Past exam of the geometry course of the University of Oxford / 2011 / 6

The genus-degree formula for a nonsingular plane curve givesA divisor is a finite formal sum $D={\sum }_p{n}_p[p]$ with integer coefficients; its degree is $\mathrm{deg}D={\sum }_p{n}_p$. An effective divisor has all $n_p\ge 0$. A hyperplane divisor is the intersection with a projective line not containing the curve, counted with local intersection multiplicities. Its degree is $d$.

Define $L(D)=\{0\}\cup \{f\text{ meromorphic}:(f)+D\ge 0\}$ and $\ell (D)={\dim }_{\mathbb{C}}L(D)$. For a canonical divisor $K$, the Riemann-Roch theorem statesHere $(f)$ records zero orders positively and pole orders negatively, and $\mathrm{deg}K=2g-2$.

A meromorphic function on a compact Riemann surface has finitely many poles, say $p_j$ of orders $r_j$. For each $p_j$, choose a projective line through it that does not contain $C$. Its hyperplane divisor contains $p_j$ with multiplicity at least one. Include $r_j$ copies of this divisor. The sum then dominates the pole divisor of $f$.

Add further arbitrary hyperplane divisors until their number $m$ satisfies $m\ge d$ and $md>2g-2$. Adding effective divisors cannot destroy the pole bound, so for $D=H_1+\cdots +H_m$ we haveFor a constant function, including zero, no initial poles need to be covered; simply add sufficiently many lines.

The divisor $D={\sum }_i{H}_i$ has degree $md$. The hypothesis gives $\mathrm{deg}(K-D)=2g-2-md<0$. A divisor of negative degree has no nonzero section: if $h\in L(K-D)\setminus \{0\}$, then $(h)+K-D$ would be effective, but its degree would be negative because the divisor of a meromorphic function has degree zero. Thus $\ell (K-D)=0$. The Riemann-Roch theorem now gives

The monomials $x^i{y}^j{z}^k$ with nonnegative integers $i+j+k=m$ form a basis for the vector space of homogeneous polynomials of degree $m$. For each $i=0,\dots ,m$, there are $m-i+1$ choices of $(j,k)$. Hence

Write $L_i=a_{i}x+b_{i}y+c_{i}z$ and $S={\prod }_i{L}_i$. The numerator and denominator have the same homogeneous degree, so their ratio is well-defined on projective points away from its poles and extends as a meromorphic function on $C$. Locally the pole order cannot exceed the zero order of $S{|}_C$, whose divisor is $D$. Thus $f_Q\in L(D)$.

Equality of two ratios is equivalent to $Q-Q^{\prime }$ vanishing on $C$. A nonsingular projective plane curve is irreducible: distinct positive-degree components would intersect by Bézout's theorem, producing a singular point. Its homogeneous vanishing ideal is therefore generated by its irreducible defining polynomial $P$. One can also see this using Bézout's theorem: a homogeneous polynomial not divisible by $P$ has only finitely many intersections with $C$, so cannot vanish on the whole curve. ConsequentlyMultiplication by the nonzero polynomial $P$ is injective. The linear map $V_m\to L(D)$ therefore has kernel $P{V}_{m-d}$ and image dimensionThe assumption $m\ge d$ makes both polynomial-space dimensions applicable directly.

Given $f$, choose the hyperplane divisors as in (a). The ratios $f_Q$ form a linear subspace of $L(D)$ by (d), and its dimension equals $md+1-g=\ell (D)$ by (b). A linear subspace of a finite-dimensional vector space with the same dimension is the whole space. Hence $f=f_Q$ for some homogeneous polynomial $Q$ of degree $m$, givingThe zero function corresponds to $Q=0$. Thus every allowed meromorphic function is a ratio of homogeneous polynomials of equal degree, with a denominator that can be chosen as a product of linear forms.