John Cremona on Tue, 05 Dec 2023 19:38:12 +0100 |
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Re: correction to previous post |
To all:
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.
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.
It is hard to learn when you are self taught and stumbling over things
in the dark and making mistakes.
I want to take the time to thank him for all the help and instruction
that he has given.
Thank you Bill !
Randall
On 12/3/23 15:49, Bill Allombert wrote:
> On Sun, Dec 03, 2023 at 02:59:52PM -0800, American Citizen wrote:
>> In regard to heights of points across isogenous curves
>>
>> On 12/2/23 11:08, Bill Allombert wrote:
>>> ? ellisomat(E)[2][,1]
>>> %5 = [1,2,4,4,2,2,4,4]
>> 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?
> 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.
>
>> 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?
> 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.