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Im looking for an algorithm that does the following in a quick way:

Input: Natural number $r \geq 2$, natural number $s \geq 3$, prime power $q$. Output:

Finds all two-sided ideals in $J^2/J^s \subseteq F_q <x_1,...,x_r>/J^s$ ,where $A=F_q <x_1,...,x_r>$ is the noncommutative polynomial ring in r variables over the finite field with q elements and $J=<x_1,...,x_r>$ is the ideal generated by $x_1, ... , x_r$.

How many ideals are there?

And how many up to isomorphism? (Two ideals $I_1$ and $I_2$ are isomorphic iff $A/ I_1$ and $A/I_2$ are isomorphic as $K$-algebras)

It might be also interesting to replace the noncommutative polynomial ring by a commutative one.

Note that the problem can also be formulated as to find all admissible ideals containing $J^2$ of a quiver algebra with one point and $r$ loops over a finite field.

As a motivation I offer a 100 Euro reward if someone can make a quick programm which works for small r and s (lets say $r=2,3$ and $s=2,3,4$ and choosing $q=2,3,5$). It should be programmed with the GAP-packet qpa.

edit: I rewrote the question to make it shorter.Easiest special case is a question on stackexchange: https://math.stackexchange.com/questions/2322214/number-of-ideals-with-gap .

Im looking for an algorithm that does the following in a quick way:

Input: Natural number $r \geq 2$, natural number $s \geq 3$, prime power $q$. Output:

Finds all two-sided ideals in $J^2/J^s \subseteq \mathbb{F}_q \langle x_1,...,x_r\rangle/J^s$ ,where $A=\mathbb{F}_q \langle x_1,...,x_r\rangle $ is the noncommutative polynomial ring in r variables over the finite field with q elements and $J=(x_1,...,x_r)$ is the ideal generated by $x_1, ... , x_r$.

How many ideals are there?

And how many up to isomorphism? (Two ideals $I_1$ and $I_2$ are isomorphic iff $A/ I_1$ and $A/I_2$ are isomorphic as $K$-algebras)

It might be also interesting to replace the noncommutative polynomial ring by a commutative one.

Note that the problem can also be formulated as to find all admissible ideals containing $J^2$ of a quiver algebra with one point and $r$ loops over a finite field.

As a motivation I offer a 100 Euro reward if someone can make a quick programm which works for small r and s (lets say $r=2,3$ and $s=2,3,4$ and choosing $q=2,3,5$). It should be programmed with the GAP-packet qpa.

edit: I rewrote the question to make it shorter.Easiest special case is a question on stackexchange: https://math.stackexchange.com/questions/2322214/number-of-ideals-with-gap .

Im looking for an algorithm that does the following in a quick way:

Input: Natural number $r \geq 2$, natural number $s \geq 3$, prime power $q$. Output:

Finds all two-sided ideals in $J^2/J^s \subseteq F_q <x_1,...,x_r>/J^s$ ,where $A=F_q <x_1,...,x_r>$ is the noncommutative polynomial ring in r variables over the finite field with q elements and $J=<x_1,...,x_r>$ is the ideal generated by $x_1, ... , x_r$.

How many ideals are there?

And how many up to isomorphism? (Two ideals $I_1$ and $I_2$ are isomorphic iff $A/ I_1$ and $A/I_2$ are isomorphic as $K$-algebras)

It might be also interesting to replace the noncommutative polynomial ring by a commutative one.

Note that the problem can also be formulated as to find all admissible ideals containing $J^2$ of a quiver algebra with one point and $r$ loops over a finite field.

As a motivation I offer a 100 Euro reward if someone can make a quick programm which works for small r and s (and choosing q to be 2 or 3...). It should be programmed with the GAP-packet qpa.

edit: I rewrote the question to make it shorter.Easiest special case is a question on stackexchange: https://math.stackexchange.com/questions/2322214/number-of-ideals-with-gap .

Im looking for an algorithm that does the following in a quick way:

Input: Natural number $r \geq 2$, natural number $s \geq 3$, prime power $q$. Output:

Finds all two-sided ideals in $J^2/J^s \subseteq F_q <x_1,...,x_r>/J^s$ ,where $A=F_q <x_1,...,x_r>$ is the noncommutative polynomial ring in r variables over the finite field with q elements and $J=<x_1,...,x_r>$ is the ideal generated by $x_1, ... , x_r$.

How many ideals are there?

And how many up to isomorphism? (Two ideals $I_1$ and $I_2$ are isomorphic iff $A/ I_1$ and $A/I_2$ are isomorphic as $K$-algebras)

It might be also interesting to replace the noncommutative polynomial ring by a commutative one.

Note that the problem can also be formulated as to find all admissible ideals containing $J^2$ of a quiver algebra with one point and $r$ loops over a finite field.

As a motivation I offer a 100 Euro reward if someone can make a quick programm which works for small r and s (lets say $r=2,3$ and $s=2,3,4$ and choosing $q=2,3,5$). It should be programmed with the GAP-packet qpa.

edit: I rewrote the question to make it shorter.Easiest special case is a question on stackexchange: https://math.stackexchange.com/questions/2322214/number-of-ideals-with-gap .

Im looking for an algorithm that does the following:

Input: Natural number $r \geq 2$, natural number $s \geq 3$, prime power $q$. Output: (Up to isomorphism) all quiver algebras (with admissible ideal) over the field $F_q$ with 1 point and $r$ loops and having the property that its Jacobsonradical $J$ satisfies $J^s=0$. Call the number of those algebras $a_{r,s,q}$. What can be said about $a_{r,s,q}$? Probably its too complicated to give a closed formula but maybe there are good bounds to get an idea how large those numbers can get. Even restricting to r=2 or r=3 would be interesting for some fixed q.

So the main question is: What is the fastest possible algorithm to get the output, when one wants the algebras up to isomorphism. It might be too complicated, since I have no idea how to check wheter two algebras are isomorphic in a quick way. So another question would be to get all algebras not up to isomorphism.

As a motivation I offer a 100 Euro reward if someone can make a quick programm which works for small r and s (and choosing q to be 2 or 3...). It should be programmed with the GAP-packet qpa.

I posted this some time ago here: https://math.stackexchange.com/questions/1592545/finding-all-admissible-ideals-of-a-given-quiver-with-gap-qpa

Im looking for an algorithm that does the following in a quick way:

Input: Natural number $r \geq 2$, natural number $s \geq 3$, prime power $q$. Output:

Finds all two-sided ideals in $J^2/J^s \subseteq F_q <x_1,...,x_r>/J^s$ ,where $A=F_q <x_1,...,x_r>$ is the noncommutative polynomial ring in r variables over the finite field with q elements and $J=<x_1,...,x_r>$ is the ideal generated by $x_1, ... , x_r$.

How many ideals are there?

And how many up to isomorphism? (Two ideals $I_1$ and $I_2$ are isomorphic iff $A/ I_1$ and $A/I_2$ are isomorphic as $K$-algebras)

It might be also interesting to replace the noncommutative polynomial ring by a commutative one.

Note that the problem can also be formulated as to find all admissible ideals containing $J^2$ of a quiver algebra with one point and $r$ loops over a finite field.

As a motivation I offer a 100 Euro reward if someone can make a quick programm which works for small r and s (and choosing q to be 2 or 3...). It should be programmed with the GAP-packet qpa.

edit: I rewrote the question to make it shorter.Easiest special case is a question on stackexchange: https://math.stackexchange.com/questions/2322214/number-of-ideals-with-gap .