When is $C^\infty(M)$ a division ring?

Question

Given a smooth manifold $M$ and its tangent bundle $TM$, the set of vector fields on $M$, denoted $\Gamma(TM)$, is a module over the commutative algebra of smooth functions on $M$, denoted $C^\infty(M)$. Now, since every nonzero module over a division ring has a basis by Zorn's lemma, I was wondering if there are any smooth manifolds $M$ with $C^\infty(M)$ as a division ring. If so, what are the conditions on $M$?

Answer 1

If $m \in M$ is any point, $C^{\infty}(M)$ has a maximal ideal given by the kernel of the evaluation map $C^{\infty}(M) \to \mathbb{R}$ at $m$. $C^{\infty}(M)$ is a field iff it has no nontrivial ideals so this ideal must be zero, and (with a little effort) this implies $m = M$; that is, $M$ must consist of a single point.

As for your motivation, vector fields form not only a module but a finitely generated projective module over $C^{\infty}(M)$ (by the smooth Serre-Swan theorem), so a less restrictive version of your question is to ask when every such module is free. This is true, for example, if $M = \mathbb{R}^n$, but otherwise almost never true, e.g. it is never true if the (even) $K$-theory $K^0(M)$ of $M$ is nontrivial, and $K$-theory is nontrivial whenever even rational cohomology $H^{2\bullet}(M, \mathbb{Q})$ is nontrivial due to the existence of the Chern character isomorphism, at least if $M$ is compact.

Finally, the least restrictive version of your question is to ask when vector fields themselves form a free module. Manifolds with this property are called parallelizable; examples include the spheres $S^1, S^3, S^7$ as well as any Lie group and any orientable $3$-manifold, but generally speaking examples are rare. A parallelizable manifold has the property that all of its characteristic classes vanish which is quite a strong condition.