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?


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."


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.

  • $\begingroup$ Thank you so much! I have some questions just to clarify some words. (1) A "degree 3" element is just a particular case, right? It works for any odd-degree element, such as $|\alpha|=1$, right? (2) Are The "algebras" you mentioned in the second paragraph free graded-commutative algebra? $\endgroup$ – Herman Chu Oct 11 '18 at 18:43
  • $\begingroup$ @Herman (1) Somehow I misread your post as $\alpha$ being of degree 3. Yes, this is all odd degree. (2) yup! $\endgroup$ – user98602 Oct 11 '18 at 19:36

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