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I am trying to follow case 2 of the procedure given in Cohen:

Procedure

for the cubic $f(x,y,z) = x^3 + 3 y^3 - 11 z^3$ using the rational point $P_0 = (2 : 1 : 1)$. The tangent at this point is $y = - \tfrac{12}{9}(x-2) + 1$ and it intersects the curve again at the point $P = (28:-19:5)$.

So I built the linear change of variables $(x:y:z) = (\bar x + 297825\bar y + 28\bar z:\bar x + 215600\bar y - 19\bar z :\bar x + 1698144\bar y + 5\bar z)$ which produces

$\begin{align} \bar{f}(\bar x,\bar y,\bar z) =& -7 \bar{x}^3 - 53204877 \bar{x}^2 \bar{y} - 252 \bar{x}^2 \bar{z} \\\\ & - 94477421044413 \bar{x} \bar{y}^2 - 584088120 \bar{x} \bar{y} \bar{z} + 4776 \bar{x} \bar{z}^2 \\\\ & - 53809704024840479199 \bar{y}^3 - 476307229568940 \bar{z} \bar{y}^2 \end{align}$.

As expected $(0:0:1)$ is a root and the tangent to the curve at that point is $\bar{x}=0$. The problem is there are no linear terms in $\bar{y}$ so $y L(0,1) + y^2 Q(0,1) + y^3 C(0,1)$ has $y=0$ as a double root and another point ($y=-8.8516976\ldots \times 10^{-6}$) contradicting the assertion, further $Q(0,1)^2-4 L(0,1) C(0,1)$ is not zero (since $L(0,1)$ is zero, it would have to be $-950232922990035300/901521983$ for that expression to be zero).

Am I making a mistake somewhere or following the procedure wrong or can I find some simpler instructions elsewhere?


I tried to take what you apply the "final trick" and I got this:

? l(1,t) + x*q(1,t) + x^2*c(1,t)
% = (165/64*t^3 + 81/64*t^2 - 81/64*t + 27/64)*x^2 + (-99/4*t^2 - 189/10*t + 189/20)*x + (297/5*t + 1764/25)
? (2*x*c(1,t)+q(1,t))^2 - q(1,t)^2 + 4*c(1,t)*l(1,t)
% = (27225/1024*t^6 + 13365/512*t^5 - 20169/1024*t^4 - 1053/256*t^3 + 10935/1024*t^2 - 2187/512*t + 729/1024)*x^2 + (-16335/64*t^5 - 20493/64*t^4 + 20331/160*t^3 + 16281/160*t^2 - 5103/64*t + 5103/320)*x + (9801/16*t^4 + 82269/80*t^3 + 22599/400*t^2 - 102789/400*t + 11907/100)

neither of which seem to be Weierstrass form.


? w(s,t)
% = s^2 + 1485/16*t^3 + 2673/16*t^2 + 8019/80*t + 11907/400
? s = (90*x^3 - 756*x^2 + (270*y^3 + 1539*y^2 + 495))/(80*x^2 + (120*y - 440)*x + (45*y^2 - 330*y + 605));
? t = (15*y + 57)/(20*x + (15*y - 55));
? w(s,t)*(1600*u)/(64*25*4*c(1,t))
% = x^3 + (3*y^3 - 11)
? c(1,'t)
% = 165/64*t^3 + 81/64*t^2 - 81/64*t + 27/64
? c(1,t)
% = (675*x^3 - 11340*x^2 + 63504*x + (2025*y^3 + 23085*y^2 + 87723*y - 7425))/(1600*x^3 + (3600*y - 13200)*x^2 + (2700*y^2 - 19800*y + 36300)*x + (675*y^3 - 7425*y^2 + 27225*y - 33275))
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1 Answer 1

up vote 3 down vote accepted

It is easier (for me at least) to work in affine coordinates rather than homogeneous coordinates, so I will write $x$ and $y$ where you would write $x/z$ and $y/z$. The affine equation for the curve you start with is $x^3 + 3 y^3 = 11$.

The instructions are to make a linear transformation (following you, I will write the new variables as $\overline{x}$ and $\overline{y}$) so that the tangent line to the point $(2,1)$ becomes the line $\overline{x} = 0$, and the point $(28/5,-19/5)$ becomes the origin.

As you note, the tangent line has the equation $y = -\dfrac{4}{3}x +\dfrac{11}{3}.$

Thus the relevant change of coordinates is $\overline{x} = \dfrac{4}{3}x + y - \dfrac{11}{3}$, $\overline{y} = $ any other line passing through $(28/5,-19/5)$. Let me choose $\overline{y} = y+19/5$.

Then we find that $x = \dfrac{3}{4}(\overline{x} - \overline{y}) + \dfrac{28}{5},$ $y = \overline{y} -\dfrac{19}{5}.$

This seems to be where your mistake is. There are many possible linear changes of coordiantes at this point. Indeed, I restricted my attention to considering affine linear changes of coordinates (I didn't change the line at infinity), and then made a particularly simple choice. There are many others, and it is also possible to consider changes of coordinates which don't preserve the line at infinity (which is what you do). But ultimately, one should be obtaining a new set of coordinates so that $\overline{x} = 0$ passes through $(0,0,1)$, and is tangent to the curve at the original point $P_0$ (i.e. the point where $x = 2$ and $y = 1$). In your new coordinates, the curve $\overline{x} = 0$ is instead tangent to the curve at the point $P$, where $\overline{x} = \overline{y} = 0$. This shows that there is a mistake in your choice of new coordinates.

If you continue with my choice of new coordinates, you find that at the original point $P_0$, one has $(\overline{x},\overline{y}) = (0,24/5).$ The curve itself has the equation $$\dfrac{27}{64} \overline{x}^3 - \dfrac{81}{64}(\overline{x}^2 \overline{y} - \overline{y}\overline{x}^2) + \dfrac{165}{64} \overline{y}^3 + \dfrac{189}{20} (\overline{x}^2 - 2 \overline{x}\overline{y}) - \dfrac{99}{4} \overline{y}^2 + \dfrac{1764}{25}\overline{x} + \dfrac{297}{5} \overline{y} = 0.$$

The intersection with the line $\overline{x} = 0$ is given by the cubic $$\dfrac{165}{64} \overline{y}^3 - \dfrac{99}{4}\overline{y}^2 + \dfrac{297}{5}\overline{y} = \dfrac{165}{64}(\overline{y}-\dfrac{24}{5})^2 \overline{y} = 0,$$ which has vanishing discriminant (as it must, by Cohen's argument).

I didn't have the courage to make the final birational transformation; I leave it to you.

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thank you very much I think this given some real progress but I still couldn't get the Weierstrass form - I added the computation to my question. –  user16697 Sep 29 '11 at 5:30
1  
@QED: Dear QED, You should read Cohen's description of the algorithm more carefully. To get Weierstrass form, you take the birationally transformed equation (i.e. coordinates are now $x$ and $t$) and then write it in the form $(2x C(1,t) + Q(1,t))^2 = \cdots$; this is the Weierstrass form. There is another change of coordinates implicit: you have to rewrite the LHS as $y^2$ (here $y$ is a new variable). The RHS is a priori a quartic, but actually it is a cubic. Cohen explains why it has to work out this way, and if you compute carefully, it will necessarily work out ... –  Matt E Sep 29 '11 at 6:38
1  
... this way in your particular case. Regards, –  Matt E Sep 29 '11 at 6:38
    
Thanks so much you must understand I find it so difficult. –  user16697 Sep 29 '11 at 7:46

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