Karim Belabas on Tue, 20 Jan 2015 10:53:59 +0100

 Re: Mixing variables in Mod expressions

```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
>
> ? 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]
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