Is parallelizability equivalent to the set of vector fields being free? We have the $C^{\infty}(M)$-module $\mathcal{D}^1(M)$ of vector fields over a $C^{\infty}$ manifold $M$. Is being parallelizable equivalent to this module being free, of dimension $n$?
I have the impression it is, since we can take as a basis the $n$ vector fields given by the assumption that $M$ is parallelizable and decompose every vector field in their components in each tangent space, and reciprocally the basis of the module would give the vector fields needed to parallelizability. Am I overlooking something?
If this is true, is it ever possible for $\mathcal{D}^1(M)$ to be a free module of a different dimension, other than $n$? 
 A: That's correct, although you need to say a few words to justify the claim that if $n$ vector fields form a basis for $\mathcal{D}^1(M)$ over $C^\infty(M)$, then they are linearly independent at every point (those words are, if they weren't linearly independent at a point, then some tangent vector at that point would be linearly independent of them, and then a vector field that takes that value at the point could not be in their span over $C^\infty(M)$).
It is impossible for $\mathcal{D}^1(M)$ to be free of any other rank.  One way to show this is to note that for any $p\in M$, the ideal $I_p\subset C^\infty(M)$ of functions vanishing at $p$ is a maximal ideal with quotient $\mathbb{R}$, and that $I_p\mathcal{D}^1(M)$ is exactly the vector fields vanishing at $p$ (this is not entirely obvious, but is easy locally and follows globally using bump functions).  Thus $\mathcal{D}^1(M)/I_p\mathcal{D}^1(M)$ can be naturally identified with the tangent space at $p$, and thus must be $n$-dimensional over $C^\infty(M)/I_p\cong\mathbb{R}$.  So if $\mathcal{D}^1(M)$ is free, it can only be of rank $n$.
