Karim Belabas on Tue, 20 Jan 2015 10:53:59 +0100 |
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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 > 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] `