Number of ideals with GAP
$begingroup$
Let $A=Klangle x,yrangle$ be the polynomial in non-commuting variables $x,y$ over a finite field $K$ with $q$ elements and $J=langle x,yrangle$ 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.
combinatorics ring-theory representation-theory gap constraint-programming
edited Jun 14 '17 at 11:54
Shaun
9,310113684
asked Jun 14 '17 at 11:42
MareMare
676417
$endgroup$
add a comment |
$begingroup$
Let $A=Klangle x,yrangle$ be the polynomial in non-commuting variables $x,y$ over a finite field $K$ with $q$ elements and $J=langle x,yrangle$ 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.
combinatorics ring-theory representation-theory gap constraint-programming
edited Jun 14 '17 at 11:54
Shaun
9,310113684
asked Jun 14 '17 at 11:42
MareMare
676417
$endgroup$
$begingroup$
What do you mean? all my A/I will be finite dimensional since i factor out something larger than J^4. In your example (xy^3) is not a ideal having the right conditions, since it does not contain J^4.
$endgroup$
– Mare
Jun 14 '17 at 12:41
$begingroup$
Yes the problem can be formulated to count the ideals of $F_q <x,y>/J^4$ contained in $J^2/J^4$, where the polynomial ring here is the noncommutative one. If you have a solution for the commutative one, that might be also interesting.
$endgroup$
– Mare
Jun 14 '17 at 12:46
$begingroup$
The problem is to count the ideals of $mathbb{F}_q[x,y]/J^4$. It is $mathbb{F}_q[M_1,ldots,M_k]$ where $M_k$ are the $4 times 4$ matrices with minimal polynomial $M^4 = 0$
$endgroup$
– reuns
Jun 14 '17 at 12:46
$begingroup$
Instead of forming collections of elements, first form the different principal ideals generated by individual elements and then form combinations of principal ideals.
$endgroup$
– ahulpke
Jun 14 '17 at 17:57
add a comment |
$begingroup$
Let $A=Klangle x,yrangle$ be the polynomial in non-commuting variables $x,y$ over a finite field $K$ with $q$ elements and $J=langle x,yrangle$ 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.
combinatorics ring-theory representation-theory gap constraint-programming
edited Jun 14 '17 at 11:54
Shaun
9,310113684
asked Jun 14 '17 at 11:42
MareMare
676417
$endgroup$
Let $A=Klangle x,yrangle$ be the polynomial in non-commuting variables $x,y$ over a finite field $K$ with $q$ elements and $J=langle x,yrangle$ 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.
combinatorics ring-theory representation-theory gap constraint-programming
combinatorics ring-theory representation-theory gap constraint-programming
edited Jun 14 '17 at 11:54
Shaun
9,310113684
asked Jun 14 '17 at 11:42
MareMare
676417
edited Jun 14 '17 at 11:54
Shaun
9,310113684
asked Jun 14 '17 at 11:42
MareMare
676417
edited Jun 14 '17 at 11:54
Shaun
9,310113684
edited Jun 14 '17 at 11:54
Shaun
9,310113684
edited Jun 14 '17 at 11:54
Shaun
9,310113684
9,310113684
asked Jun 14 '17 at 11:42
MareMare
676417
asked Jun 14 '17 at 11:42
MareMare
676417
asked Jun 14 '17 at 11:42
MareMare
676417
676417
$begingroup$
What do you mean? all my A/I will be finite dimensional since i factor out something larger than J^4. In your example (xy^3) is not a ideal having the right conditions, since it does not contain J^4.
$endgroup$
– Mare
Jun 14 '17 at 12:41
$begingroup$
Yes the problem can be formulated to count the ideals of $F_q <x,y>/J^4$ contained in $J^2/J^4$, where the polynomial ring here is the noncommutative one. If you have a solution for the commutative one, that might be also interesting.
$endgroup$
– Mare
Jun 14 '17 at 12:46
$begingroup$
The problem is to count the ideals of $mathbb{F}_q[x,y]/J^4$. It is $mathbb{F}_q[M_1,ldots,M_k]$ where $M_k$ are the $4 times 4$ matrices with minimal polynomial $M^4 = 0$
$endgroup$
– reuns
Jun 14 '17 at 12:46
$begingroup$
Instead of forming collections of elements, first form the different principal ideals generated by individual elements and then form combinations of principal ideals.
$endgroup$
– ahulpke
Jun 14 '17 at 17:57
add a comment |
$begingroup$
What do you mean? all my A/I will be finite dimensional since i factor out something larger than J^4. In your example (xy^3) is not a ideal having the right conditions, since it does not contain J^4.
$endgroup$
– Mare
Jun 14 '17 at 12:41
$begingroup$
Yes the problem can be formulated to count the ideals of $F_q <x,y>/J^4$ contained in $J^2/J^4$, where the polynomial ring here is the noncommutative one. If you have a solution for the commutative one, that might be also interesting.
$endgroup$
– Mare
Jun 14 '17 at 12:46
$begingroup$
The problem is to count the ideals of $mathbb{F}_q[x,y]/J^4$. It is $mathbb{F}_q[M_1,ldots,M_k]$ where $M_k$ are the $4 times 4$ matrices with minimal polynomial $M^4 = 0$
$endgroup$
– reuns
Jun 14 '17 at 12:46
$begingroup$
Instead of forming collections of elements, first form the different principal ideals generated by individual elements and then form combinations of principal ideals.
$endgroup$
– ahulpke
Jun 14 '17 at 17:57
$begingroup$
What do you mean? all my A/I will be finite dimensional since i factor out something larger than J^4. In your example (xy^3) is not a ideal having the right conditions, since it does not contain J^4.
$endgroup$
– Mare
Jun 14 '17 at 12:41
$begingroup$
What do you mean? all my A/I will be finite dimensional since i factor out something larger than J^4. In your example (xy^3) is not a ideal having the right conditions, since it does not contain J^4.
$endgroup$
– Mare
Jun 14 '17 at 12:41
$begingroup$
Yes the problem can be formulated to count the ideals of $F_q <x,y>/J^4$ contained in $J^2/J^4$, where the polynomial ring here is the noncommutative one. If you have a solution for the commutative one, that might be also interesting.
$endgroup$
– Mare
Jun 14 '17 at 12:46
$begingroup$
Yes the problem can be formulated to count the ideals of $F_q <x,y>/J^4$ contained in $J^2/J^4$, where the polynomial ring here is the noncommutative one. If you have a solution for the commutative one, that might be also interesting.
$endgroup$
– Mare
Jun 14 '17 at 12:46
$begingroup$
The problem is to count the ideals of $mathbb{F}_q[x,y]/J^4$. It is $mathbb{F}_q[M_1,ldots,M_k]$ where $M_k$ are the $4 times 4$ matrices with minimal polynomial $M^4 = 0$
$endgroup$
– reuns
Jun 14 '17 at 12:46
$begingroup$
The problem is to count the ideals of $mathbb{F}_q[x,y]/J^4$. It is $mathbb{F}_q[M_1,ldots,M_k]$ where $M_k$ are the $4 times 4$ matrices with minimal polynomial $M^4 = 0$
$endgroup$
– reuns
Jun 14 '17 at 12:46
$begingroup$
Instead of forming collections of elements, first form the different principal ideals generated by individual elements and then form combinations of principal ideals.
$endgroup$
– ahulpke
Jun 14 '17 at 17:57
$begingroup$
Instead of forming collections of elements, first form the different principal ideals generated by individual elements and then form combinations of principal ideals.
$endgroup$
– ahulpke
Jun 14 '17 at 17:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $A = Klangle x, yrangle/J^3$ 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 $J^3subseteq I subseteq I^2$ 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 $J^4subseteq I subseteq J^2$, 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.
answered Dec 12 '18 at 23:38
Oeyvind SolbergOeyvind Solberg
1064
$endgroup$
add a comment |
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1 Answer
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oldest
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active
oldest
votes
$begingroup$
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 $A = Klangle x, yrangle/J^3$ 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 $J^3subseteq I subseteq I^2$ 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 $J^4subseteq I subseteq J^2$, 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.
answered Dec 12 '18 at 23:38
Oeyvind SolbergOeyvind Solberg
1064
$endgroup$
add a comment |
$begingroup$
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 $A = Klangle x, yrangle/J^3$ 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 $J^3subseteq I subseteq I^2$ 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 $J^4subseteq I subseteq J^2$, 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.
answered Dec 12 '18 at 23:38
Oeyvind SolbergOeyvind Solberg
1064
$endgroup$
add a comment |
$begingroup$
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 $A = Klangle x, yrangle/J^3$ 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 $J^3subseteq I subseteq I^2$ 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 $J^4subseteq I subseteq J^2$, 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.
answered Dec 12 '18 at 23:38
Oeyvind SolbergOeyvind Solberg
1064
$endgroup$
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 $A = Klangle x, yrangle/J^3$ 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 $J^3subseteq I subseteq I^2$ 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 $J^4subseteq I subseteq J^2$, 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.
answered Dec 12 '18 at 23:38
Oeyvind SolbergOeyvind Solberg
1064
answered Dec 12 '18 at 23:38
Oeyvind SolbergOeyvind Solberg
1064
answered Dec 12 '18 at 23:38
Oeyvind SolbergOeyvind Solberg
1064
answered Dec 12 '18 at 23:38
Oeyvind SolbergOeyvind Solberg
1064
1064
add a comment |
add a comment |
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$begingroup$
What do you mean? all my A/I will be finite dimensional since i factor out something larger than J^4. In your example (xy^3) is not a ideal having the right conditions, since it does not contain J^4.
$endgroup$
– Mare
Jun 14 '17 at 12:41
$begingroup$
Yes the problem can be formulated to count the ideals of $F_q <x,y>/J^4$ contained in $J^2/J^4$, where the polynomial ring here is the noncommutative one. If you have a solution for the commutative one, that might be also interesting.
$endgroup$
– Mare
Jun 14 '17 at 12:46
$begingroup$
The problem is to count the ideals of $mathbb{F}_q[x,y]/J^4$. It is $mathbb{F}_q[M_1,ldots,M_k]$ where $M_k$ are the $4 times 4$ matrices with minimal polynomial $M^4 = 0$
$endgroup$
– reuns
Jun 14 '17 at 12:46
$begingroup$
Instead of forming collections of elements, first form the different principal ideals generated by individual elements and then form combinations of principal ideals.
$endgroup$
– ahulpke
Jun 14 '17 at 17:57