Re: Computation with two algebraic integers

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

On Thu, Jan 22, 2026 at 03:17:55PM +0100, Bill Allombert wrote:
> On Thu, Jan 22, 2026 at 02:56:58PM +0100, Ewan Delanoy wrote:
> > I have two algebraic integers g and h, and
> > 
> > - I know the minimal polynomial of g, stored in a variable called minpoly_for_g, which has degree 18 in g
> > - I know the minimal polynomial of h, stored in a variable called minpoly_for_h, which has degree 72 in h
> > - I also know an extra relation between g and h, in the form of a polynomial with integer coefficients called
> > relator, and which has degree 17 in g, 71 in h. This relation ensures that g is unique in terms of h.
> > 
> > 
> > My goal is to compute the (polynomial) expression of g in terms of h from those data.
> 
> So you have three polynomials
> minpoly_for_g(x), minpoly_for_h(y) and relator(x,y) and you define g,h by
> 
> minpoly_for_g(g) = 0
> minpoly_for_h(h) = 0
> relator(g,h) = 0
> 
> What is the equation satisfied by the polynomial you are looking for ? 
> Is it 
> g = pol(h)

In that case, I suggest you do:

? G=nfisincl(minpoly_for_g, minpoly_for_h);
? #G
%10 = 2
\\ There are two solutions, we need to check that relator(g,h) = 0!
? substvec(relator,[g,h],[G[1],h]*Mod(1,minpoly_for_h))==0
%11 = 1
G[1] is the right one.

There is a better way to do that, which is used internally by nfisincl.

Cheers,
Bill