I have a question regarding a proof of the "surjectivity of morphisms of projective varieties" (a whole mouthfull). Though there are proofs using completeness of varieties, I am interested in an elementary proof. I found an exercise which guided me to such a proof:
Let $C,C' \in \mathbb{P}$ be two smooth, irreducible plane curves over an algebraically closed field $k$ and let $\phi : C \rightarrow C'$ be a morphism.
(a) - Suppose $P \in C'(k)$ does not lie in the image of $\phi$. Show that there is a non-constant rational function $f \in k(C')$ having no poles except at $P$, and deduce that the image of $\phi$ is contained in a strict algebraic subset of $C'$
(b) - A strict algebraic subset of $C'$ is a finite set of points
(c) - Show that the image of $\phi$ cannot consist of $n>1$ distinct points of $C'$, and deduce that $\phi$ is either constant or surjective.
My main problem lies in (a). A hint was given that every rational function on $C$ with no poles must be constant. And thus if such an $f$ exists, we have that $f \circ \phi \in k(C)$ is constant. My thought was to apply Riemann-Roch theorem to show that such a function $f$ must exist.
Someone got any suggestion/ideas?
Partial answer:
To apply Riemann-Roch: let $D=P$ be the divisor corresponding to the point $P$. Then Riemann-Roch says that $$ l(D) -l(K-D) = 1+n-g $$
For $n \gg 0$ the second term vanishes, hence $l(D) = 1+n-g > 0$. Thus there is some $f \in l(D)$. But such an $f$ satisfies by definition $(f) + n(P) \geq 0$. This is equivalent to $(f) \geq -nP$, which happens only if $f$ have no poles except possibly one of order up to $n$ at $P$.
(I'm not sure why this implies that the image is algebraic yet)