Re: Computation with two algebraic integers

Hi Erwan,

you can also use

g;h;

minpoly_for_h=h^6 - 3*h^4 + 3*h^2 - 3

minpoly_for_g=g^3-2

relator=(15*h^4 - 18*h^2 - 15)*g^2 + (6*h^5 - 12*h^3 + 18*h^2 - 30*h - 30)*g + (36*h^3 - 18*h^2 - 48*h - 6)

H=Mod(h,minpoly_for_h)
R=subst(relator,h,H);
gcd(minpoly_for_g,R)

Denis.

De: "Ewan Delanoy" <ewan.delanoy@zoho.com>
À: "pari-users" <pari-users@pari.math.u-bordeaux.fr>
Envoyé: Jeudi 22 Janvier 2026 16:24:57
Objet: Re: Computation with two algebraic integers

> Maybe provide a small example as a tested for experiment.

On a small example one can proceed as follows :

minpoly_for_h=h^6 - 3*h^4 + 3*h^2 - 3
minpoly_for_g=g^3-2
relator=(15*h^4 - 18*h^2 - 15)*g^2 + (6*h^5 - 12*h^3 + 18*h^2 - 30*h - 30)*g + (36*h^3 - 18*h^2 - 48*h - 6)

polrem(pol1,pol2,var)=divrem(pol1,pol2,var)[2]

relator_lead_inverse=lift(Mod(1/pollead(relator,g),minpoly_for_h))
relator2=polrem(relator_lead_inverse*relator,minpoly_for_h,h)
relator3=polrem(minpoly_for_g,relator2,g)
relator4=polrem(relator3,minpoly_for_h,h)
g_in_terms_of_h=lift(Mod(-polcoeff(relator4,0,g)/polcoeff(relator4,1,g),minpoly_for_h))

Cheers,
Ewan