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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).

azerbajdzan
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  • @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. – azerbajdzan Dec 22 '19 at 21:12
  • @Somos: I have just appended it at the end of my question. – azerbajdzan Dec 22 '19 at 21:19
  • 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 $[X:Y:Z]=[0:1:0]$. – Jyrki Lahtonen Dec 22 '19 at 21:26
  • @Jyrki Lahtonen: So this answers questions 2 and 3 - meaning that $Q_{\infty}$ corresponds to $P_1$. But how the conclusion can be derived from the substitutions (without knowing what point was used for transformation)? And what about question 1 and 4? – azerbajdzan Dec 22 '19 at 21:31
  • Thinking... I'm not sure. IIRC the point at infinity of the quartic curve splits into two points or something. When viewed as a projective curve it is not smooth at $[0:1:0]$, and something exceptional happens. – Jyrki Lahtonen Dec 22 '19 at 21:50
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    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. – Jyrki Lahtonen Dec 22 '19 at 22:39
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    @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 – reuns Dec 22 '19 at 23:39
  • @JyrkiLahtonen,@reuns: I do not understand. Are you trying to say, that rational point at infinity $P_{\infty}=(0,1,0)$ on $E_1$ becomes an irrational point on $E_2$ even though using birational transformations? – azerbajdzan Dec 23 '19 at 09:10
  • I think that it means that there is no single point $P_\infty$ on $E_1$. The group law on an elliptic curve always is on the non-singular model. There is a singularity at $[0:1:0]$ of $$Y^2Z^2=7X^4+X^3Z+X^2Z^2+XZ^3+3Z^4.$$ When we blow up the singularity, it becomes two points, algebraic conjugates of each other. I'm afraid I'm not very conversant with what's going on. But you do see this happening with real locus. When $x\to\infty$, the $y$-coordinate can follow either the $\sqrt7x^2$ branch or its negative. And the two branches approach at distinct points in the $(u,v)$ domain. – Jyrki Lahtonen Dec 23 '19 at 10:18
  • When working with the Weierstrass equation, the cubic curve on the projective plave is smooth, and has been set up to have a single point at infinity. With other curves this need not happen. There may be several points at infinity, or a singularity (as is the case here). If you look at the Fermat curve $X^3+Y^3=Z^3$, there will be three distinct points on the line at infinity. The Weierstrass equation is very special in the sense that the line at infinity makes degree three contact with the point $[0:1:0]$, so e.g. Bezout holds. – Jyrki Lahtonen Dec 23 '19 at 10:25
  • I know that there may be more than one point at infinity, but as I am working over rationals I count only rational points so points $[1:\sqrt{7}:0],[1:-\sqrt{7}:0]$ are out of my interest. They would be relevant if I was working over $\mathbb{Q}\left(\sqrt{7}\right)$. So then point $[0:1:0]$ on quartic has no practical usage because of its singularity in contrast with the same point on cubic which can be used as identity element. Then I want to ask how many rational points at infinity there are on quartic. We already mentioned $[0:1:0]$, which is singular, and what are others? – azerbajdzan Dec 23 '19 at 12:47

1 Answers1

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

azerbajdzan
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