Solutions to Hartshorne's Algebraic Geometry/Riemann-Roch Theorem

Exercise IV.1.1

Let $X$ have genus $g$. Since $X$ is dimension 1, there exists a point $Q\in X$, $Q\neq P$. Pick an $n>\max(g,2g-2,1)$. Then for the divisor $D=n(2P-Q)$ of degree $n$, $l(K-D)=0$(Example 1.3.4), so Riemann-Roch gives $l(D)=n+1-g>1$. Thus there is an effective divisor $D'$ such that $D-D'=(f)$. Since $(f)$ is degree 0 (II 6.10), $D'$ has degree $n$, so $D'$ cannot have a zero of order large enough to kill the pole of $D$ of order $2n$. Therefore, $f$ is regular everywhere except at $P$.