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 for instance if $D=P-Q$ on $\mathbb{P}^1$ then $D$ induces an ismorphism of $\mathbb{P}^1$, which is obviously degree 1, which isn't the degree of $D$.

Thanks.

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.

Thanks