Sufficient condition for a quartic curve to be birationally equivalent to an elliptic curve

Question

I have come across a few sources which talks about transforming a quartic curve of the form $y^2=Ax^4+Bx^3+Cx^2+Dx+E$ to an elliptic curve such as the Joseph H. Silverman's book- Rational Points on Elliptic Curves, Lawrence Washington's Elliptic curves, Number theory and Cryptography and also Lectures on Elliptic curves by J.W.S. Cassels. The minimum requirement mentioned in the first two books is the existence of a rational point and the lack of singular points while Cassels transforms it just assuming the existence of rational points. I understand that elliptic curves must be non-singular. My question is whether it is sufficient for us to check the existence of rational points on the quartic curve in order to transform it to a cubic curve. Because if so, we can then check the singularity of the cubic to conclude if it is an elliptic curve or not.

Answer 1

If the curve $X=V(y^2=Ax^4+Bx^3+Cx^2+Dx+E)\subset\Bbb A^2_k$ is nonsingular (equivalent to $Ax^4+Bx^3+Cx^2+Dx+E$ having no repeated roots), then it defines a genus-one curve: the morphism $X\to\Bbb A^1_k$ by projecting to the $x$-axis extends to a degree-two morphism from the smooth compactification $\overline{X}\to\Bbb P^1$ which is ramified at the four points $X\cap V(y)$. By Riemann-Hurwitz, $$2g(\overline{X})-2=2(2g(\Bbb P^1)-2)+\sum_{x\in \overline{X}} (e_x-1),$$ so as $g(\Bbb P^1)=0$, $e_x=1$ for all non-ramified points, and $\sum_{x\mapsto y} e_x = 2$, we see that $g(\overline{X})=1$ and the morphism is unramified at infinity.

To determine if we get an elliptic curve, we need to check for the presence of a rational point. Let's see if we get one at infinity: embedding $X\subset\Bbb P^2$ in the standard way, we get the homogeneous equation $y^2z^2=Ax^4+Bx^3z+Cx^2z^2+Dxz^3+Ez^4$, and we need to look at points on $z=0$ which are contained in the chart $D(y)$. Dehomogenizing with respect to $y$, we get the equation $$z^2=Ax^4+Bx^3z+Cx^2z^2+Dxz^3+Ez^4,$$ which is singular at $x=0$, $z=0$. Blowing up via $z=tx$, this becomes $$t^2=Ax^2+Bx^2t+Cx^2t^2+Dx^2t^3+Ex^2t^4$$ which is still singular at $x=0,t=0$; blowing up again via $t=ux$ gives $$u^2=A^2+Bux+Cu^2x^2+Du^3x^3+Eu^4x^4$$ which is nonsingular on the line $x=0$. There's a rational point on $x=0$ (i.e. a rational point on $\overline{X}$ over $\infty\in\Bbb P^1$) iff $A$ is a square in $k$.

In summary, you get an elliptic curve iff $Ax^4+Bx^3+Cx^2+Dx+E$ has no repeated roots and at least one of the following is true: $A$ is a square in $k$ or $V(y^2=Ax^4+Bx^3+Cx^2+Dx+E)\subset\Bbb A^2_k$ has a $k$-rational point.