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.

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    $\begingroup$ I’ve edited the title, although the original was much better! Hope you don’t mind. $\endgroup$ Jun 28 '18 at 17:11
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    $\begingroup$ @JeremyRickard, what is this? Commutativity for ants? $\endgroup$
    – LSpice
    Jun 28 '18 at 17:24
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    $\begingroup$ I thought ants commute all the time (e.g. for food). $\endgroup$
    – M.G.
    Jun 28 '18 at 17:49
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    $\begingroup$ 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 $i < j$ ? $\endgroup$ Jun 28 '18 at 18:56
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    $\begingroup$ That is, it is still quite interesting as an Abelian group, but the multiplication is kind of boring. $\endgroup$
    – Vincent
    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 $d\ge 3$ .


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