Bill Allombert on Sat, 02 Dec 2023 20:08:54 +0100

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Re: correction to previous post

On Sat, Dec 02, 2023 at 10:20:49AM -0800, American Citizen wrote:
> = Heights:
> =   1  69.462661097568452675010358103720853484
> =   2 138.92532219513690535002071620744170697
> =   3 277.85064439027381070004143241488341394
> =   4 277.85064439027381070004143241488341394
> =   5 138.92532219513690535002071620744170697
> =   6 138.92532219513690535002071620744170697
> =   7 277.85064439027381070004143241488341394
> =   8 277.85064439027381070004143241488341393
> =   relative: 1,2,4,4,2,2,4,4

Note that this matches the degree of the isogenies themselves as expected:

? ellisomat(E)[2][,1]
%5 = [1,2,4,4,2,2,4,4]~

> =  which does NOT match the above for points in Q

The relative should be the same.

Note that by extending the base field this way, you likely increase the rank of
the curve, so the height matrix has a larger dimension, making things more

> However I have noticed from time to time when mapping points to an isogenous
> curve, that sometimes a smaller point (1/2 height) might exist.

Yes: let f be an K isogeny from E/K to F/K that is not an idomorphism. 
f is a rationale function of degree > 1.
The preimage of a point in F/K might be in E/L where L is an algebraic extension of K.
So if P is in E(K) and Q in F(K) it is possible
that f(P)=n.Q for some n and that if R=f^-1(Q) in E(L) then n.R = P

However, you can find Q by using ellisdivisible, so you can fix that!

If we accept BSD, this it is linked with the Tate-Shafarevich group as follow:
If f is of degree d, it should increase the canonical height by d.
If your curve is of rank 1 with E(K) generated by P,
then if f(P) generates F(K) then
BSD=h(f(P))*Sha_F = d*h(P)*Sha_F so
Sha_E = d*Sha_F

More generally, if G is the generator of F(K) and f(P)=n.G, then h(f(P)) = n^2*h(G),
BSD=h(G)*Sha_F = d/n^2 * h(P) * Sha_F
so Sha_E = d/n^2 * Sha_F

So yes, this can help you find the curve with the smallest Sha.

(I hope I did not mess up anything)