}$ of zero sets is said to be a **z-filter** iff it contains $X$, and is closed under enlargement (i.e., if $Z \subseteq Z'$ are zero-sets and $Z\in \mathscr{F}$ then $Z'\in \mathscr{F}$) and finite intersection.
* A z-filter is said to be a **z-ultrafilter** iff it is proper (i.e., does not contain $\varnothing$), and maximal for inclusion among proper z-filters.


Now let $R$ be either the ring $C(X)$ of continuous real-valued functions on $X$ or the ring $C^\infty(X)$ of smooth real-valued functions on $X$. Define two maps $\mathscr{Z}\colon \{\text{ideals of $R$}\} \to \{\text{z-filters}\}$ and $\mathscr{I}\colon \{\text{z-filters}\} \to \{\text{ideals of $R$}\}$ as follows:


* If $I$ is an ideal of $R$, then $\mathscr{Z}(I) := \{Z(f) : f\in I\}$. This is indeed a z-filter, as is easy to check (for enlargement, note that if $Z(f) \subseteq Z(f')$ with $f\in I$ then $Z(f') = Z(ff')$; and for finite intersection, note that $Z(f\_1) \cap Z(f\_2) = Z(f\_1^2 + f\_2^2)$).
* If $\mathscr{F}$ is a z-filter, then $\mathscr{I}(\mathscr{F}) := \{f\in R : Z(f)\in \mathscr{F}\}$. This is indeed an ideal as is easy to check (note that $Z(g\_1 f\_1+\cdots+g\_r f\_r) \supseteq Z(f\_1)\cap \cdots \cap Z(f\_r)$ if $f\_1,\ldots,f\_r$ are in $\mathscr{I}(\mathscr{F})$).


These functions are not bijections, but still, they preserve inclusions and we trivially have $\mathscr{Z}(I) \subseteq \mathscr{F}$ iff $I \subseteq \mathscr{I}(\mathscr{F})$ (when $I$ is an ideal of $R$ and $\mathscr{F}$ a z-filter); from this follows $\mathscr{Z}(\mathscr{I}(\mathscr{F})) \subseteq \mathscr{F}$ for any z-filter $\mathscr{F}$, and $\mathscr{I}(\mathscr{Z}(I)) \supseteq I$ for any ideal $I$ of $R$, but in fact it is clear that the former is easily seen to be an equality: $\mathscr{Z}(\mathscr{I}(\mathscr{F})) = \mathscr{F}$ (the latter inclusion, $\mathscr{I}(\mathscr{Z}(I)) \supseteq I$ can be proper in general as evidenced by the ideal of functions vanishing to order $2$ at a point). Also note that an ideal $I$ is the unit ideal $R$ iff $\mathscr{Z}(I)$ is the improper z-filter $\mathscr{Z}(R)$ (the one consisting of every zero-set) because $Z(f)$ is empty iff $f$ is invertible in $R$.


At this point, it is easy to check that: if $M$ is a maximal ideal of $R$ then $\mathscr{Z}(M)$ is a z-ultrafilter, and if $\mathscr{U}$ is a z-ultrafilter then $\mathscr{I}(\mathscr{U})$ is a maximal ideal. So $\mathscr{Z}$ and $\mathscr{I}$ restrict to bijections between maximal ideals of $R$ and z-ultrafilters.


But because the above worked just as well for $R$ being the ring $C(X)$ of continuous real-valued or the ring $C^\infty(X)$ of smooth real-valued functions, we can conclude that their maximal ideals are in bijection by the compositions of the corresponding $\mathscr{Z}$ and $\mathscr{I}$ functions, and in fact since $\mathscr{I}\_{C^\infty(X)}(\mathscr{F}) = \mathscr{I}\_{C(X)}(\mathscr{F}) \cap C^\infty(X)$ the bijection from maximal ideals of $C^\infty(X)$ to those of $C(X)$ can more simply be described as $M \mapsto M \cap C^\infty(X)$. Furthermore, the Zariski topologies correspond under this bijection, because (since every z-filter is of the form $\mathscr{Z}(I)$) they both correspond to the Zariski topology on the set of z-ultrafilters whose closed sets are given by the sets of ultrafilters containing a given z-filter.


(It might be worth while examining exactly which properties are required for a ring $R$ of “functions” on $X$ to make the above reasoning ensure that $\operatorname{Specm}(R) = \beta X$.)


**PS:** It might also be interesting to consider what happens for *bounded* functions. The maximal ideals of the ring $C^\*(X)$ of bounded continuous functions on $X$ are also naturally in bijection with the Stone-Čech compactification, but the bijection between ideals of $C(X)$ and $C^\*(X)$ is *not* the naïve $M \mapsto M \cap C^\*(X)$. So I'm not sure what happens to bounded smooth functions (I didn't give it thought).


