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the default behaviour of reduce() for ideals
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More options Jul 30 2012, 4:18 am
From: Thomas Feulner <thomas.feul...@uni-bayreuth.de>
Date: Mon, 30 Jul 2012 01:18:02 -0700 (PDT)
Local: Mon, Jul 30 2012 4:18 am
Subject: the default behaviour of reduce() for ideals

Hi,

in the definition of a QuotientRing there is the following assumption

ASSUMPTION:

I has a method I.reduce(x) returning the normal form
of elements x\in R. In other words, it is required that
I.reduce(x)==I.reduce(y) \iff x-y \in I, and
x-I.reduce(x) in I, for all x,y\in R.

On the other hand, the default definition of reduce in
sage/rings/ideal.py says
def reduce(self, f):
return f       # default

Wouldn't it be better to raise a NotImplementedError?

These are the consequences:

sage: Z16x.<x> = Integers(16)[]
sage: GR.<y> =  Z16x.quotient(x**2 + x+1 )
sage: I = GR.ideal(4)
sage: I.reduce(GR(6))
6 # should be reduced mod 4

another example is
sage: J = Z16x.ideal([x+1, x+1]) # just to avoid that the ideal is
identified as a principal ideal
sage: R.<z> = Z16x.quotient(J)
sage: R(x) # should be 15

Best,
Thomas

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More options Jul 30 2012, 5:36 am
From: Marco Streng <marco.str...@gmail.com>
Date: Mon, 30 Jul 2012 11:36:06 +0200
Local: Mon, Jul 30 2012 5:36 am
Subject: Re: [sage-devel] the default behaviour of reduce() for ideals
2012/7/30 Thomas Feulner <thomas.feul...@uni-bayreuth.de>:

It would be safer, but it sounds like it would break a lot of code.
Right now, reduce does exactly what it says:
"""
Return the reduction of the element of f modulo the ideal
I (=self). This is an element of R that is
equivalent modulo I to f.
"""

If you don't get any objections, you could change this documentation
and add a deprecation warning to this default implementation. Then
later it can be changed to NotImplementedError.

> These are the consequences:

> sage: Z16x.<x> = Integers(16)[]
> sage: GR.<y> =  Z16x.quotient(x**2 + x+1 )
> sage: I = GR.ideal(4)
> sage: I.reduce(GR(6))
> 6 # should be reduced mod 4
> another example is
> sage: J = Z16x.ideal([x+1, x+1]) # just to avoid that the ideal is
> identified as a principal ideal
> sage: R.<z> = Z16x.quotient(J)
> sage: R(x) # should be 15

I'm fine with all the outputs above, though they can be improved.
However, I think the following is a bug:

sage: R(x+1) == 0
False

The only correct answer is "True". This goes wrong because the
"ASSUMPTION" that you quotes above is not satisfied. However, I would
trust R to be correct, because I never get to see the assumption, even
when I type

Z16x.quotient??
R??

Suggested fixes:
- add the assumption to the documentation of the quotient method of Z16x, or
- add some checks to the equality tests of QuotientRing, or
- kill the default implementation of reduce as you suggest (but do
deprecation first).

(or some combination of the above)

Best,
Marco

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More options Aug 6 2012, 5:50 pm
From: Marco Streng <marco.str...@gmail.com>
Date: Mon, 6 Aug 2012 14:50:27 -0700 (PDT)
Local: Mon, Aug 6 2012 5:50 pm
Subject: Re: [sage-devel] the default behaviour of reduce() for ideals

I'm just letting the list know that there is a patch
at http://trac.sagemath.org/sage_trac/ticket/13345 deprecating the default
implementation of "Ideal_generic.reduce" in favour of NotImplementedError.

Op maandag 30 juli 2012 11:36:06 UTC+2 schreef Marco Streng het volgende:

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