Let $X$ be a manifold. Is it true that $\beta X\cong \operatorname{Specm}(C^\infty(X))$? Let $X$ be a (smooth) manifold. It's well known that its Stone-Cech compactification $\beta X$ is homeomorphic to $\operatorname{Specm}(C(X))$, with its Zariski topology.
Is $\beta X$ also homeomorphic to $\operatorname{Specm}(C^\infty(X))$?
 A: 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).  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).
