American Citizen on Tue, 05 Dec 2023 16:59:06 +0100

 Re: correction to previous post

• To: pari-users@pari.math.u-bordeaux.fr
• Subject: Re: correction to previous post
• Date: Mon, 4 Dec 2023 12:44:42 -0800
• Delivery-date: Tue, 05 Dec 2023 16:59:06 +0100
• References: <8035c594-5909-438a-b937-9de90d709112@gmail.com> <ZWuAwYjCfBkTBLd+@seventeen> <d24c305a-6079-4eee-a1a1-9fb44a1ed2e4@gmail.com> <ZW0T86WSQnrugphn@seventeen>
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```To all:

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The past 2 or 3 weeks, I have received numerous replies from Bill, and he's cleared up questions that I have had about elliptic curves, the past 20-25 years or so.
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This is much appreciated, and using a selected point on a number field initialized elliptic curve to obtain the relative heights and isogeny degrees is something which I have sought the last 15 years or so, when working with face cuboids and the special Z2xZ4 curves, when I was unable to obtain the L-series coefficients due to the very large size of the conductor.
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It is hard to learn when you are self taught and stumbling over things in the dark and making mistakes.
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I want to take the time to thank him for all the help and instruction that he has given.
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Thank you Bill !

Randall

On 12/3/23 15:49, Bill Allombert wrote:
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```On Sun, Dec 03, 2023 at 02:59:52PM -0800, American Citizen wrote:
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```In regard to heights of points across isogenous curves

On 12/2/23 11:08, Bill Allombert wrote:
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```? ellisomat(E)[2][,1]
%5 = [1,2,4,4,2,2,4,4]
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```As stated this should be the same in Q as in a number field.

I think that one of my pari programs was checking for saturation by a prime
and when it used 2, it was able to find a smaller 1/2 height point on some
of the isogenous curves, hence my relative vector did not match, but was
altered from the normal result in %5.

This begs the question, given all the curves in the ellisomat() command can
we ferret out those who have a 2-degree function such that we can reliably
locate where the situation occurs and the relative height vector is altered?
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```I am not sure I understand. On one hand, ellisomat(E)[2] return the degrees
of the isogenies between all the curve, on the other hand, by using
ellisdivisible and elltors (or over Q, ellsaturation) you can saturate the
points to get the generator.

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```I have a follow up question, suppose we select a point on an elliptic curve
in a number field, and in my case I picked a rational x-coordinate, and
deriving y as a quadratic surd,  which gave me the number field to use. Are
we able to find another independent point in the same number field also on
the curve and I am NOT talking about doubling the given point and adding or
subtracting multiples to give ht, 4ht, 9ht, 16ht etc, but a truly
independent point in the same NF. I tried a naive approach and quickly
realized the chances of hitting this was incredibly low.

Or is this as hard as trying to find a rational point on a curve in Q?
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```Yes. If P is a point in a number fields, the sum of its conjugates (that you can
compute with elltrace) is a rational point.
So the only easy points P are the ones which give a torsion point (or the point
at infinity).

In particular, if you pick x in Q and y in a quadratic extension, you will get
the point at infinity (this is easy to prove).

Cheers,
Bill.
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