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 $E_2$.


  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 $E_1$?


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).

  • $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ @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


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


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