Re: Mixing variables in Mod expressions

[Aurel.Page@math.u-bordeaux1.fr]

What about a warning when having to take gcd of moduli ? It would keep  
the current behaviour but a student should understand he is doing  
something wrong.

Aurel

Quoting John Cremona <john.cremona@gmail.com>:

> The link does explain the first output in a way that makes sense to
> me:   the two moduli are coprime so the common quotient is mod 1 hence
> 0.
>
> For a beginner, a run-time error might be more informative though,
> along the lines of "invalid summands" or "incompatible moduli" -- but
> I think that the policy might be to always allow addition between
> objects of the same type?
>
> John
>
> On 20 January 2015 at 09:53, Karim Belabas
> <Karim.Belabas@math.u-bordeaux.fr> wrote:
> > Dear John,
> >
> > * John Cremona [2015-01-20 10:20]:
> > > I am trying to understand the following.  The answer might well be
> > > something like "if you try to do something stupid then you must expect
> > > a stupid answer"
> > >
> > > ? Mod(x,x^2-3) + Mod(x,x^2-5)
> > > %1 = 0
> >
> > This is by design. And (IMHO) impossible to fix to get "expected"
> > results (e.g. POLMOD variables being treated as "mute" variables).
> >
> > Does the following FAQ explain the situation in a satisfactory way ?
> >
> >   http://pari.math.u-bordeaux1.fr/faq.html#modular
> >
> > > -- but I was pretending to be a student trying to find the polynomial
> > > satisfied by sqrt(3)+sqrt(5) and doing what seemed natural.
> >
> > There are various ways to achieve this:
> >
> > (10:42) gp > algdep(sqrt(3)+sqrt(5), 4)
> > %1 = x^4 - 16*x^2 + 4
> >
> > (10:43) gp > polcompositum(x^2-3, x^2-5)
> > %2 = [x^4 - 16*x^2 + 4]
> >
> > (10:43) gp > rnfequation(y^2-3, x^2-5)
> > %3 = x^4 - 16*x^2 + 4
> >
> > [ all 3 methods can break in various ways, but they can all be made to
> > work (provably) with extra effort ]
> >
> > > This version works:
> > >
> > > ? Mod(x,x^2-3) + Mod(y,y^2-5)
> > > %2 = Mod(x + Mod(y, y^2 - 5), x^2 - 3)
> > > ? a = Mod(x,x^2-3) + Mod(y,y^2-5)
> > > %3 = Mod(x + Mod(y, y^2 - 5), x^2 - 3)
> > > ? a^2-8
> > > %4 = Mod(Mod(2*y, y^2 - 5)*x, x^2 - 3)
> > > ? (a^2-8)^2
> > > %5 = Mod(Mod(60, y^2 - 5), x^2 - 3)
> > > ? (a^2-8)^2-60
> > > %6 = 0
> > > but that is not the point;  result %1 is surely going to confuse people.
> >
> > Cheers,
> >
> >     K.B.
> > --
> > Karim Belabas, IMB (UMR 5251)  Tel: (+33) (0)5 40 00 26 17
> > Universite de Bordeaux         Fax: (+33) (0)5 40 00 69 50
> > 351, cours de la Liberation    http://www.math.u-bordeaux1.fr/~kbelabas/
> > F-33405 Talence (France)       http://pari.math.u-bordeaux1.fr/  [PARI/GP]
> > `