Re: question on converting a decimal back into an algebraic number

[Kurt Foster <drsardonicus@earthlink.net>]

On Jun 27, 2025, at 3:14 AM, American Citizen wrote:

> Suppose I have a decimal value for a point
>
> [48.524409093877418991299749586199724677,  
> -358.16199519959579719705531742590169930]
>
> How do I recover
>
> [Mod(x^2 - 3, x^4 - 1752*x^2 + 87616), Mod(-591/592*x^3 + 57/37*x,  
> x^4 - 1752*x^2 + 87616)]

If you have a numerical x-value, you can use algdep to find a  
candidate for q with x = Mod(x, q), and then use the following  
rigmarole to express y as a polmod.

I assume your elliptic curve has an expression of the form

y^2 = f(x)

where f(x) is a cubic polynomial with rational coefficients.

If x is an algebraic number, f(x) will be algebraic, and the solutions  
to y^2 = f(x)  will also be algebraic.

If we assume x is an algebraic number given by a polmod, say Mod(x,  
q), where q = q(x) is an irreducible polynomial of degree greater than  
1, then we have

y^2 = subst(f, x, Mod(x, q)) or

y^2 = Mod(g, q) where g = g(x) is a polynomial of degree less than the  
degree of q.

Let P = charpoly(Mod(g,q)).

Then P(y^2) = 0; that is, y is a zero of the polynomial P(x^2).

Unfortunately, I don't have a cubic polynomial f for the curve in your  
example.

I note that if q is an irreducible quadratic [i.e. x is a quadratic  
irrational] then P(x^2) will either be reducible (in this case, y  
could be in the quadratic field determined by q); or (as in your  
example) P(x^2) could be irreducible with a square discriminant, in  
which case y is in a 4-group extension of Q which contains the  
quadratic field determined by q as a subfield; or (AFAIK) P(x^2) could  
be irreducible with non-square discriminant, in which case its Galois  
group will be D_{4}, the dihedral group with 8 elements.