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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

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  • $\begingroup$ It is always the case. A very similar question is here: mathoverflow.net/questions/9703 $\endgroup$
    – t3suji
    Jan 29, 2010 at 20:23
  • $\begingroup$ Is the degree of $\phi$ (which is the phrase used in the question) the same thing as the degree of the image? $\endgroup$
    – Emerton
    Jan 29, 2010 at 21:27
  • $\begingroup$ @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 image has degree $1$. $\endgroup$
    – Pete L. Clark
    Jan 29, 2010 at 21:40
  • $\begingroup$ My question was one of terminology: does the questioner really mean the degree of the map $X \to \phi(X)$, or rather the degree of the image $\phi(X)$. If the former, the answer to the questions is no, in almost all cases, which makes the assertion in the first clause of the last sentence odd. I presume this is why t3suji interpreted the question in the latter manner. $\endgroup$
    – Emerton
    Jan 30, 2010 at 4:04
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    $\begingroup$ It makes sense to talk about the degree of a map to projective space. If the map is an embedding, it is the degree of the projective variety; if the map is surjective, it is the degree of the dominant map; if the map is neither, it is the product of both. The reference I gave is to a special case of embedding (which is why I said "very similar" and not "here's the answer")... $\endgroup$
    – t3suji
    Jan 30, 2010 at 14:14

2 Answers 2

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

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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$.

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  • $\begingroup$ Sorry for the stupid question .. But I have seen the topological degree only in the case of maps from sphere to sphere .. As the image of $1$ in the map from $H_n$ to $H_n$. Where can I find a definition of the degree you are referring to? In this case, the homology of these curves have multiple generators. $\endgroup$
    – Anweshi
    Jan 29, 2010 at 23:06
  • $\begingroup$ To define the topological degree just take a general (closed) point at the target and count the number of pre-images. Or, if you prefer, interpret the map as defining a field extension from the function field of the source over the function field of the target and look at its degree. $\endgroup$
    – Jorge Vitório Pereira
    Jan 29, 2010 at 23:11
  • $\begingroup$ There is no need to use topology here. The degree of the map is equal to degree of the field extension [K(X):K(C)], which agrees with the topological degree when one works over C. $\endgroup$
    – Andrea Ferretti
    Jan 30, 2010 at 2:33

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