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I have the following elliptic curve that I want to look up in the LMFDB-database:

$$\text{k}:\space\space\space y^2-3y=x^3+x^2$$

Using the Weierstrass form of my elliptic curve, I wrote my equation in the form:

$$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$

Which gives $a_1=0,a_2=1,a_3=-3,a_4=0$ and $a_6=0$. Then I will get two parameters of this elliptic curve:

  • Discriminant: $$\Delta=-b_2^2b_8-8b_4^3-27b_6^2+9b_2b_4b_6$$

Where $b_2=a_1^2+4a_2,b_4=2a_4+a_1a_3,b_6=a_3^2+4a_6$ and $b_8=a_1^2a_6+4a_2a_6-a_1a_3a_4+a_2a_3^2-a_4^2$.

Using my values I get:

$$\Delta=-2331$$

  • j-invariant:

$$\text{j}=\frac{\left(b_2^2-24b_4\right)^3}{\Delta}$$

Using my values I get:

$$\text{j}=-\frac{4096}{2331}$$

But when I tried to look up my elliptic curve using the discriminant I didn't find mine.

Where did I go wrong?

Joiryu
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1 Answers1

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I put the sequence of Weierstrass coefficients $[a_1,a_2,a_3,a_4,a_6]=[0,1,-3,0,0]$ into the LMFDB search engine, and it gave the link to the curve $$ y^2+y=x^3+x^2+2.\qquad (*) $$ Apparently this is the minimal Weierstrass model. If you substitute $y\mapsto y-2$ into $(*)$ you get your equation back. Anyway inverting that subsitution gives you the list of integer points on your curve $$ (-1,3), (0,3), (9,30) $$ and the negatives of those points.

My very limited experience with using LMFDB suggests that typing in the Weierstrass coefficients is the simplest way. It will automatically show the minimal Weierstrass model isomorphic to the curve I input.

Jyrki Lahtonen
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  • Your link does lead to the general website, not to a specific curve? – Joiryu Jan 14 '20 at 18:19
  • Sorry about that. I copy/pasted from the wrong tab. Should work now, @Joiryu. – Jyrki Lahtonen Jan 14 '20 at 18:20
  • I see. But this does not correspond to my equation, because it has different coefficients? Or am I wrong about that? – Joiryu Jan 14 '20 at 18:22
  • I described how I got there. Working out the transformation bringing this curve to yours is not difficult. LMFDB search engine automatically takes you to a standard (=minimal?) model. Try it a couple times more. – Jyrki Lahtonen Jan 14 '20 at 18:24
  • Yes I see now, thanks. Are the integral points, that are given by the database the only ones (is that proven)? – Joiryu Jan 14 '20 at 18:25
  • I don't know. Click the link Reliability of the data in top right corner for all the details. They claim that the list of integer points has been verified by other means, so I would trust it, but I'm not an expert. – Jyrki Lahtonen Jan 14 '20 at 18:27
  • And what will the curve be if I have the coefficients: [0,3,−3,0,0] because if I use the search at the website I get a errormessage? So I've the equation: $$y^2-3y=x^3+3x^2$$ – Joiryu Jan 14 '20 at 20:20
  • When I entered [0,3,-3,0,0] I was taken to this curve They seem to depress the cubic polynomial on $x$ with a linear substitution and do the same to the quadratic on $y$. – Jyrki Lahtonen Jan 14 '20 at 20:28
  • Thank you very much. What substitution do I've to take to go back to my previous equation $$y^2-3y=x^3+3x^2$$? – Joiryu Jan 14 '20 at 20:29
  • $(y+2)^2-3(y+2)=y^2+y-2$ and $(x-1)^3+3(x-1)^2=x^3-3x+2$. – Jyrki Lahtonen Jan 14 '20 at 20:32
  • If I use that I get: $$y^2+y=x^3-3x+4$$ which is not my original equation , right? – Joiryu Jan 14 '20 at 20:35
  • It is exactly the variant of your curve LMFDB gave me in response to [0,3,-3,0,0]$. Use it. – Jyrki Lahtonen Jan 14 '20 at 20:43
  • Yes I will, but I've to trace it back to some original equation and therefore I need to know the substitution that is done by LMFDB. So that's why I'm aksing – Joiryu Jan 14 '20 at 20:44