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Updated this question, just focusing on the relevant part

$f_2(u,v) = \tfrac{27}{64} u^3 - \tfrac{81}{64} u^2 v + \tfrac{189}{20} u^2 + \tfrac{81}{64} u v^2 - \tfrac{189}{10} u v + \tfrac{1764}{25} u + \tfrac{165}{64} v^3 - \tfrac{99}{4} v^2 + \tfrac{297}{5} v$

f2(u,v) = 27/64*u^3 + (-81/64*v + 189/20)*u^2 + (81/64*v^2 - 189/10*v + 1764/25)*u + (165/64*v^3 - 99/4*v^2 + 297/5*v) 

$f_3(s,t) = s^2 + \tfrac{1485}{16} t^3 + \tfrac{2673}{16} t^2 + \tfrac{8019}{80} t + \tfrac{11907}{400}$

f3(s,t) = 1485/16*t^3 + 2673/16*t^2 + 8019/80*t + (s^2 + 11907/400) 

Given the substitution

$t = \frac{v}{u}, s = \tfrac{1}{160 u^2}\left[135 u^2 - 405 u^2 v + 1512 u^2 + 405 u v^2 - 3024 u v + 825 v^3 - 3960 v^2\right]$

t = v/u s = (135*u^3 + (-405*v + 1512)*u^2 + (405*v^2 - 3024*v)*u + (825*v^3 - 3960*v^2))/(160*u^2) 

I have checked that $f_2(u,v) = \frac{16 u}{165 t^3 + 81 t^2 - 81 t + 27} f_3(s,t)$

also we have the inverse

u = (3960*t^2 + 3024*t + (160*s - 1512))/(825*t^3 + 405*t^2 - 405*t + 135) v = t*u 

So in what way are the rational solutions of $f_2$ related to those of $f_3$? I don't think they are in bijection because of the problem of zero denominators. Although, it does seem that maybe I can get around it since $165 t^3 + 81 t^2 - 81 t + 27$ is not zero for any rational $t$.. is that the general way to deal with it? I wonder what happens if the denominator had a rational root. On the other hand, if $u=0$ then the LHS might is not zero (except for $v=0,24/5$) but the RHS is...


old version of the question, ignorable:

I have started with the elliptic curve $f$:

$f(x,y) = x^3 + 3y^3 - 11$

and using the techniques here transformed it into the Weierstrass form

$w(s,t) = s^2 + 37125 t^3 + 66825 t^2 + 40095 t + 11907$

using the change of variables

$ \begin{eqnarray} s &=& \frac{360 x^3 - 3024 x^2 + 1080 y^3 + 6156 y^2 + 1980}{16 x^2 + 24 x y - 88 x + 9 y^2 - 66 y + 121} \\ t &=& \frac{15 y + 57}{20 x + 15 y - 55}. \end{eqnarray}$

The problem, we don't have $f(x,y) = w(s,t)$, instead we have the identity $ \frac{f(x,y)}{4x+3y-11} = \frac{w(s,t)}{12375 t^3 + 6075 t^2 - 6075 t + 2025}.$

So my question is how can I be sure that I have found a birational transformation? If I want to study the rational solutions of $f$ via $w$ I need to know the exact relation between the two and how to avoid problems related to zero denominators. Thanks very much.

  • 1
    However, there is a general result that a birational map between *nonsingular*, *projective* curves always extends to an isomorphism. So if you do a sequence of birational transformations starting and ending with such curves, you know that the result extends to a bijection, regardless of the intermediate steps. "Projective" is key here: The points resulting from division by zero become points at infinity when you consider the projective rather than the affine curve. (I'm reasonably confident of all this... hopefully someone will correct me if I've slipped up here.)2011-09-30

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