Past exam of the geometry course of the University of Oxford / 2009 / B

For every $q\in L$, the line $pq$ meets the nonsingular conic in two points counted with multiplicity by the Bézout theorem. One is $p$; define $\alpha (q)$ as the second, using $\alpha (q)=p$ for the unique tangent line at $p$. Conversely, for $r\in C\setminus \{p\}$, intersect the line $pr$ with $L$, and send $p$ itself to the intersection of its tangent with $L$. Since $p\notin L$, each intersection with $L$ is unique. These constructions are inverse, so $\alpha :L\to C$ is a bijection preserving the stated collinearity. This is parametrization of a conic by projection.

For the particular conic take $p=[1:0:1]$ and $L=\{x=0\}$. If $q=[0:s:t]$, points on $pq$ have coordinates $[a:bs:a+bt]$. Substitution into the conic givesTaking the second intersection gives the explicit mapThe three coordinates do not vanish simultaneously. If $s+2t=0$, the formula gives $p$, as required for the tangent direction.

For rational points, put $r=s/t$ when $t\ne 0$. The full affine list istogether with the parameter $r=\infty$, which gives $(x,y)=(1,1)$. The denominator is $(r+1/2{)}^2+3/4>0$ for rational $r$. Conversely, every rational point other than $p$ determines a rational joining line and hence a rational point of $L$; the point $p$ is obtained at $r=-2$. Thus this lists all rational solutions.

An inflection point of an algebraic plane curve is a smooth point where the tangent has intersection multiplicity at least three. For a homogeneous defining polynomial $P$ of degree $d$, its Hessian curve is defined by the determinant of the matrix of second partial derivatives, of degree $3(d-2)$. In characteristic zero, the smooth inflection points are exactly the smooth points of $C\cap \mathrm{Hess}(C)$.

For a local verification, move a smooth point to $[0:0:1]$ and its tangent to $y=0$, and scale so $P(x,y,1)=y+a{x}^2+bxy+c{y}^2+\cdots$. The tangent contact is at least three exactly when $a=0$. The homogeneous Euler identities give the Hessian matrix at that point as $\left(\begin{array}{ccc}2a& b& 0\\ b& 2c& d-1\\ 0& d-1& 0\end{array}\right)$, whose determinant is $-2a(d-1{)}^2$. This proves the criterion for $d>1$.

For the given cubic, the singularity equations are $x^2+\lambda yz=y^2+\lambda xz=z^2+\lambda xy=0$. No coordinate can vanish at a projective solution, since that would force all three to vanish. Multiplying the equations therefore gives $1=(-\lambda {)}^3$, so singularity requires ${\lambda }^3=-1$. Conversely, under that condition the point $[1:1:-1/\lambda ]$ satisfies all three equations, and the Euler identity places it on the cubic. Thus

The second-derivative matrix isIts determinant is $54[(4+{\lambda }^3)xyz-{\lambda }^2(x^3+y^3+z^3)]$. Hence the Hessian curve has equationIf $\lambda =0$, this is the union $xyz=0$ of three lines and is singular. If $\lambda \ne 0$, divide by ${\lambda }^2$ and write it as a Hesse cubic with parameter $\mu =-({\lambda }^3+4)/(3{\lambda }^2)$. The preceding singularity criterion gives ${\mu }^3=-1$, equivalentlyThus the complete condition isThe other values beyond the source's first family are zero and the three cube roots of eight.

On the nonsingular cubic, $x^3+y^3+z^3=-3\lambda xyz$. Substituting this into the Hessian equation leaves $-4({\lambda }^3+1)xyz=0$. Since ${\lambda }^3+1\ne 0$, a flex has a zero coordinate. The remaining cubic equation then gives the nine pointswhere $\omega =e^{2\pi i/3}$. All are distinct, smooth, and lie on both the cubic and its Hessian curve, so they are exactly the inflection points. In particular the nine flexes are independent of $\lambda$ within the nonsingular family. This is the flex configuration of the Hesse pencil.

A divisor on an algebraic curve is a finite formal integer sum $D={\sum }_p{n}_{p}p$. Its degree is ${\sum }_p{n}_p$. The Riemann-Roch space $L(D)$ consists of zero and the meromorphic functions $h$ with $(h)+D\ge 0$, and $\ell (D)=\dim L(D)$. The Riemann-Roch theorem stateswhere $K$ is a canonical divisor, of degree $2g-2$. In genus one, $\mathrm{deg}(K-D)=-n<0$ for $\mathrm{deg}D=n>0$. A negative-degree divisor on an algebraic curve has no nonzero section, since a principal divisor has degree zero and an effective divisor has nonnegative degree. HenceA nonsingular plane cubic has genus one by the genus-degree formula. Therefore $\ell (p+q)=2$, and there is a nonconstant $f\in L(p+q)$. It has at most simple poles at $p,q$ and no other poles. It cannot have no poles, since a holomorphic function on a compact Riemann surface is constant. It cannot have just one simple pole either, by the allowed fact, or by $\ell (p)=\ell (q)=1$. Thus $f$ has simple poles at both $p$ and $q$, and no others.

Apply the preceding argument to $q+r$ to obtain a nonconstant meromorphic function $g$ with simple poles exactly at $q,r$. Both $f,g$, and the constant function one, belong to $L(p+q+r)$.

If $a+bf+cg=0$, examination at $p$, where only $f$ has a pole, forces $b=0$. Examination at $r$, where only $g$ has a pole, forces $c=0$, and then $a=0$. Thus the three functions are linearly independent. Since the Riemann-Roch theorem gives $\ell (p+q+r)=3$,

Away from $p,q,r$, all three coordinate functions are holomorphic and the first is one, so they define a projective point. Near $p$, take a local coordinate $z$ vanishing there. Write $f=a/z+O(1)$ with $a\ne 0$, while $g$ is holomorphic. ThenThe scaled coordinates are holomorphic and do not all vanish, proving a holomorphic extension. Similarly $F(r)=[0:0:1]$. At $q$, write $f=b/z+O(1)$ and $g=c/z+O(1)$ with $b,c\ne 0$. The same scaling gives $F(q)=[0:b:c]$, again well-defined. Hence

Every degree-three monomial evaluated at $(1,f,g)$ is $f^a{g}^b$ with $a+b\le 3$. Its possible pole orders at $p,q,r$ are at most $a$, $a+b$, $b$, respectively, and hence at most three each. Therefore $Q(1,f,g)\in L(3p+3q+3r)$.

The vector space of homogeneous degree-three polynomials in three variables has dimension $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. The target Riemann-Roch space has dimension nine, because its divisor on an algebraic curve has degree nine and genus is one. Thus the evaluation linear maphas nontrivial kernel. Consequently

Away from the three poles, the identity from the preceding part puts $F(u)$ on the cubic $Q=0$. The local extensions and continuity put the three pole images there too. Thus $F(C)\subseteq \{Q=0\}$, a cubic curve.

The local calculations gaveThey are three distinct points on the line with first coordinate zero. Hence $F(p),F(q),F(r)$ are collinear. This is three-point Riemann-Roch construction of a plane cubic.