Let $R$ be a commutative ring and $T$ a graded commutative $R$-algebra. This means that $\,T$ consists of a collection $\{T_n\}_{n\geq 0}$ of $\,R$-algebras, where the elements of $R_n$ are called homogeneus of degree $n$. The graded commutativity is expressed by the following axiom:

  • if $a\in R_s$ and $\,b\in R_t$ then $a\cdot b = (-1)^{s\cdot t} \;b\cdot a$

Now consider the polynomial ring in one variable $R[X]$, and declare a grading by $$ |\; X^n\,| = d\cdot n $$ for a chosen $d\in \mathbb{Z}$. If $d$ is even gives $R[X]$ the structure of a graded commutative $R$-algebra. On the other hand if $d$ is odd we need to mod out the ideal $(2X^2)$ from $R[X]$ to obtain such a structure, since we have the relation $$X\cdot X = (-1)^{d^2} X\cdot X \implies 2X^2 = 0. $$

It's easy to prove that for $d$ even there is a bijection $$ \{ \text{Morphisms } f:R[X] \to T \text{ of graded commutative R-algebras}\} \leftrightarrow \{ \text{Elements of } T^d \}, $$ given by the assignment $f\mapsto f(X)$.

My question is: Does this still hold when $d$ is odd? I think it does, but I'm not convinced, I have the feeling that I'm missing something...

  • $\begingroup$ If $a\in R_s$, is $aX^q$ in degree $s+dn$? $\endgroup$ – Avitus Jun 21 '13 at 15:23
  • $\begingroup$ If $a\in R_s$ is of degree $s$ and $b\in R_t$ is of degree $t$ then $a\cdot b \in R_{s+t}$ is of degree $s+t$. So in particular the degree of $aX^q$ is $s+d\cdot q$. $\endgroup$ – Abramo Jun 21 '13 at 15:44
  • $\begingroup$ I do not fully understand the need of the extra grading on the polynomials. If $d\neq 0$, then the evaluation at zero $ev: R[X]\rightarrow R$ is not a degree $0$ morphism of $R$-modules. Consider for example the case $\mathbb K=R$ concentrated in degree $0$, with $\mathbb K[e_1,\dots,e_n]$ also concentrated in degree $0$ and $\{e_i\}$ basis of a finite dim. vector space $X$ over $\mathbb K$. Then the Koszul complex $K^{-i}:=\mathbb K[e_1,\dots,e_n]\otimes \wedge^{i}X$ has differential of degree $+1$. This is a natural choice $\endgroup$ – Avitus Jun 21 '13 at 16:55

If $d$ is odd, then $R[X]/(2X^2)$ (with $|X|=d$) is commutative but also graded-commutative: It suffices to prove that $X^n X^m = (-1)^{nm} X^m X^n$. This is clear if $n=0$ or $m=0$. But if $n,m \geq 1$, it follows from $0 = 2 X^{n+m}$.

Now if $T$ is graded-commutative and $t \in T^d$, there is a unique homomorphism of graded $R$-algebras $R[X] \to T$ mapping $X \mapsto t$. But since we have $2 t^2=0$ (this follows by applying graded-commutativity to $t$ with itsself), it follows that $2X^2$ lies in the kernel. Hence, it factors uniquely through $R[X]/(2X^2)$.

Hence, $R[X]/(2X^2)$ satisfies the desired universal property: It is the universal graded-commutative $R$-algebra containing an element of degree $d$. Note that when $2=0$ in $R$, this simplifies to the polynomial algebra $R[X]=\mathrm{Sym}(R)$, and when $2 \in R^*$, it simplifies to the exterior algebra $R[X]/(X^2)=\Lambda(R)$, aka the algebra of dual numbers. This gives a complete description in the case of fields. However, for $R=\mathbb{Z}$, I don't know any better description. Note that there is an isomorphism of $R$-modules $R[X]/(2X^2) \cong R \cdot 1 \oplus R \cdot X \oplus \bigoplus_{i \geq 2} R/(2) \cdot X^i$.


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