Degree of divisors and degrees of the corresponding maps to projective space - MathOverflow

Degree of divisors and degrees of the corresponding maps to projective space

2010-01-29T20:10:33Z

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

Answer by jvp for Degree of divisors and degrees of the corresponding maps to projective space

2010-01-29T23:01:28Z

Edit: I am working over $\mathbb C$ here, but a similar answer work over an arbitrary algebraically closed field. See my comment below as well as Andrea Ferreti's.

The degree of the divisor is equal to the degree of the image of $\varphi$, let's call it $C$, times the topological degree of the map $ \varphi : X \to C$.

Answer by Fei YE for Degree of divisors and degrees of the corresponding maps to projective space

2010-01-30T17:53:31Z

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 $D\cdot A^{\dim D}$.