Past exam of the algebraic curves course of the University of Oxford / 2023 / 2

A nonsingular point of an algebraic curve $p$ is an inflection point of an algebraic plane curve, or flex, if its projective tangent line has contact of order at least three:Here the projective tangent line must not be a component of $C$. Ordinary tangency has contact order two.

Let $C=V(F)$, where $F$ is a homogeneous polynomial of degree $d\ge 2$. Its Hessian curve is defined byEach entry has degree $d-2$, so a nonzero $H_F$ has degree $3(d-2)$. By the Hessian criterion for a flex, the flexes are exactly the points of $C\cap V(H_F)$. The stated no-common-component assumption and Bézout's theorem therefore giveThis counts distinct points; higher contact can use more than one unit of intersection multiplicity at a single point.

For completeness, the Hessian criterion for a flex follows by putting $p=[0:0:1]$ and $T_{p}C=\{y=0\}$, with $F_y(p)\ne 0$. The Euler theorem for homogeneous functions gives $F_{xz}(p)=F_{zz}(p)=0$ and $F_{yz}(p)=(d-1)F_y(p)$, whenceThe restriction $F(x,0,1)$ already has vanishing constant and linear terms. Its quadratic term vanishes exactly when the intersection multiplicity with the tangent is at least three, proving the criterion.

For $d\ge 3$ the bound is the actual maximum number of flexes of a smooth plane curve: a general nonsingular curve has only contact of order three, and the corresponding intersections with its Hessian curve are simple. One way to see genericity is to vary a point and a line through it, a three-dimensional family. Contact of order at least four imposes four independent linear conditions on the coefficients of a degree-$d$ equation for $d\ge 3$. Thus the locus of equations with such contact is a proper exceptional locus. Locally, if the curve is $y=h(x)$, its Hessian curve restricted to $C$ is a nonvanishing factor times $h^{\prime \prime }(x)$; contact exactly three gives a simple zero.

A nonsingular plane conic has no flexes, agreeing with the formula at $d=2$. A line is excluded from this formulation: its tangent is itself and its Hessian matrix is zero, so the no-common-component hypothesis fails. The formula is therefore intended for $d\ge 2$, rather than giving the meaningless negative value at $d=1$.

For $F=x^3+y^3+3x{z}^2$, the first partial derivatives areTheir simultaneous vanishing would force $y=0$, $xz=0$ and $x^2+z^2=0$, hence $x=y=z=0$, which is not a projective point. This also verifies that the given plane cubic is nonsingular.

The Hessian matrix and its determinant areThus the Hessian curve is the union of $y=0$, $x=z$ and $x=-z$. On $y=0$, the equation of $C$ reduces to $x(x^2+3z^2)=0$, givingOn $x=z$ one must have $z\ne 0$, so put $z=x=1$ and obtain $y^3=-4$. On $x=-z$ similarly put $z=1$, $x=-1$, obtaining $y^3=4$. Choose $r\in \mathbb{C}$ with $r^3=4$ and let $\omega =\exp (2\pi \mathrm{i}/3)$. The remaining points areAll nine listed projective points are distinct. They exhaust the intersection with the Hessian curve, so the Hessian criterion for a flex proves that these, and only these, are the flexes in this example. There are none at infinity: $z=0$ and $H_F=0$ would force $x=y=0$ on $C$.

First make the invertible linear change of homogeneous coordinatesThe original equation becomesThis takes the flex $[0:0:1]$ to the point at infinity $[0:1:0]$ of a Weierstrass model.

Put $h=\omega -1$ and choose $s\in \mathbb{C}$ with $s^2=h^3$. Both are nonzero. DefineSince $U=hX+Z$, the three factors on the right are $hX$, $h(X-Z)$ and $h(X-\lambda Z)$, whereThe resulting equation is preciselythe Legendre form of an elliptic curve. In particular, $\lambda \ne 0,1$.

Combining the two changes gives an explicit projective transformation from the original coordinates:Its determinant is $-\sqrt{3}/(hs)\ne 0$, so this is a valid transformation of the whole projective plane, including the points where an affine coordinate would be undefined.

Let $(r_0,r_1,r_2)$ be any ordering of the three distinct numbers $0,1,\lambda$. Set $h=r_1-r_0$ and choose $s\ne 0$ with $s^2=h^3$. The projective transformationhas the required matrix form. As $x=hX+r_{0}Z$ and $y=sY$, substitution givesThe choice $s^2=h^3$ is exactly what cancels the coefficient introduced by the change in $x$.

The six orderings give the Legendre parameter orbit:Every entry is different from $0,1$ when $\lambda \ne 0,1$.

There are no further possibilities for a matrix of the stated form. Its determinant is $acd$, so $a,c,d$ must be nonzero. On $z=1$ it induces the affine change $u\mapsto (a/d)u+b/d$ and scales $v=y/z$ by $c/d$. It preserves the line $y=0$, whose three intersections with the plane cubic have $u$-coordinates $0,1,\lambda$. Their images must have coordinates $0,1,\tilde{\lambda }$. Choosing the two roots sent to $0,1$ determines the affine change and leaves exactly the six orderings above.

For general $\lambda$ the six values are distinct. They coincide in some special cases: $\lambda \in \{-1,2,1/2\}$ gives three values, and ${\lambda }^2-\lambda +1=0$ gives two. In particular, for the value obtained in part (c), the two possible values are $(1\pm \mathrm{i}\sqrt{3})/2$. This does not contradict the paper's qualification that there are six for general $\lambda$.