Past exam of the geometry course of the University of Oxford / 2009 / B / 6

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.