Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 14 / 2

Let $F$ define the nonsingular plane cubic and let $H_F=\det \mathrm{Hess}F$ be its Hessian curve of a plane cubic equation. The Hessian criterion for a flex applies in the allowed field characteristic. Here is the local calculation, so no characteristic-zero argument is needed.

Move any point $P$ to $[0:1:0]$ and its tangent line to $Z=0$. Smoothness gives a nonzero coefficient $g$ of $Y^{2}Z$, while the coefficients of $Y^3$ and $X{Y}^2$ vanish. ThusThe intersection multiplicity with the tangent line is at least three precisely when $b=0$, which is precisely $H_F(P)=0$ because the field characteristic is not two. The tangent cannot be a component of a nonsingular plane cubic.

If $H_F$ vanishes identically, every point is a flex by this calculation. Otherwise it is a degree-three homogeneous polynomial. Its zero locus meets the cubic by Bézout's theorem; if they share a component, that also supplies a point of intersection. Every such point is smooth and satisfies the criterion, so a flex exists.

Choose a flex as $O=[0:1:0]$, with tangent line $Z=0$. The flex coordinates for a smooth plane cubic giveThe coefficient $a$ is nonzero because $Z=0$ is not a component, and $b\ne 0$ expresses smoothness at $O$. Since the field characteristic is not two, the projective linear changecompletes the square. The equation becomes $Y^{\prime 2}Z=P_3(X,Z)$ after dividing by a nonzero constant, where $P_3$ is a degree-three homogeneous polynomial with nonzero $X^3$ coefficient.

On $Z=1$, the polynomial $P_3(x,1)$ has three distinct roots $r_1,r_2,r_3$ in the algebraically closed field. A repeated root $r$ would make $(r,0)$ a singular point of $y^{\prime 2}=P_3(x,1)$, since both first partial derivatives vanish there. PutA further nonzero scaling of $y^{\prime }$ absorbs the leading coefficient and $(r_2-r_1{)}^3$; the required square root exists in the algebraically closed field. These are projective linear changes of the original coordinates. Renaming the new coordinates gives the Legendre form of an elliptic curveIts projective completion is $Y^{2}Z=X(X-Z)(X-\lambda Z)$, with the chosen flex at infinity.

The genus-degree formula givesbecause the nonsingular plane cubic is already its own normalization of an algebraic curve. If it were birational to the projective line, the two smooth projective curves would have isomorphic function fields. The uniqueness of the finite normalization model of a one-variable function field would then make them isomorphic. But the projective line has genus of a smooth projective curve zero. Therefore the cubic is not rational. This obstruction works in the stated positive characteristics as well as in characteristic zero.