Detecting compactness from the ring of smooth functions

Question

Given a smooth manifold $M$, is there some ring-theoretic property (preferably not mentioning $M$) such that $C^{\infty}(M)$ has this property if and only if $M$ is compact?

Answer 1

As Dmitri Pavlov claims $M$ is compact iff all maximal ideals of $C^\infty(M)$ of codimension $1$. For a more about algebraic properties of $C^\infty(M)$ see this and this MO discussions.

Answer 2

The following is an answer courtesy of a brighter friend:

$M$ is compact iff every maximal ideal in $C^{\infty}(M)$ is finitely-generated

If $M$ is compact, one can show that every maximal ideal is the kernel of some evaluation map and use Hadamard's Lemma to show that it is finitely-generated, as was done in the answer here. Conversely, if $M$ is not compact one can find a maximal ideal not finitely generated, as indicated in Exercise 8.20 in Nestruev's 'Smooth Manifolds and Observables.'