Is there a computer programm that can list all finite dimensional commutative local (nonsemisimple) k-algebras $K[x_1,x_2,...,x_n]/I$ over a finite field k (of order $q$) with $J^l=0$ for a given $q, n$ and $l$ , when $J=\langle x_1,...,x_n\rangle$ denotes the Jacobson radical and $I \subseteq J^2$?

One way might be as follows: Let $A$ be the algebra $K[x_1,x_2,...,x_n]/J^l$, then one just has to find all ideals $I$ (submodules) of $A$ and look at the quotients.

Is there an estimate how many there are (up to isomorphism or not)?

How and in what time can a computer check if two such algebras are isomorphic?

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    $\begingroup$ These are parametrized by the Hilbert scheme and they can be very complicated. I doubt whether there are any reasonable way you can write them all down. $\endgroup$ – Mohan Jul 29 '17 at 0:46
  • $\begingroup$ Yes but maybe choosing q to be 2 or 3 and l at most 4 one might hope that a computer can list all algebras to do some experiments with them. $\endgroup$ – Mare Jul 29 '17 at 0:58
  • $\begingroup$ I doubt whether these can be written down for large $\dim_K K[x_1,\ldots, x_n]/I$. $\endgroup$ – Mohan Jul 29 '17 at 2:36
  • $\begingroup$ Maybe we can restrict at first to $n \leq 3$ and $l \leq 4$ at first with $q=2$ or $q=3$. That sounds doable for a computer. $\endgroup$ – Mare Jul 29 '17 at 8:48
  • $\begingroup$ It is not exactly a Hilbert scheme. Hilbert schemes (of zero-dimensional subschemes of affine space) parametrize algebras $A=k[x_1,\dotsc,x_n]/I$ where $\dim_k(A)$ is a fixed integer. But the power $l$ so that $J^l=0$ can vary as $A$ varies across the Hilbert scheme. That power—more precisely, the value $l-1$, the maximum so that $J^{l-1} \neq 0$—is called the *socle degree*, and $n$ is called the embedding dimension of the algebra. You're asking for the algebras of given embedding dimension and socle degree. $\endgroup$ – Zach Teitler Jul 2 '18 at 8:34

The paper https://arxiv.org/pdf/1201.3529.pdf shows there are an astronomical number of commutative semigroups satisfying the product of any 3 elements is 0. The contracted semigroup algebra (the zero of the semigroup is identified with the zero of the algebra) is then an algebra of your sort. So there is no way to enumerate them all.

They also show there is still an astronomical number of these semigroups up to isomorphism. I would be very surprised if two such semigroups can have isomorphic algebras without being isomorphic since $J^3=0$ doesn't give you much room but I have not tried to prove this.


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