Re: Computation with two algebraic integers

Hi Erwan, 

Just before your computation, write 
g;h; 
Then after having defined minpoly_for_h and minpoly_for_g and relator, 
you can try 
nffactor(minpoly_for_h,relator) 
or 
nffactor(minpoly_for_h,minpoly_for_g) 

Denis. 

> De: "Ewan Delanoy" <[email protected]>
> À: "pari-users" <[email protected]>
> 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