# A morphism from a projective curve $X$ to a curve $Y$ is either constant or surjective

How to prove with riemann roch "A morphism from a projective curve $X$ to a curve $Y$ is either constant or surjective, if it is surjective then $Y$ must be projective"?.

• Would you be happy with arguments that avoided Riemann-Roch? I really don't see the connection, but my eyes are not very good. If you don't need to use Riemann-Roch, then you can try to use the fact that the image of a projective curve under a regular map is closed. This is sometimes known as properness of projective curves (and varieties). It is very handy. (Also, your curves should be irreducible.) – Lorenzo Najt Apr 18 '16 at 2:53

We should start by assuming that $Y$ is irreducible and $X$ is connected, otherwise the claim is not true. Furthermore, it suffices to also assume that $X$ is in fact irreducible.
So say the morphism is $f : X \rightarrow Y$. Then, since $X$ is projective, it is (in particular) complete (i.e. proper over the base field). This is the algebraic analogue of compactness and the analogy carries so far as for us to have that the image of a complete variety is a closed, complete subvariety (here is a reference for this fact), so $f(X)$ is a complete variety which is a closed irreducible subvariety of $Y$. Since $Y$ is a curve, its closed irreducible subvarieties are single (reduced) points and the irreducible components of $Y$. But $Y$ is irreducible, thus $f(X)$ is either a single point (i.e. $f$ is constant) or all of $Y$ itself.
To see that $Y$ is subsequently projective, assume first that it is smooth. Then, as a smooth complete curve, Riemann-Roch tells us that any divisor of positive degree is ample, so $Y$ is projective.
(If, however, $Y$ is not smooth, then take its normalization $\widetilde{Y}$ (which is smooth since $Y$ is a curve). You can prove (using the cohomology criterion for ampleness and the fact that $\widetilde{Y} \rightarrow Y$ is finite) that the pullback of a divisor on $Y$ to one on $\widetilde{Y}$ is ample if and only if the divisor downstairs on $Y$ was ample. So you could choose any non-zero effective divisor away from the singular points of $Y$ and it will be ample, so $Y$ is projective.)