#20722: Comparison in polynomial quotient rings
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Reporter: nbruin | Owner:
Type: defect | Status: needs_info
Priority: major | Milestone: sage-7.3
Component: algebra | Resolution:
Keywords: | Merged in:
Authors: | Reviewers:
Report Upstream: N/A | Work issues:
Branch: | Commit:
u/nbruin/comparison_in_polynomial_quotient_rings|
45d7f4e294356ef1a5fc7a391314506c66aaf061
Dependencies: | Stopgaps:
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Comment (by nbruin):
Replying to [comment:5 SimonKing]:
> I thought that the letterplace implementation of free algebras does
provide a normal form. Thus, there shouldn't be a problem involved here.
Or am I missing something?
I don't think that `normalform` with respect to an ideal is defined in all
text as something that is unique. Clearly, in the example in this ticket,
this fails. Whether this is a bug in the relevant grobner basis/reduction
code or whether this is an unfortunate feature of the definition of
groebner basis used there, I don't know.
The test branch I wrote avoids depending on a normal form by testing if
the difference of representatives lies in the ideal. On the FreeAlgebra
test case, this leads to subtracting elements of unequal degree in the
standard powering algorithm, which leads to an error.
The fact that you cannot subtract any two elements in FreeAlgebra is a bug
in itself: it means its ring structure isn't fully implemented. It would
be good to change this. If FreeAlgebras are always graded (and we only
consider homogeneous ideals in them) then perhaps elements should be
implemented as vectors of homogeneous elements, with component-wise
addition and subtraction. Then the thing is actually a ring. Now it's just
a disjoint union of modules tied together with multiplication morphisms.
Or are FreeAlgebras certifiably only useful with the partial operations?
If we can regain unique NormalForm with respect to all ideals in all rings
this code gets used for, we could hide the problem again by reverting the
definition of equality testing. However, I think the "is difference in
ideal" test is preferable anyway:
- you only need one reduction instead of two (although perhaps the
elements are already in reduced form and hopefully the reduction algorithm
detects that quickly)
- I expect positive answers to have a better chance of being quick,
because showing an element does lie in an ideal often allows shortcuts
(that I hope many reduction algorithms would find)
--
Ticket URL: <http://trac.sagemath.org/ticket/20722#comment:6>
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