Why is $R[\alpha]/2\alpha^2$ a free graded commutative algebra?

Question

In Hatcher's book, on page 227, he said

Polynomial algebras are examples of free graded commutative algebras, where ‘free’ means loosely ‘having no unnecessary relations.’ In general, a free graded com- mutative algebra is a tensor product of single-generator free graded commutative algebras. The latter are either polynomial algebras $R[α]$ on even-dimension generators α or quotients $R[α]/(2α^2)$ with α odd-dimensional. Note that if $R$ is a field then $R[α]/(2α^2)$ is either the exterior algebra $Λ_R[α]$ if the characteristic of $R$ is not 2, or the polynomial algebra $R[α]$ otherwise. Every graded commutative algebra is a quotient of a free one, clearly.

My question is:

Why is $R[\alpha]/2\alpha^2$ a free graded commutative algebra? I can understand that when $R$ is a field and $R[\alpha]/2\alpha^2$ is either a exterior algebra or a polynomial algebra. Therefore, $R[\alpha]/2\alpha^2$ is a free graded commutative algebra in this case. However, if $R$ is not a field, can we say it is a free graded commutative algebra?

Answer 1

The argument doesn't depend on whether $R$ is a field or not. The point is that if $\alpha$ has odd degree then the Koszul sign rule demands that

$$\alpha \alpha = (-1)^{|\alpha| |\alpha|} \alpha \alpha$$

so we always get $\alpha^2 = - \alpha^2$; this relation is always "necessary."

Answer 2

If $\alpha$ is odd-dimensional, then the graded-commutativity relations force $\alpha \cdot \alpha = - \alpha \cdot \alpha$, so that relation is always true. So he has added no unnecessary relations.

More precisely, given any degree 3 element of a graded commutative $R$-algebra, there is an induced homomorphism from your algebra to mine sending $\alpha$ to my element. So choosing a homomorphism out of your algebra is the same as choosing a degree 3 element in the target. This is what 'free on a basis' means: specifying a map out is precisely the same as specifying where the basis goes.