| 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 complex. > 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(P)*Sha_E 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) Cheers, Bill