# How many commutative local algebras are there?

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?

• 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. – Mohan Jul 29 '17 at 0:46
• 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. – Mare Jul 29 '17 at 0:58
• I doubt whether these can be written down for large $\dim_K K[x_1,\ldots, x_n]/I$. – Mohan Jul 29 '17 at 2:36
• 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. – Mare Jul 29 '17 at 8:48
• 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. – Zach Teitler Jul 2 '18 at 8:34

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.