# Why is this curve nonsingular?

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.

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 $k$.

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 $E$.

• Dear Alvaro, If by isomorphism you mean the same thing that most other people mean (i.e. regular isomorphism of varieties) then in fact any curve isomorphic to an everywhere non-singular curve will again be everywhere non-singular (since non-singularity is an intrinsic geometric concept, and so is preserved under isomorphism). What you mean by isomorphism is rather birationally equivalent. Also, your maps $\phi$ and $\phi^{-1}$ don't quite seem to make sense: e.g. $\phi(x,y)$ is a map from the $(x,y)$-plane to the $(X,Y)$-plane, so its formula should be expressed in terms of $(x,y)$, ... – Matt E Jun 13 '13 at 13:49
• ... not $(X,Y)$. (I think you should switch $\phi$ and $\phi^{-1}$, so that $\phi(x,y) = (x, y/x)$.) Regards, – Matt E Jun 13 '13 at 13:52
• @Matt E, you are correct, fixed it! Thanks. – Álvaro Lozano-Robledo Jun 13 '13 at 14:11
• And yes, I did mean birational equivalence. – Álvaro Lozano-Robledo Jun 13 '13 at 14:19

Every irreducible projective curve $C$ (i.e. irreducible one-dimensional variety in some projective space) has a canonical desingularization, i.e. there is a non-singular curve $C'$, and a morphism $C' \to C$, which is an isomorphism over the complement of the singular locus in $C$.

Projective curves of the form $y^2 = f(x)$ (dehomogenized) with $f(x)$ a quartic with distinct roots are singular at infinity, but their desingularization (which is easily computed, since the singularity at infinity is not complicated) is genus one. This is why Magma is calling your curve an elliptic curve; the "true" elliptic curve that it is referring to is actually the desingularization of your curve.

The answer of Alvaro Lozano-Robledo gives one example (his $E$ is the desingularization of his $C$).

By the way, to get the desingularization, you can perform repeated blow-ups, or alternatively you can take a normalization. The first chapter of Kollar's book on resolution of singularities has a very nice discussion of desingularizing curves, although the case of $y^2 =$ a quartic is much simpler than the general theory, and is discussed in lot's of places. (Searching for "Jacobi quartic" might be one way to find more information.)

• Do you mention the curves $y^2 =$ a quartic because my curve can be put in that form? Because I don't see how. – Ricardo Buring Jun 21 '13 at 0:30
• @RicardoBuring: Dear Ricardo, I more mentioned it because it – Matt E Jun 21 '13 at 1:24
• @RicardoBuring: Dear Ricardo, I mentioned it because that it is the most common form of a singular quartic giving rise to an elliptic curve upon desingularization, and for some reason I didn't pay much attention to the explicit form of your curve (sorry about that!). Your curve is not obviously of this form, but it is a singular quartic. I'll have to think a bit before I can say anything more useful; I'll report back in comments if I can find anything helpful to say. Regards, – Matt E Jun 21 '13 at 1:27