Re: New GP function ellrank (2-descent)

[John Cremona <john.cremona@gmail.com>]

On Tue, 16 Feb 2021 at 13:30, Bill Allombert
<Bill.Allombert@math.u-bordeaux.fr> wrote:
>
> On Tue, Feb 16, 2021 at 12:04:39PM +0000, John Cremona wrote:
> > Bill,
> >
> > Can you add some documentation for the parameter 'effort'?  (That name
> > sounds very Magma-like, by the way).   I assume that it is related to
> > a bound on a search for rational points on 2-covers (quartics), but is
> > it linear or exponential?
>
> It is a bound on the number of Selmer classes to inspect, a bound of the number
> of quartic models to try for each Selmer classes, and a bound (* 10000) on
> the naive height of the points for hyperellratpoints.
>
> For example you can try:
> ? \g2
> ? E=ellinit("2170b1");
> ? setrand(2);ellrank(E,10)
>
> The selmer rank is 1, so there is a single Selmer classes, however
> we can generate different quartics by taking different representatives,
> for example
>
> Y^2 = 5812437140379431936*x^4+1347241111842961644544*x^3+117156424752666733221120*x^2+4530122383197205883049984*x+65719549423500847171260416
>
> Y^2 = 221275160067200000*x^4-475741594144480000*x^3+756678069244798800*x^2-644795816435820800*x+377052872754508800
>
> This second one has a small rational point:
> [48513/7192,14958243512261070/808201]
> while the point on the first is much (~10000 time) harder to find:
> [-4298555/68258,83178143745675062784/1164788641]

There is something wrong with this example.  The first one is a
2-cover of 2170b1 but the second is not.  So rational points on the
second cannot help you find points on that elliptic curve.

Some other comments: using quartic reduction, the first quartic q1 you
gave reduces to the quartic with coefficients
(1364,-292,12837,6916,41748) -- this is q1(x-58) / 65278768^2.  On
this, x=-339591/68258 gives a rational point.

The second quartic q2 has Jacobian of conductor 319377950753934190 if
I copied correctly, so something is not right.  Did you really find a
rational point on 2170b1 from your second quartic?

For this curve since, as you say, the Selmer rank is 1, there is
exactly one equivalence class of (everywhere locally soluble)
quartics.  Your only choice is to pick a representative.  These will
all be obtained from each other by a linear fractional transform of x
followed by multiplication by a square.


>
> It is more efficient to try more quartics than to use a large bound for
> hyperellratpoints (as an analogy, MPQS is more efficient than QS).

I really think that implementing minimization and reduction is the
best strategy, rather than finding random quartics in the equivalence
class.

>
> I came up with this trick by experimenting with Denis program.
> If you have a geometrical interpretation, I would be happy to hear it.

I have written about this at great length in various papers and a book...

John

>
> > Also what does the output of ellrankinint() contain?   (If you say
> > that it is very technical and I should not worry about it, I will feel
> > like a Mathematica user!)   In particular, does it contain the
> > quartics which represent elements of (or generators of) the 2-Selmer
> > group?
>
> No, it only contains the bnfinit of the 2-division polynomial.
> This way you can also use it for twists.
>
> Cheers,
> Bill.
>