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.Tue, 04 May 2021 18:38:22 +0200how to calcualte the basis of quotient module over the steenrod algebra in Sagehttps://ask.sagemath.org/question/56943/how-to-calcualte-the-basis-of-quotient-module-over-the-steenrod-algebra-in-sage/ A(2) is steenrod algebra generated by sq1, sq2,sq4 with Adem relations, we knew A(2) has dimension 64, how to calculate the basis like quotient algebra A(2)/A(2){sq1} and A(2)/A(2){sq1,sq2sq3} and A(2)/A(2){sq1, sq7,sq3sq7+sq4sq6} in Sage? Thanks !Tue, 04 May 2021 14:28:08 +0200https://ask.sagemath.org/question/56943/how-to-calcualte-the-basis-of-quotient-module-over-the-steenrod-algebra-in-sage/Answer by John Palmieri for <p>A(2) is steenrod algebra generated by sq1, sq2,sq4 with Adem relations, we knew A(2) has dimension 64, how to calculate the basis like quotient algebra A(2)/A(2){sq1} and A(2)/A(2){sq1,sq2sq3} and A(2)/A(2){sq1, sq7,sq3sq7+sq4sq6} in Sage? Thanks !</p>
https://ask.sagemath.org/question/56943/how-to-calcualte-the-basis-of-quotient-module-over-the-steenrod-algebra-in-sage/?answer=56951#post-id-56951Quotients like this are not well implemented in Sage, and in particular, there is no simple way to compute a basis. Examples of ways in which the implementation is not great:
sage: A2 = SteenrodAlgebra(profile=(3,2,1))
sage: s1 = A2.Sq(1)
sage: I = A2.ideal(s1)
sage: I
Twosided Ideal (Sq(1)) of sub-Hopf algebra of mod 2 Steenrod algebra, milnor basis, profile function [3, 2, 1]
sage: s1 in I
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NotImplementedError Traceback (most recent call last)
....
NotImplementedError:
sage: R = A2.quotient(I)
sage: R(s1) == 0
False
Tue, 04 May 2021 18:38:22 +0200https://ask.sagemath.org/question/56943/how-to-calcualte-the-basis-of-quotient-module-over-the-steenrod-algebra-in-sage/?answer=56951#post-id-56951