> 2012/7/30 Thomas Feulner:
> > 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?
> 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
