 Solution

A divisor on a complex manifold, specialized to a compact Riemann surface, is a finite sum $D={\sum }_p{n}_{p}p$ with integer coefficients, and $\mathrm{deg}D={\sum }_p{n}_p$. Two divisors are linearly equivalent when $D^{\prime }=D+(h)$ for a nonzero meromorphic function $h$, whose principal divisor on a complex manifold records zero orders minus pole orders. DefineMultiplication by $1/h$ gives an isomorphism $L(D)\to L(D^{\prime })$, because $(f/h)+D^{\prime }=(f)+D$. Its inverse is multiplication by $h$, proving $\ell (D^{\prime })=\ell (D)$.

On the Riemann sphere move $P$ to infinity by a Möbius coordinate. Meromorphic functions on the sphere are rational; those with no finite poles and a pole of order at most one at infinity are exactly $az+b$. Thus $\ell (P)=2$, without using Riemann–Roch.

The Riemann-Roch theorem for a compact connected genus-$g$ surface is $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$, where $K$ is a canonical divisor. Choose $D=(g+1)P$. Since the second dimension is nonnegative, $\ell (D)\ge 2$. Constants span only one dimension, so a nonconstant $f$ exists whose only possible pole is at $P$ with order at most $g+1$. The degree of its holomorphic map to the sphere is the total pole order. Therefore