Graded commutative $R$-algebras 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...
 A: 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 $R[X]/(2X^2) \cong R \cdot 1 \oplus R \cdot X \oplus \bigoplus_{i \geq 2} R/(2) \cdot X^i$.
