Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
The transformation is birational because each step of the procedure you followed was a birational transformation, hence the composition is. If you want the inverse explicitly, then invert the procedure you followed, step-by-step; I don't recommend trying to do it in one step from the formulas you wrote above :) –  Ted Sep 30 '11 at 4:01
    
@Ted, good idea, I did this substitution in three parts.. I've edited my question so that it is focused just on the third part (which is the bit where I have the trouble). –  user16697 Sep 30 '11 at 5:31
1  
When you have a birational map between 2 curves, the points won't necessarily be in perfect bijection with one another, because of the division by zero problems that you mention. Instead, there will be a bijection covering "most" of the points, because the problematic division by 0 points will be a "small" portion of both curves. (Precisely: There are dense subsets of each curve such that the restriction of the birational map is a bijection.) –  Ted Sep 30 '11 at 6:16
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.) –  Ted Sep 30 '11 at 6:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.