Classification of real algebra with zero divisors (subquotient of Clifford algebra?) Consider the even-dimensional real vector space $\Bbb R^{2N}$. We construct the algebra as follows:


*

*Pick a basis in this space.

*Partition the $2N$ basis elements into $N$ pairs of zero divisors, called $e_1, \hat e_1, e_2, \hat e_2, ..., e_N, \hat e_N$.

*For all $n$, $e_n \hat e_n = 0$.

*For all $n$, $e_n^2 = e_n$, and $\hat e_n^2 = \hat e_n$.

*For all $m, n$, $e_m e_n = e_n e_m$, $\hat e_m \hat e_n = \hat e_n \hat e_m$, and $e_n \hat e_m = \hat e_m e_n$.


In other words, the basis elements are all idempotent, come in zero-divisor pairs, and commute. The full algebra is the construction then generated by the above relations, given N generators.
This has the structure of a graded algebra, so that for products $e_m e_n$, these exist in a larger space that the original vector space is a subspace of, in a similar to the exterior powers of the exterior algebra. Assuming we throw in the field of scalars as grade-0 vectors, you end up with the full algebra being finite-dimensional.
My question: how do you classify this algebra? I can see this possibly being a subquotient of the Clifford algebra, but it seems messy to look at it that way.
An interesting case is the 3-dimensional real algebra yielded by this construction - you get the field of real numbers, plus two additional elements $i$ and $j$ which have the property that $ij = ji = 0$, $i^2 = i$, $j^2 = j$, and also $(i+j)^2 = i+j$.

The algebra is a quotient of a polynomial algebra: 
$$
A_N=\Bbb{R}[x_1,y_1,x_2,y_2,…,x_N,y_N]/\langle x^2_i=x_i,y^2_i=y_i,x_iy_i=0\rangle,
$$
 and $\dim(A_N)=3^N$. In fact, the non zero monomials in that algebra can be reduced to monomials where every variable has at most power one, and if $x_i$ is in that monomial, then $y_i$ is not. So given a monomial, for each $i$ we have three choices: Either $x_i$ is a factor, or $y_i$ is a factor or none of them is a factor. This leaves us with $3^N$ monomials that give the linear basis of the algebra. If for all $i$ the choice is that none of $x_i$, $y_i$ is a factor, then we obtain the scalars with basis element "1". The grading is simply the usual degree of polynomials.
