Re: New GP function ellrank (2-descent)

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

Thanks for the example, I have been looking at those quartics.  I see
that they are all from the curve 2170b1 which has rank 1 and trivial
torsion so all of them represent the non-trivial element of the
2-Selmer group.  They are (therefore, but I also checked) all
equivalent.  However the 5th in your list (with the largest
coefficients) is not minimal, specifically it is not minimal at p=29.
That is a small detail, but Cremona-Stoll reduction is about
unimodular equivalence, i.e. with respect to GL(2,Z), but there is
also the question of minimization -- making the discriminant and I, J
invariants as small as possible.  Your 5th quartic has discriminant
29^12 times the others.

One rather obvious point to make is that if you have any soluble
quartic then you can use the rational point's x-value to construct an
equivalent quartic on which the rational point has x=0/1, i.e. the
constant term is square and finding a rational point is then trivial.
Similarly, you can find a unimodular transform to take your
x-coordinate to any rational at all -- just because SL(2,Z) acts
transitively on Q.   So it is certainly true that within the
equivalence class, some quartics have smaller rational points than
others!  I cannot see how that helps, really -- it will depend on how
you are constructing the different quartics.  If it is random (which I
do not expect) then it is not so different from trying random rational
x and hoping it gives a point.  Not exactly -- it is more like, take a
random transform and then look for rational points corresponding to
transforms near that one.

The motivation behind our reduction method was to try to make the
coefficients as small as possible (though we do not ever prove that
"reduced" quartics are smallest in that sense).  We cannot design a
reduction which makes the rational point as small as possible, at
least not without knowing the rational point in advance.

I am a little puzzled by one thing about your example:  you have 9 or
10 quartics, all reduced (after minimization, the 5th one transforms
to (12005,-18130,27117,-17648,7488)), and all equivalent.  According
to the theory of reduction, each has a covariant point in the upper
half-plane which is in the fundamental region for SL(2,Z), and as they
are equivalent, these points are SL(2,Z)-equivalent, hence equal.  (I
ignore equivalence between points on the boundary of the fundamental
region, for simplicity.)   It is a long time since I did that work on
minimization, but this surprises me.   I think that for the first two
quartics in your list the points in the upper half-plane are

w1 = -0.144945794164345 + 2.45094355681874*I, |w1| = 2.45522577413102
w2 = 0.0951889821025962 + 2.34443097032587*I, |w2| = 2.34636261411931

and these are not SL(2,Z)-equivalent.  So I am remembering something
wrong here.  I was using my C++ code for these computations.  Since I
wrote that I worked out (with Tom Fisher) a rather nicer equivalence
test for quartics (with exactly the same invariants) which also gives
you the 2x2 matrix taking one to the other.  You can look that up, but
from memory: let the (homogenised) quartics be q1(x,y), q2(x,y);  set
g1 and g2 to be their quartic covariants;  compute
q1(x1,y1)*g2(x2,y2)-q2(x1,y1)*g1(x2,y2), which is homogeneous of
degree 4 in each pair of variables.  Factor this.  Then each factor
will be of bi-degree (1,1) and its 4 coefficients are the entries in
the matrix.  (If no linear factors, they are not equivalent.)  I do
not have code for that to hand.

I'll come back to this tomorrow.

John

On Wed, 17 Feb 2021 at 15:45, Bill Allombert
<Bill.Allombert@math.u-bordeaux.fr> wrote:
>
> On Tue, Feb 16, 2021 at 02:47:51PM +0000, John Cremona wrote:
> > 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.
>
> I am looking again at this example.
> Denis program find the following set of quartics
> (reduced using Cremona-Stoll reduction):
>
> 1277*x^4+202*x^3+11877*x^2+6680*x+46800
> 1364*x^4+292*x^3+12837*x^2-6916*x+41748
> 1973*x^4+2874*x^3+18681*x^2+7892*x+24212
> 468*x^4+668*x^3+11877*x^2+2020*x+127700
> 4745552*x^4+7271768*x^3+19077981*x^2+12368046*x+8955533
> 5108*x^4+9812*x^3+18933*x^2+10788*x+13268
> 6053*x^4+3946*x^3+18681*x^2+5748*x+7892
> 7488*x^4+12304*x^3+19101*x^2+13112*x+10832
> 8000*x^4+14800*x^3+23757*x^2+11802*x+8477
> 8525*x^4+14370*x^3+21897*x^2+10036*x+7892
>
> The smallest rational points are:
>
> 1:[[-69957/407849,35677573217664/166340806801]]->407849
> 2:[[339591/68258,1274199043488/1164788641]]->  ->339591
> 3:[[-612623/271333,25483980869760/73621596889]]->612623
> 4:[[-4078490/69957,118925244058880/1631327283]]->4078490
> 5:???
> 6:[[-168946/407849,17838786608832/166340806801]]->407849
> 7:[[-542666/612623,50967961739520/375306940129]]->612623
> 8:[[984644/69957,29731311014720/1631327283]]    ->984644
> 9:[[41321/7192,5688388587/1616402]]             ->41321
> 10:[[271333/170645,2548398086976/5823943205]]   ->271333
>
> The quartic #9 has a much smaller rational points than the others.
> Given that ratpoints is quadratic in the height of x, and
> that generating new quartics is fast, it makes sense to try several
> of them.
> In my example, I did a search of bound 70000 and find #9.
> To find the second best one, I would need a bound of 280000
> which would take 16 more times, which is more than the time
> to run hyperellratpoints(,70000) on the 10 quartics.
>
> Does that make more sense now ?
> Do you have a better reduction procedure ?
>
> Cheers,
> Bill.
>