I was skimming through Taylor's book "Several complex variables with connections to algebraic geometry and Lie groups" (first chapter was the best introduction i have ever read!), glimpsed to section 7.7 on de Rham cohomology and was stuck by the definition of differential forms, which are something that i have yet to grasp.
{Personal recap, jump to the end for the actual question}
Let $U \subseteq \mathbb{R}^n$ be an open set and consider the ring/algebra $A = C^{\infty}(U, \mathbb{R})$, one is interested in the definition of differential forms.
Given an $A$-module $M$ one defines the exterior algebra $\Lambda(M)$ as the universal exterior algebra; in particular if $M$ is free of finite rank then its exterior algebra is free of finite rank with a known description.
My naive understanding of differential forms was the following.
If $f \in C^{\infty}(U, \mathbb{R})$ then there is a best linear approximation, pointwise given as
$$
f(x) - f(x_0) = \partial_if(x_0) \, (x^i - x_0^i)
$$
with $C^{\infty}$ derivatives. Since there are $n$ directions of linearization, one considers
$$M = A[\mathrm{d}x^1, \ldots, \mathrm{d}x^n] = A^n$$
and its exterior algebra $\Omega(U)$. Then there is a unique graded morphism $\mathrm{d} \colon \Omega(U) \to \Omega(U)$ such that
- $\mathrm{d} f = \partial_i f \, \mathrm{d} x^i$;
- $\mathrm{d}^2 = 0$;
- $ \mathrm{d}(\alpha \wedge \beta) = \mathrm{d} \alpha \wedge \beta + (-1)^r \alpha \wedge \mathrm{d} \beta $.
The first requirement is the linearization, the second is a way to make this a chain complex, and the third is Leibniz rule (it should be enough to have it on $1$-forms, since a morphism $M \to M$ induces a unique morphism on the free exterior algebra?).
Actual question:
Taylor considers the same (sheaf of) ring $A$ and then defines the sheaf of $C^{\infty}$-modules $D^1$ of derivations, which is (uniformly) free of rank $n$.
Then he considers the exterior algebra and so on.
What is the reason to define differential forms starting from derivations?
Could it be that given any ring $A$ one can formally define derivations, consider the exterior algebra, define an associated chain complex and so on? ("de Rham cohomology of rings"?)
Then the case $U \subseteq \mathbb{R}^n$ is special because the derivations form a free f.g. module?
Any thought and / or reference is well appreciated. Also, sorry for my ignorance.