Take the tour ×  Suppose I have a divisor $D$ on a curve $X$
(Hartshorne curve - smooth, projective, dimension one over an
algebraically closed $k$). If the complete linear system $|D|$
is basepoint free then I get a map
$\varphi:X\rightarrow\mathbb{P}^n_k$. My question is, say for
simplicity our map ends up being to $\mathbb{P}^1_k$, what if
anything is the relationship between the degree of the divisor
$D$ and the degree of the morphism $\varphi$?  It seems for many
cases that we have $deg(\varphi)=deg(K)$, however I can't find
anywhere that proves that this is always the case.
share|improve this question It is always the case. A very
similar question is here: mathoverflow.net/questions/9703 – 
t3suji Jan 29 '10 at 20:23 Is the degree of $\phi$ (which is the
phrase used in the question) the same thing as the degree of the
image? –  Emerton Jan 29 '10 at 21:27 @Emerton: no, not always,
especially when the image is P^1. E.g. take D = 2[O] on an
elliptic curve. Then the degree of $\varphi$ is $2$, but the im