I am getting myself acquainted to superalgebras. One often comes across odd polynomial rings of the form


$$k\langle x\_i \rangle\_{i\in I} / (x\_ix\_j -(-1)^{|x\_i||x\_j|} x\_jx\_i)$$


for some index set $I$, where the relation divided out sometimes is required for all choices of $i, j$ (which implies $x\_i^2=0$ if $x\_i$ is odd), and sometimes this is only required for $i\neq j$ (for instance in the definition 2.1.1 of the odd NilHecke Algebra in <https://arxiv.org/abs/1111.1320>).


**Question:** How do I know when $x\_i^2$ should be assumed to vanish? 


**Edit:** Maybe I should clarify why I care. For perfectly supersymmetric algebras, I have nice identities how to drag tensor factors and morphisms past elements. It seems that this gets highly problematic if some elements are not assumed to anticommute with themselves. So why should one consider this?



Generators of odd polynomial superalgbras
Why should one consider both cases? Just as Lie algebras are "built out of copies of $\mathfrak{sl}(2)$", (basic classical) Lie superalgebras are essentially built out of copies of three elementary algebras: $\mathfrak{sl}(2)$, $\mathfrak{sl}(1|1)$, and $\mathfrak{osp}(1|2)$. (This essentially corresponds to the three flavors of roots: even, odd isotropic, and odd anisotropic.) If we look at the positive parts of the enveloping algebras of direct sums of these pieces over some index set $I$, we have


$$U^+(\mathfrak{sl}(2)^{\oplus I})\cong k[e\_i\ |\ i\in I]\qquad U^+(\mathfrak{sl}(1|1)^{\oplus I})\cong k[e\_i\ |\ i\in I]/(e\_ie\_j=-e\_je\_i)\qquad U^+(\mathfrak{osp}(1|2)^{\oplus I})\cong k[e\_i\ |\ i\in I]/(e\_ie\_j=-e\_je\_i\text{ for }i\neq j)$$


where $k$ is the base field, and the $e\_i$ in the skew-commuting versions are odd.


In other words, all three flavors (non-super; super with all odd generators anticommuting; and super with all distinct odd generators anticommuting) arise naturally from the elementary Lie superalgebras, so one is led to consider all three possibilities.


Note, however, that odd elements which do not anticommute with themselves **should** still anticommute when living in different tensor factors: that is,
if $e$ is odd, then $$(1\otimes e)(e\otimes 1)=-(e\otimes 1)(1\otimes e)=-e\otimes e$$
regardless of whether $e^2$ is zero or not.
One way to see that this ought to be the case is that we would still like e.g.
$U^+(\mathfrak{osp}(1|2)^{\oplus I})\cong U^+(\mathfrak{osp}(1|2))^{\otimes I}$ to be true.


