# Curious anti-commutative ring

Has anyone seen the ring $$\Lambda[x_0, x_1, x_2, \ldots]/(x_i x_j - (i+1) x_0 x_{i+j})$$ in some natural context?

Here $$\Lambda[x_0, x_1, x_2, \ldots]$$ is the (graded-)commutative algebra (either over the integers or the integers localized at 2) freely generated by elements $$x_0,x_1,x_2,\ldots$$ of odd homological degrees, so that $$x_i x_j = - x_j x_i$$. In particular, we only get $$2 x_i^2 = 0$$, not $$x_i^2 = 0$$. I probably shouldn't write $$\Lambda$$ here: in characteristic $$2$$ or integrally, $$\Lambda$$ usually adds the relation $$x_i^2 = 0$$.

ADDED NOTE: In the meantime, I have found a better way of solving the problem in which this arose, so it is merely a curiosity now. I am happy to delete it if people wish.

Forgive me for posting this to 'Commutative algebra', but as a topologist, commutative means $$x y = (-1)^{(\deg x)(\deg y)} yx$$, and my $$x_i$$ are in odd degrees.

SECOND NOTE: This algebra has now shown up in another context, so Vladimir's answer below has been quite useful. Thanks to Vladimir and MO.

• I’ve edited the title, although the original was much better! Hope you don’t mind. Jun 28 '18 at 17:11
• @JeremyRickard, what is this? Commutativity for ants? Jun 28 '18 at 17:24
• I thought ants commute all the time (e.g. for food).
– M.G.
Jun 28 '18 at 17:49
• What exactly is $x_i x_j - \left(i+1\right) x_0 x_{i+j}$ quantified over? All pairs of nonnegative $i$ and $j$ ? Only those with $i < j$ ? Jun 28 '18 at 18:56
• That is, it is still quite interesting as an Abelian group, but the multiplication is kind of boring. Jun 29 '18 at 8:14

I noticed this now, and I want to remark that the underlying abelian group can in fact be described very precisely. To do that, note that:

(1) the defining relations easily imply that the abelian group of elements of degree $$d\ge 2$$ in this algebra is certainly generated by $$x_0^{d-1}x_k$$, $$k\ge 0$$, and

(2) as discussed in the comments, there are the relations $$(i+j+2)x_0x_{i+j}=0$$ that follow from the defining relations and anticommutativity; effectively, these give just one relation for each $$n$$, namely $$(n+2)x_0x_n=0$$.

Now let me (inspired by typical Gröbner bases calculations) consider the following two chains of equalities: $$x_ix_jx_k=(i+1)x_0x_{i+j}x_k=(i+1)(i+j+1)x_0^2x_{i+j+k}$$ and $$x_ix_jx_k=(j+1)x_ix_0x_{j+k}=-(j+1)x_0x_ix_{j+k}=-(j+1)(i+1)x_0^2x_{i+j+k}.$$ They imply that $$(i+1)(i+2j+2)x_0^2x_{i+j+k}=0$$ for each choice of $$i$$ and $$j$$ with $$i+j\le n$$. In particular, if $$n\ge 1$$, we may take $$i=n-1$$, $$j=1$$, obtaining $$n(n+3)x_0^2x_n=(n-1+1)(n-1+2+2)x_0^2x_n=0.$$ But $$(n+2)x_0x_n=0$$ implies $$(n+1)(n+2)x_0^2x_n=0$$, so by subtraction we see that $$2x_0^2x_n=0$$. Moreover, no further relations can be obtained in a similar way, because once we have the 2-torsion property, we have $$(i+1)(i+2j+2)x_0^2x_n=(i+1)ix_0^2x_n=0,$$ since $$(i+1)i$$ is always even.

In fact, using a version of Gröbner bases (or rewriting systems) for ideals in free anticommutative algebras, one can see that the system of all the defining relations thus obtained, namely $$\begin{cases} x_ix_j=(i+1)x_0x_{i+j},\\ (n+2)x_0x_n=0,\\ 2x_0^2x_n=0 \end{cases}$$ is complete, and so your ring as an abelian group :

is freely generated by $$1$$ in degree $$0$$,

is freely generated by $$x_0,x_1,\ldots$$ in degree $$1$$,

is the product of cyclic groups of orders $$2,3,\ldots$$ generated by $$x_0^2, x_0x_1, x_0x_2, \ldots$$ respectively in degree $$2$$,

is a product of countably many cyclic groups of order $$2$$ generated by $$x_0^d, x_0^{d-1}x_1, x_0^{d-1}x_2, \ldots$$ in each degree $$d\ge 3$$ .