Number of ideals with GAP Let $A=K\langle x,y\rangle$ be the polynomial in non-commuting variables $x,y$ over a finite field $K$ with $q$ elements and $J=\langle x,y\rangle$ the ideal generated by $x$ and $y$.
I want to find all ideals I having the property that $J^4 \subseteq I \subset J^2$.
It is a finite problem and the most stupid idea goes as follows:
We can write $I=J^4+X$ , where $X$ is the ideal generated by a subset of the $K$-vector space $V$ spanned by the elements $x^2, xy, yx ,y^2, xyx, xy^2, x^3, x^2y, yx^2, yxy, y^2x, y^3$.
Now $V$ has $q^{12}$ elements and $X$ can be any subset.
So I have $2^{q^{12}}$ possibilities, which are $2^{4096}$ when the field has $2$ elements :(( .

Is there any good idea to improve finding all possible $X$ using GAP?

For example when $X$ contains $x^2$ , then it is not necessary that there is also the relation $x^2y$ or $x^3$ since they are in the ideal automatically then.

Is it realistic that GAP can produce all ideals and how many are there?

It would be also interesting to see how many ideals there are up to isomorphism, where two ideals $I_1$ and $I_2$ are isomorphic when $A/I_1$ and $A/I_2$ are isomorphic as $K$-algebras.
For simplicity we can assume first that $K$ is the field with two or three elements.
 A: QPA can in principle find all the different ideals in question through the command 
AllSubmodulesOfModule

(in the latest version on GitHub, https://github.com/gap-packages/qpa). 
Constructing the algebra $A = K\langle x, y\rangle/J^3$ and considering the ideal $J^2/J^3$ as a module $M$ over the enveloping algebra of $A$, then finding all ideals $I$ with the property that $J^3\subseteq I \subseteq I^2$ is the same as finding all submodules of $M$. Here is how you can do it in QPA:
Q := Quiver(1,[[1,1,"a"],[1,1,"b"]]);;
KQ:= PathAlgebra(GF(2), Q);;
AssignGeneratorVariables(KQ);;
rels := AddNthPowerToRelations( KQ, [], 3 );;
A:= KQ/rels;;
M := AlgebraAsModuleOverEnvelopingAlgebra(A);;
radM := RadicalOfModule(M);;
rad2M := RadicalOfModule(radM);;
test := AllSubmodulesOfModule( rad2M );;
Length(Flat(test));
27;

The number of ideals are 27 in this case.  For the case $J^4\subseteq I \subseteq J^2$, the number of ideals are quite a bit larger.  It took long time to complete, but I lost the result (maybe corresponding number was 573).
I hope that this is a helpful comment.
Best regards, The QPA-team.
