smooth Gelfand-duality Assume $M$ is a compact smooth manifold (without boundary). What can we say about the spectrum of the $\mathbb{R}$-algebra $A=C^{\infty}(M)$? The elements of $M$ give rise to rational points of $A$, are there other ones? Does the smooth structure of $M$ endow $A$ with additional structure such that $M$ can be completely recovered from $A$? In other words, is there some kind of smooth Gelfand-duality?
 A: You don't need any additional (topological or whatsoever) structure on the $\mathbb{R}$-algebra $A=C^\infty(M)$. As an algebra alone it allows you to reconstruct the manifold completely. (And not just as a set or topological space, but with it's smooth structure).
This had been answered in other questions on MO, unfortunately I don't remember where. Anyway a good reference is the book: Jet Nestruev, Smooth manifolds and observables, as well as the above mentioned book C-infinty differentiable spaces.
A: I suspect you may be interested in the following nlab page: http://ncatlab.org/nlab/show/smooth+algebra.  In particular, note the section on smooth function algebras on smooth manifolds and the remark immediately preceding it:

By the properties of $C^∞(X)$ for $X$ a smooth manifold discussed below, the $ℝ$-points of $C^∞(X)$ are precisely the ordinary points of the manifold $X$.

A: Perhaps I should post this as an answer (even if I don't really know that theory): in 
Juan A. Navarro González & Juan B. Sancho de Salas, C∞-Differentiable Spaces, LNM 1824 Google Books Preview
a theory of "$C^{\infty}$ -differentiable spaces" is developped, and it would be something like the smooth analog to (possibly singular and nonreduced) complex analytic spaces.
A: The functor from the category of smooth manifolds to to the category of real algebras
that sends a manifold M to C^∞(M) is fully faithful, hence it is an equivalence
of categories of smooth manifolds and real algebras of certain type.
The inverse functor sends a real algebra A to the real spectrum of A (homomorphisms
of algebras from A to R) equipped with the standard Zariski topology (every ideal corresponds
to a closed set) and the obvious structure sheaf of smooth functions
(every element of A gives a function on the real spectrum of A).
The construction also works for manifolds with boundaries and/or corners.
Here is a link to a related question:
How much of differential geometry can be developed entirely without atlases?
A: If I am not mistaken the algebraic data distinguishing $C(M)$ from $C^\infty(M)$ is that $C^\infty(M)$ is equipped with a space of derivations which is a module over the algebra $C^\infty(M)$.
A derivation in this case is an $\mathbb R$-linear map $D$ of $C^\infty(M)$ to itself satisfying Leibniz's product rule: $D(fg) = D(f)g + fD(g)$ for all $f,g\in C^\infty(M)$.
I don't think the Gelfand duality itself is different from what you'd expect. In fact, the point of the Gelfand duality in this case would be to prove that $C(M)$ is the closure of $C^\infty(M)$ under the compact-open topology. The differentiable manifold structure is given by the derivations.
