Rational Maps Between Curves Let $F:C'\rightarrow C$ be a rational map. Then either $F$ is dominating, or $F$ is constant. Furthermore, if $F$ is dominating, then $k(C')$ is a finite algebraic extension of $\tilde{F}(k(C))$. Why is this? I can't seem to grasp what should be a simple question. I feel like there's a result about curves that could be useful, but I'm not seeing it. Any response is much appreciated.
 A: Let us suppose that the base field is algebraically closed.  The field $k(C)$ is a finite extension of a transcendental field of one variable, and any transcendental element over $k$, $T$, is such that $k(C)|k(T)$ is finite.
Apply this to the extension of fields of rational functions, and you'll get your answer.  In fact, if the curves are nonsingular you get more (the local degree will be constant).
Let me add a nice reference.  Silverman's The Arithmetic of Elliptic Curves has a lovely first chapter of introduction to Algebraic Geometry, and in this chapter or the second there is a thorough treatment of nonconstant maps of curves defined over a perfect field.
A: While working with $\textit{Algebraic curves}$ by Fulton, I came accross this problem. Here is my (simple) solution for the first part.
Let $F:C'\rightarrow C$ be a rational map between curves. Assume $F$ is not dominating and take $f:U\rightarrow C$ a morphism representing $F$ ($U$ open in $C'$) such that $f(U)$ is  not dense in $C$. Then $\overline{f(U)}$ (closure of $f(U)$ in $C$) is a proper closed subvariety of $C$, hence a point by (4) of proposition 10 of chapter 6. So $f(U)$ is a point (assuming it is not empty).
