# Fulton 8.17 ¿$\Gamma(X) = k$?

Problem 8.17 Let $X, Y$ be nonsingular projectives curves, $f:X\to Y$ a dominating morphism. Prove that $f (X)=Y$.

Im trying to solve the problem 8.17 of Algebraic Curves Book of Fulton, there is a hint "If $P \in Y \setminus f(X)$, then $\widetilde{f}(\Gamma(Y\setminus\{P\}))\subset\Gamma(X)=k$", the part i don't understand is $\Gamma(X)=k$, Why?

Perhaps the rational functions defined on a nonsingular variety is isomorphic to a field , so the evaluation homomorphism and that .. ? or why?

• Can you please reproduce the problem? Not all of us are lucky enough to have the book at hand when viewing your question. As far as the hint, without knowing what $X$ is, I would guess that $X$ is complete and thus only has constant global functions. – KReiser Oct 23 '14 at 4:46
On any projective variety there are no nonconstant functions, so $\Gamma(X)$ just consists of constants and is therefore isomorphic to $k$.