Help with algebraic manipulation to prove that $\Omega (M)$ constructed from $\Omega^1 (M)$ forms an algebra over $C^\infty (M)$ In Baez´s Gauge Theories, Knots and Gravity he states that the differential forms on a n dimensional manifold M, $\Omega (M) = {\bigoplus}_p  \Omega^p (M)$, constructed from $\Omega^1(M)$ and the wedge product between 1-forms is indeed an algebra over $C^\infty (M)$ i.e., that for $\omega, \mu, \nu \in \Omega (M)$ and $f,g \in C^\infty (M)$ we have the 9 properties of an algebra: 


*

*$\omega + \mu = \mu + \omega$

*$\omega + (\mu + \nu) = (\omega + \mu) + \nu$

*$\omega \wedge (\mu \wedge \nu) = (\omega \wedge \mu)\wedge \nu$ , etc. 


My question is, for example, if I want to prove properties 1 and 3, because they are differential forms, do I need to feed them a vector field to prove the equalities? and it still confuses me how do I prove the properties if $\omega, \mu, \nu \in \Omega (M)$ are not p-forms with a given p, with $0 \leq p \leq n$ ? 
More concretly, my first guess to prove property 1 is to note that we can write the forms as $$\omega= \sum_{{i}_1...{i}_p =1}^{n} {\omega}_{{i}_1...{i}_p} dx^{{i}_1} \wedge...\wedge dx^{{i}_p}$$ $$\mu= \sum_{{i}_1...{i}_p =1}^{n} {\mu}_{{i}_1...{i}_p} dx^{{i}_1} \wedge...\wedge dx^{{i}_p}$$ 
so $\omega+\mu = \sum_{{i}_1...{i}_p =1}^{n} {\omega}_{{i}_1...{i}_p} dx^{{i}_1} \wedge...\wedge dx^{{i}_p} + \sum_{{i}_1...{i}_p =1}^{n} {\mu}_{{i}_1...{i}_p} dx^{{i}_1} \wedge...\wedge dx^{{i}_p} = \sum_{{i}_1...{i}_p =1}^{n} {\mu}_{{i}_1...{i}_p} dx^{{i}_1} \wedge...\wedge dx^{{i}_p} +\sum_{{i}_1...{i}_p =1}^{n} {\omega}_{{i}_1...{i}_p} dx^{{i}_1} \wedge...\wedge dx^{{i}_p} = \mu + \omega$ am I right?
 A: Yes, you could feed the forms appropriate $p$-tuples of tangent vectors to prove those properties. The fact that $p$ is not given explicitly simply indicates these properties hold across all appropriate degrees. I'm not sure your last equation is anything more than changing notation, I wouldn't consider it a proof unless you already have other results to which I'm unaware.
I suppose the base of your question is really how to construct the exterior algebra. The long, very concrete, way is to begin by defining the tensor product:
$$ (T \otimes S)(x,y) = T(x)S(y) $$
Then use that to build the wedge:
$$ T \wedge S = T \otimes S - S \otimes T $$
In this view you get to prove some properties of the wedge-product as derived from the tensor product. In particular, properties 1 and 3 follow for $\wedge$ because they're true for $\otimes$ and the properties filter nicely through the construction:
$$ T_1 \wedge T_2 \wedge \cdots \wedge T_k = \sum_{\sigma} sgn(\sigma) T_{\sigma(1)} \otimes T_{\sigma(2)} \otimes \cdots \otimes T_{\sigma(k)} $$
The better way to construct the algebra is by taking a quotient and using a universal property. In that way of thinking these properties are essentially automatic and characterize what makes the exterior algebra what it is. I suspect you'll find the universal construction if you start hunting through the posts linked to your post.
