Re: tossing out an idea to use heights of algebraic points on elliptic curves to help sort isogenies for point searches

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

On Fri, Dec 01, 2023 at 05:26:34PM -0800, American Citizen wrote:
> Bill
> 
> Can you please explain some more on how you came up with P1?

You gave me

  P1 = [-2641, (2641+sqrt(373230717))/2]

by definition, sqrt(373230717) is a root of x^2-373230717
(which is irreducible over Q).
Consider the map
Q[x] -> Q(sqrt(373230717))
P(x) -> P(sqrt(373230717))  (P is a polynomial in Q[x])
its kernel is the principal ideal generated by x^2-373230717
so this define an isomorphism from 
Q[x]/(x^2-373230717) to Q(sqrt(373230717))
which send x mod (x^2-373230717) to sqrt(373230717)

Mod(x, x^2-373230717) is just the GP syntax for "x mod (x^2-373230717)"

In gp I use the variable 'a because it is safer to keep x and y for true
polynomials, due to variable priority.

So P1 = [-2641, (2641+sqrt(373230717)/2] algebraic form is 
   P1 = [-2641, (2641+Mod(x, x^2-373230717))/2]

Of course this is an easy case. When there are several algebraic numbers,
it is more involved.

Note that, since (2641+sqrt(373230717))/2 minimal polynomial is
x^2-2641*x-91563959, you could also write
P1 = [-2641,Mod(b,b^2-2641*b-91563959)]

However, again if you start from scratch, you will directly find the equation
satified by y since your curve is given by
y^2+x*y=x^3-114223080*x -283150929600
and x=-2641, you get y^2-2641*y=91563959 directly.

Cheers,
Bill