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.)
Over $\mathbf{Q}$ this is apparently an elliptic curve (Magma code):
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);
P0:= C![-2,0,1];
E, phi:=EllipticCurve(C,P0); // works
However, it is my understanding that an elliptic curve $Z(F)$ is supposed to be nonsingular, and that nonsingularity on such a projective curve is equivalent to $\partial{F}/\partial{x}$, $\partial{F}/\partial{y}$, and $\partial{F}/\partial{z}$ never simultaneously vanishing at a point (Hartshorne exercise I.5.9). In this case (Sage code):
sage: var('x,y,z')
(x, y, z)
sage: F = 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
sage: F.subs(x=sqrt(2),y=sqrt(2),z=1)
0
sage: diff(F,x).subs(x=sqrt(2),y=sqrt(2),z=1)
0
sage: diff(F,y).subs(x=sqrt(2),y=sqrt(2),z=1)
0
sage: diff(F,z).subs(x=sqrt(2),y=sqrt(2),z=1)
0
Hence $[\sqrt{2} : \sqrt{2} : 1]$ is a singular point of $C$, and similarly $[-\sqrt{2} : -\sqrt{2} : 1]$ is. What gives? I know these are not rational points, but the curve $C$ lies in $\mathbf{P}^2(K)$ where $K$ is an algebraic closure of $\mathbf{Q}$, doesn't it?
If the curve is indeed singular, then I would like to know why Magma can turn it into an elliptic curve. If the curve is in fact nonsingular then I would like to know how to approach a proof of this fact, preferably in a way that might be generalized to the rest of my family of curves.
Thanks in advance.
Note. For this question to make sense, let all affine and projective spaces be over algebraically closed fields. An elliptic curve over a field $k$ is then a smooth curve in $\mathbf{P}^2(K)$ where $K$ is an algebraic closure of $k$, with genus 1 and a given point on it with coordinates in $k$.
Best Answer
An elliptic curve $E$ over a field $K$ is a projective smooth curve of genus $1$ defined over $K$ with at least one $K$-rational point. This means that there is a non-singular model for $E$, but this does not mean that every curve that is birationally equivalent to $E$ is necessarily non-singular everywhere.
For example, take $E:y^2=x^3+1$ and consider $C: X^2Y^2 = X^3+1$. The curves $E$ and $C$ are birationally equivalent via $\phi: E\to C$ such that $\phi(x,y)=(x,x/y)$. Clearly, $\phi$ is such an equivalence, with inverse $\phi^{-1}: C\to E$ given by $\phi^{-1}(X,Y)=(X,XY)$. However, the curve $E$ is non-singular, but $C$ is not. The curve $C$ is singular at $[0,1,0]$, because in projective coordinates $C: F(X,Y,Z)=0$ with $F(X,Y,Z)=X^2Y^2-X^3Z-Z^4$ and $$\left(\frac{\partial F}{\partial X},\frac{\partial F}{\partial Y},\frac{\partial F}{\partial Z}\right)=(2XY^2-3X^2Z,2X^2Y,-X^3-4Z^3),$$ which vanishes at $[0,1,0]\in C$.
I believe Magma is simply saying that your curve $C$ is genus $1$ and has a non-singular model $E$.