The exterior algebra is a superalgebra? 
Can someone explain how the exterior algebra of a vector space or a module over a commutative ring is a superalgebra? 

The exterior algebra has an obvious $\mathbb{Z}$-grading, but I don't see where the $\mathbb{Z}_2$-grading comes from. I'm not familiar with superalgebras and supervector spaces. I'm just curious because I've seen this mentioned a few times.
 A: The $\mathbb{Z}_2$ grading is easy enough to anticipate. Given an integer it is either even or odd. So, there's your grading. A one-form is odd. A two-form is even. Even elements commute with all other elements under the wedge product whereas the product of odd elements anticommute. All of this is plainly seen in:
$$ \alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha $$
Notice, for $p,q \in 2\mathbb{Z}+1$ we have $pq \in 2\mathbb{Z}+1$ and so $(-1)^{pq}=-1$. Otherwise, if either $p$ or $q$ is in $2\mathbb{Z}$ then $pq \in 2 \mathbb{Z}$ hence $(-1)^{pq}=1$. In my experience, something similar equally well-applies to supernumbers, superfunctions, supervectorspaces, super Lie groups you name it. In my work, the odd/even is usually inherited from the corresponding parity of the supernumbers. However, I think the "odd" or "even" in other work is essentially assigned from the outset so understanding where it came from may not be a wise move. Rather, the more general question to ask, is why do we look to place a $\mathbb{Z}_2$ grading on objects. I think the natural answer to that is in physics. Classical field theory initiated the use of commuting and anticommuting Grassmann variables for bosons and fermions in the 1960's. The other, perhaps more honest, but, perhaps less satisfying answer: because we can.
A: Let $V$ be an (free) module over a ring $R$ of finite rank. Then, 
$$ \bigwedge_R^{\bullet} V = \mathcal{T}(V)/\mathcal{A}(V)= R \oplus (V \otimes V) \oplus (V \otimes V \otimes V) \oplus \cdots \oplus V^{\otimes n} / ( v \otimes v).   $$
Where $\mathcal{T}V$ denotes the tensor algebra and $\mathcal{A}(V) \subseteq \mathcal{T}V$ denotes the ideal generated by $ v \otimes v$ for all $v \in V$. By definition of the quotient,
$$ v \wedge v' = - v' \wedge v$$
$$ (v \wedge v') \wedge (w \wedge w') =  (w \wedge w') \wedge (v \wedge v')$$
That is element $ v \wedge v' \in \bigwedge^1 V$ anticommute while elements $(v \wedge v') \wedge (w \wedge w') \in \bigwedge^2 V$ commute. This extends to the following: Elements in $\bigwedge^{2n+1} V$ anticommute while elements in $\bigwedge_R^{2n} V$ commute. 
We can decompose $\bigwedge^{\bullet} V$ as ,  $$ \bigwedge_R^\bullet(V) := (\bigoplus_n \bigwedge^{2n} V) \oplus (\bigoplus_n \bigwedge^{2n+1} V) $$ 
where the decomposition is written as to denote the obvious superalgebra structure over $R$. 
That is,  if $m, m' \in \oplus \bigwedge^{2n+1}V$ we see that $m \wedge m'= (-1)^{|m||m'|} m' \wedge m = - m' \wedge m$ and if $m, m' \in \oplus \bigwedge^{2n}V$ then $m \wedge m' = (-1)^{|m||m'|} m' \wedge m = m' \wedge m$. Therefore it follows the rule of signs. 
