2
$\begingroup$

Elliptic curve defined by $$E_1: y^2=7 x^4+x^3+x^2+x+3, P_1=(-1,3)$$ can be transformed to $$E_2: v^2=u^3-\frac{250 u}{3}-\frac{1249}{27}$$ Substitutions used are: $$\left(x\to \frac{15 u-9 v+217}{39 u+9 v+209},y\to \frac{9 \left(54 u^3+639 u^2-27 v^2+592 v-16501\right)}{(39 u+9 v+209)^2}\right)$$ $$\left(u\to \frac{2 \left(20 x^2+x+9 y+8\right)}{3 (x+1)^2},v\to -\frac{81 x^3+3 x^2+26 x y-3 x-10 y-33}{(x+1)^3}\right)$$

Then we can check, as an example, that point $P_2=(\frac{6}{7},-3)$ on $E_1$ correcponds with point $Q_2=(-\frac{2}{3},3)$ on $E_2$.

Questions:

  1. Point $P_{\infty}=(0,1,0)$ on $E_1$ corresponds with what point on $E_2$?
  2. Point $Q_{\infty}=(0,1,0)$ on $E_2$ corresponds with what point on $E_1$?
  3. Point $P_1=(-1,3)$ on $E_1$ corresponds with what point on $E_2$?
  4. Point $Q_1=(-\frac{71}{9},\frac{296}{27})$ on $E_2$ corresponds with what point on $E_1$?

EDIT:

Maybe it was not clear, but for points $P_2=(\frac{6}{7},-3)$ and $Q_2=(-\frac{2}{3},3)$ I used the substitutions to verify they correspond to each other. For points in my question the same method did not work for me because of singularities (division by zero).

$\endgroup$
11
  • $\begingroup$ @Somos: This is what I did with points $P_2=(\frac{6}{7},-3)$ and $Q_2=(-\frac{2}{3},3)$. But for the points in my questions it did not work because of singularities. $\endgroup$ Dec 22, 2019 at 21:12
  • $\begingroup$ @Somos: I have just appended it at the end of my question. $\endgroup$ Dec 22, 2019 at 21:19
  • $\begingroup$ I think that the transformation you used specified $(-1,3)$ as the neutral element. In the Weierstrass form the neutral element is the unique point at infinity. That is, the one with homogeneous coordinates $[X:Y:Z]=[0:1:0]$. $\endgroup$ Dec 22, 2019 at 21:26
  • 2
    $\begingroup$ When $x$ is very large we see from the equation of $E_1$ that we have, roughly, $y=\pm\sqrt{7}x^2$. This means that in your formula for $u$, the terms $20x^2$ and $9y$ are the boss terms. Similarly, in the formula for $v$, the terms $81x^3$ and $26xy$ dominate. So when $x\to\infty$, we are approaching the points $$(u,v)=(\frac{40\pm18\sqrt{7}}3,-(81\pm 26\sqrt7))$$ the choice of sign depending on which branch you follow. $\endgroup$ Dec 22, 2019 at 22:39
  • 1
    $\begingroup$ @JyrkiLahtonen That's why we need the weighted projective plane $\Bbb{P}^2(1,2,1)$ to complete the quartic curve, it becomes $\{ [x:y:z],y^2=7 x^4+x^3z+x^2z^2+xz^3+3z^4\}/ ([x:y:z]\sim [rx:r^2y:rz])$ and this time we obtain two missing points $[1:\sqrt{7}:0],[1:-\sqrt{7}:0]$ and the birational map does extend to them $\endgroup$
    – reuns
    Dec 22, 2019 at 23:39

1 Answer 1

1
$\begingroup$

enter image description here

$$E_1(-1,3)\to E_2(0,1,0)$$ $$E_1(-1,-3)\to E_2(-\frac{71}{9},-\frac{296}{27})$$ $$E_1(-\frac{5413}{16069},-\frac{434267883}{258212761})\to E_2(-\frac{71}{9},\frac{296}{27})$$ $$E_1(1,-\sqrt{7},0)\to E_2(\frac{40}{3}-6 \sqrt{7},-81+26 \sqrt{7})$$ $$E_1(1,\sqrt{7},0)\to E_2(\frac{40}{3}+6 \sqrt{7},-81-26 \sqrt{7})$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.