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



Degree of divisors and degrees of the corresponding maps to projective space
Here's how I think about it.


Let's assume we are in the case that $\dim\varphi(X)=\dim X$. Then $\varphi : X\to \varphi(X)$ is an generic finite map. Let $d$ be the degree of this map which is defined as the degree of field extension $[k(X):k(\varphi(X))]$. The degree of $\varphi(X)$ is given by $\varphi(X)\cdot H^{\dim X}$ where $H$ is a general hyperplane of $\mathbb{P}^n$. Pulling $H$ back to $X$, we get $D$. Then, by projection formula, $D^{\dim X} = X\cdot D^{\dim X}=d\cdot(\varphi(X)\cdot H^{\dim X})$. In the case that $X$ is a curve, $D^{\dim X}$ is noting but the degree of $D$. So, the degree of $D$ equals that the degree of image times the degree of the map. 


However, in higher dimension, $D^{\dim X}$ may not be the degree of $D$. For example, $D$ is a irreducible degree 2 curve in $\mathbb{P}^2$. The degree of $D$ is 2 which is not equal to $D\cdot D=4$ by Bézout's theorem.


Edit: I think in higher dimension, to define the degree of a divisor $D$, we need to choose a very ample divisor $A$ at first and then define the degree as the intersection number $D\cdot A^{\dim D}$. 


