ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sat, 03 Oct 2020 14:29:14 +0200Compute radical and idempotents of a quotient algebrahttps://ask.sagemath.org/question/10350/compute-radical-and-idempotents-of-a-quotient-algebra/I tried the following:
R.<x, y> = PolynomialRing(QQ, 2)
I = Ideal(x^2, y^2)
S = R.quotient(I)
I have the following question:
**I would like to compute with Sage the Jacobson radical of the algebra S,
all primitive orthogonal idempotents and the central idempotents.**
Of course, you can compute this by hand, but I am interested in more
complicated examples, too (also in matrix algebras), but wanted to
start with this simple example.
Since I am relatively new to Sage, I unfortunately do not know how to compute this.
I would be grateful for any help.Fri, 12 Jul 2013 10:50:17 +0200https://ask.sagemath.org/question/10350/compute-radical-and-idempotents-of-a-quotient-algebra/Answer by Bern for <p>I tried the following:</p>
<pre><code>R.<x, y> = PolynomialRing(QQ, 2)
I = Ideal(x^2, y^2)
S = R.quotient(I)
</code></pre>
<p>I have the following question:</p>
<p><strong>I would like to compute with Sage the Jacobson radical of the algebra S,
all primitive orthogonal idempotents and the central idempotents.</strong></p>
<p>Of course, you can compute this by hand, but I am interested in more
complicated examples, too (also in matrix algebras), but wanted to
start with this simple example.</p>
<p>Since I am relatively new to Sage, I unfortunately do not know how to compute this.</p>
<p>I would be grateful for any help.</p>
https://ask.sagemath.org/question/10350/compute-radical-and-idempotents-of-a-quotient-algebra/?answer=40714#post-id-40714In the meantime I have learned that these things can be done **for admissible quotients of path algebras** with the aid of the GAP package QPA. This also works in Sage via letting GAP be the intermediator.
It works as follows:
gap> LoadPackage("qpa");;
gap> k:=Rationals;
Rationals
gap> Q:=Quiver(1,[[1,1,"a"],[1,1,"b"]]);
<quiver with 1 vertices and 2 arrows>
gap> kQ:=PathAlgebra(k,Q);
<Rationals[<quiver with 1 vertices and 2 arrows>]>
gap> AssignGeneratorVariables(kQ);
\#I Assigned the global variables [ v1, a, b ]
gap> rels:=[a^2,b^2, a*b-b*a];
[ (1)*a^2, (1)*b^2, (1)*a*b+(-1)*b*a ]
gap> A:=kQ/rels;
<Rationals[<quiver with 1 vertices and 2 arrows>]/
<two-sided ideal in <Rationals[<quiver with 1 vertices and 2 arrows>]>,
(3 generators)>>
gap> IsAdmissibleQuotientOfPathAlgebra(A);
true
gap> RAD:=RadicalOfAlgebra(A);
<algebra of dimension 3 over Rationals>
gap> Display(RAD);
Algebra( Rationals, [ [(1)*a], [(1)*b], [(1)*a*b] ] )
gap> Dimension(RAD);
3
gap>
If we let $k$ be a finite field, then the command
IdempotentsForDecomposition(A);
works, but I don't have a computational solution for the case $k=\mathbb{Q}$.
I do not yet have a solution for the non-commutative case.Sat, 20 Jan 2018 15:42:39 +0100https://ask.sagemath.org/question/10350/compute-radical-and-idempotents-of-a-quotient-algebra/?answer=40714#post-id-40714Comment by slelievre for <p>In the meantime I have learned that these things can be done <strong>for admissible quotients of path algebras</strong> with the aid of the GAP package QPA. This also works in Sage via letting GAP be the intermediator.</p>
<p>It works as follows:</p>
<pre><code>gap> LoadPackage("qpa");;
gap> k:=Rationals;
Rationals
gap> Q:=Quiver(1,[[1,1,"a"],[1,1,"b"]]);
<quiver with 1 vertices and 2 arrows>
gap> kQ:=PathAlgebra(k,Q);
<Rationals[<quiver with 1 vertices and 2 arrows>]>
gap> AssignGeneratorVariables(kQ);
\#I Assigned the global variables [ v1, a, b ]
gap> rels:=[a^2,b^2, a*b-b*a];
[ (1)*a^2, (1)*b^2, (1)*a*b+(-1)*b*a ]
gap> A:=kQ/rels;
<Rationals[<quiver with 1 vertices and 2 arrows>]/
<two-sided ideal in <Rationals[<quiver with 1 vertices and 2 arrows>]>,
(3 generators)>>
gap> IsAdmissibleQuotientOfPathAlgebra(A);
true
gap> RAD:=RadicalOfAlgebra(A);
<algebra of dimension 3 over Rationals>
gap> Display(RAD);
Algebra( Rationals, [ [(1)*a], [(1)*b], [(1)*a*b] ] )
gap> Dimension(RAD);
3
gap>
</code></pre>
<p>If we let $k$ be a finite field, then the command</p>
<pre><code>IdempotentsForDecomposition(A);
</code></pre>
<p>works, but I don't have a computational solution for the case $k=\mathbb{Q}$.
I do not yet have a solution for the non-commutative case.</p>
https://ask.sagemath.org/question/10350/compute-radical-and-idempotents-of-a-quotient-algebra/?comment=40723#post-id-40723Thanks for reporting that you found a solution. I turned your comment into an answer.
Could you edit your answer to include code to perform these computations?
It would be interesting to see how you do it in GAP, but also how to do it in Sage.Sun, 21 Jan 2018 06:33:21 +0100https://ask.sagemath.org/question/10350/compute-radical-and-idempotents-of-a-quotient-algebra/?comment=40723#post-id-40723Comment by Bern for <p>In the meantime I have learned that these things can be done <strong>for admissible quotients of path algebras</strong> with the aid of the GAP package QPA. This also works in Sage via letting GAP be the intermediator.</p>
<p>It works as follows:</p>
<pre><code>gap> LoadPackage("qpa");;
gap> k:=Rationals;
Rationals
gap> Q:=Quiver(1,[[1,1,"a"],[1,1,"b"]]);
<quiver with 1 vertices and 2 arrows>
gap> kQ:=PathAlgebra(k,Q);
<Rationals[<quiver with 1 vertices and 2 arrows>]>
gap> AssignGeneratorVariables(kQ);
\#I Assigned the global variables [ v1, a, b ]
gap> rels:=[a^2,b^2, a*b-b*a];
[ (1)*a^2, (1)*b^2, (1)*a*b+(-1)*b*a ]
gap> A:=kQ/rels;
<Rationals[<quiver with 1 vertices and 2 arrows>]/
<two-sided ideal in <Rationals[<quiver with 1 vertices and 2 arrows>]>,
(3 generators)>>
gap> IsAdmissibleQuotientOfPathAlgebra(A);
true
gap> RAD:=RadicalOfAlgebra(A);
<algebra of dimension 3 over Rationals>
gap> Display(RAD);
Algebra( Rationals, [ [(1)*a], [(1)*b], [(1)*a*b] ] )
gap> Dimension(RAD);
3
gap>
</code></pre>
<p>If we let $k$ be a finite field, then the command</p>
<pre><code>IdempotentsForDecomposition(A);
</code></pre>
<p>works, but I don't have a computational solution for the case $k=\mathbb{Q}$.
I do not yet have a solution for the non-commutative case.</p>
https://ask.sagemath.org/question/10350/compute-radical-and-idempotents-of-a-quotient-algebra/?comment=53710#post-id-53710Commutativity is not necessary - sorry that I wrote sth. wrong earlier.Sat, 03 Oct 2020 14:29:14 +0200https://ask.sagemath.org/question/10350/compute-radical-and-idempotents-of-a-quotient-algebra/?comment=53710#post-id-53710