Let $C$ be the projective closure of $Z(f) \subset \mathbf{A}^2$ where $f$ is an irreducible polynomial of degree 4 in $x$ and degree 2 in $y$, so $C = Z(f^*) \subset \mathbf{P}^2$ where $f^*$ is the homogenization of $f$.
For a particular example, let $f = x^4y – 2x^3y^2 + 6x^3y – 6x^2y^2 – 4x^3 + 12x^2y – 4xy^2 – 12x^2 + 12xy – 8x + 4y$. (I have a family of these, all of a particular form.)
Using MAGMA I find that this curve has a nonsingular model of genus 1:
> R<x,y,z> := ProjectiveSpace(Rationals(),2);
> C:= Curve(R, x^4*y - 2*x^3*y^2 + 6*x^3*y*z - 4*x^3*z^2 - 6*x^2*y^2*z + 12*x^2*y*z^2 - 12*x^2*z^3 - 4*x*y^2*z^2 + 12*x*y*z^3 - 8*x*z^4 + 4*y*z^4);
> Degree(C);
5
> Genus(C);
1
> P0:= C![-2,0,1];
> E, phi:=EllipticCurve(C,P0);
> E;
Elliptic Curve defined by y^2 - 4*x*y = x^3 - 11*x^2 + 12*x over Rational Field
How would one go about showing this by hand? I suspect that the algorithms used by MAGMA are not practical for this. Can we prove that the degree of the defining homogeneous polynomial of the nonsingular model is 3? In that case we would be done after an application of the genus-degree formula $g = (d-1)(d-2)/2$.
Note that I'm not asking for a construction of the nonsingular model, just a proof of the fact that it has genus 1. This may or may not be helpful.
Many thanks in advance.
Note: The singular points of my example $C$ that I know of are $[\sqrt{2} : \sqrt{2} : 1]$ and $[-\sqrt{2} : -\sqrt{2} : 1]$, and I believe they are ordinary singularities (multiplicity 2, two distinct tangents).
Note 2: Actually, $[0 : 1 : 0]$ is also a singular point, but I don't know what kind. If it's ordinary with multiplicity 3, then we are done by the generalized genus-degree formula discussed in the comments below. But how do I prove that these are all the singularities?
Best Answer
I don't know how Magma procedes, but in your particular example, it is easy. Namely you want to compute the genus of the projective smooth curve with function field equal to $k(x,y)$, where $k$ is the ground field (say of characteristic different from $2$) and $x,y$ are linked by the relation $f(x,y)=0$.
Write $f$ as $a_0(x)y^2+a_1(x)y+a_2$ with $a_i(x)\in k[x]$. Then you have $$ (2a_0y+a_1)^2+ ({4a_0a_2-a_1^2})=0. $$ This is the equation of a hyperelliptic. Let $4a_0a_2-a_1^2=g^2h$ with $g, h\in k[x]$ and $h(x)$ without square factor, then a non-singular equation of the function field is $$z^2+h(x)=0.$$ And it is known that the genus is $[(d-1)/2]$ if $d=\deg h(x)$. In you particular example, I found $h(x)=-(x^4 + 12x^3 + 32x^2 + 24x + 4)$ and the function field has genus $1$. It is a bad idea to write a projective equation and solve singularities.
To get an elliptic equation, you have to make a change of variables and send a rational point (say $x=0$, $z=2$) to $\infty$. I didn't check.