Exercise 5.7.5 (Cubics $cx(x^2-1) = y(y^2-1)$)

Proof. The homogeneous equation of this curve is

where

Then $(x,y,z)\in {ℂ}^3$ is a singular point of the curve if $f(x,y,z)=0$ and

• Suppose that $c\notin \{0,1,-1\}$. Then (1) is equivalent to

  $$\begin{array}{lll}\hfill \{\begin{array}{cc}3x^2-z^2\hfill & =0,\hfill \\ -3y^2+z^2\hfill & =0,\hfill \\ -\mathit{cxz}+\mathit{yz}\hfill & =0.\hfill \end{array}& & \hfill \text{(2)}\end{array}$$

  Then $3x^2=3y^2=z^2$ and $z(y-\mathit{cx})=0$.

  • If $z=0$, then $x=y=0$.
  • If $z\ne 0$, then $y=\mathit{cx}$, so $3(1-c^2)x^2=0$. Since $c^2\ne 1$, we obtain $x=y=z=0$.

  The only solution of the system is $(0:0:0)$ which is not a point of the projective plane. The curve ${𝒞}_f(ℂ)$ has no singularity, so the curve $\mathit{cx}(x^2-1)=y(y^2-1)$ is an elliptic curve.

• If $c=0$, the projective point $(1:0:0)$ is solution of the system (1), so the curve $0=y(y^2-1)$ is not an elliptic curve (it is the union of three lines passing by $(1:0:0)$).
• If $c=1$, then $(1,1,\sqrt{3})$ is a singular point (in fact $f(x,y)=(x^2+y^2+\mathit{xy}-1)(x-y)$, so the curve is the union of an ellipse with a secant line)
• If $c=-1$, then $(1,-1,\sqrt{3})$ is a singular point (and $f(x,y)=-(x^2+y^2+\mathit{xy}-1)(x+y)$).

${𝒞}_f(ℂ)$ is an elliptic curve if and only if $c\notin \{0,1,-1\}$. □