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Take the equation $$y^2 = x^4 - 2x^3 - 2x - 1$$

I found that this is a genus 1 curve, because it is well known that for $y^2 = f(x)$ where $f$ is of even degree, the genus is $\frac{\deg{f} - 2}{2}$, which is 1 in our case.

Additionally, it can be easily shown that this curve has one rational point, namely $(-1, 2)$.

Does this mean that the equation is an elliptic curve?

I am conflicted because I thought all elliptic curves were of the form $y^2 = x^3 + Ax^2 + Bx + C$, but I also thought that all genus 1 curves with at least one rational points are elliptic curves.

Lastly, if this is an elliptic curve, can we find its minimal Weierstrass Equation?

Jyrki Lahtonen
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Ravikanth Athipatla
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    Here David E. Speyer explains the process of converting a curve of the form $y^2=f(x)$, $\deg f=4$ into a Weierstrass form. In addition, there is an algorithm for finding the minimal Weierstrass form. Alternatively you can just type in the cubic into LMFDB. – Jyrki Lahtonen Jul 20 '24 at 19:38
  • And, for example, the cubic Fermat curve $x^3+y^3=1$ is also an elliptic curve (non-singular cubic). It was a fun exercise once to figure out how to put it into Weierstrass form, but I don't remember the steps :-) – Jyrki Lahtonen Jul 20 '24 at 19:40
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    @JyrkiLahtonen, I find the method that David E. Speyer used works primarily for $y^2 = x^4 + ax^2 + b$, which doesn't have a cubic or linear term. – Ravikanth Athipatla Jul 20 '24 at 20:21
  • Yeah, it seems to me, too, that the process is simpler with a biquadratic instead of a more general quartic. – Jyrki Lahtonen Jul 21 '24 at 03:56
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    @Jyrki for the experts in the room, maybe I can point to why this case may (heuristically) be a little simpler. If $C : y^2 = f(x)$ and $f(x)$ is biquadratic then the Jacobian $E$ of $C$ is an elliptic curve which admits a $2$-isogeny. Indeed, there exists a map $\phi : C \to E'$ which factors the $2$-isogeny (i.e., the $\bar{K}$-isomorphism $\iota : C \cong E$ can be chosen so that $\phi \iota^{-1}$ is the $2$-isogeny on $E$). For a general binary quartic you can only find a map $\phi$ so that $\phi \iota^{-1}$ is mulitplication by $2$ on $E$. – Mummy the turkey Jul 21 '24 at 06:39
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    Linking a few other threads where $y^2=f(x)$, $\deg f=4$ is discussed: 1, 2,3. Mostly to improve interconnectivity, and help me find relevant threads next time somebody asks about this. – Jyrki Lahtonen Jul 21 '24 at 10:05

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Yes this is an elliptic curve, and yes we can put it in Weierstrass form. An unhelpful simple answer is that computer algebra will achieve this for you (EllipticCurve(E, O) in Magma, for example). The following is essentially a walkthrough of what goes on in the background.

It is useful to note that the proof that a pointed curve of genus $1$ is isomorphic to a plane cubic in Weierstrass form is constructive. Recall what we do. If $E$ is a projective curve of (geometric) genus $1$ and $O \in E(k)$ is a point, the Riemann-Roch theorem tells us that $H^0(E, 2O)$ and $H^0(E, 3O)$ have dimensions $2$ and $3$ respectively, and we may take $k$-bases $\{ 1, \xi \}$ and $\{ 1, \xi, \nu\}$. It then follows by construction that the image of the morphism $E \to \mathbb{P}^2$ given by $P \mapsto [\xi(P), \nu(P), 1]$ is a Weierstrass equation for $E$. So the problem is essentially reduced to computing the spaces $H^0(E, 2O)$ and $H^0(E, 3O)$. This is possible with Groebner basis machinery in modern computer algebra languages (if you were brave you could do it by hand often).

Consider $E$, the curve given by $$y^2 = x^4 - 2x^3 - 2x - 1$$ in $\mathbb{A}^2$ with $O = (-1,2)$. We wish to determine rational functions $\xi$ and $\nu$ which resp. have a pole of degree $2$ and $3$ at $O$. Take $$\xi = 4\frac{y + x^2 - x}{(x+1)^2}$$ $$\nu = 4\frac{(3x - 1)y + 3x^3 - 3x^2 + 3x + 1 }{(x + 1)^3}.$$ It's a good exercise to come up with this by hand, but I must admit I used the Riemann-Roch function Basis in Magma. In any case, this gives you a rational map $\phi : E \to \mathbb{P}^2$, and it is easy to check (even by hand) that (in the function field of $E$) we have the relation $$\nu^2 = \xi^3 + 8\xi$$ and therefore the image of $\phi$ is the elliptic curve with Weierstrass equation given by the relation above.

Also, in your case the classical invariant theory of binary quartic forms gives formulae for the Jacobian of a curve $y^2 = f(x)$ where $f(x)$ has degree $4$ -- this is also true for plane cubics, quadric intersections, and (maybe?) for the more general 'elliptic normal curve of degree $n$' (the cases above being $n = 2, 3, 4$). See e.g., the Magma documentation of GenusOneModel.

Mummy the turkey
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  • I just noticed Jyrki's comment above linking to David E. Speyer's answer to a previous quesiton. I'll leave my answer here in case it is useful, his answer gives an outline of how you might construct the functions $\xi$ and $\nu$ above; with a little less language (though maybe a little more mystery as to why it works). – Mummy the turkey Jul 20 '24 at 21:30
  • Thank you. I am currently working through how to construct the above functions as you mentioned, but I definitely understand it now. – Ravikanth Athipatla Jul 20 '24 at 23:34
  • Ok. Basically following a standard procedure in the argument that (by Riemann-Roch) the function field of a curve of genus one and a rational point has rational functions with a single of the prescribed order. Why did I fail to remember that when looking at D.E.Speyer's post :-) – Jyrki Lahtonen Jul 21 '24 at 09:58
  • @Mummytheturkey ,what are references to transform Quartic=$y^2$ to Elliptic curve. you used different transformation so I need read it. – Guruprasad Feb 27 '25 at 14:49