Degree of divisors and degrees of the corresponding maps to projective space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T18:23:34Z http://mathoverflow.net/feeds/question/13410 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13410/degree-of-divisors-and-degrees-of-the-corresponding-maps-to-projective-space Degree of divisors and degrees of the corresponding maps to projective space Randy Reddick 2010-01-29T20:10:33Z 2010-01-30T17:53:31Z <p>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 $\varphi$?</p> <p>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.</p> <p>Thanks</p> http://mathoverflow.net/questions/13410/degree-of-divisors-and-degrees-of-the-corresponding-maps-to-projective-space/13433#13433 Answer by jvp for Degree of divisors and degrees of the corresponding maps to projective space jvp 2010-01-29T23:01:28Z 2010-01-30T02:40:32Z <p><strong>Edit:</strong> 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.</p> <p>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 $\varphi : X \to C$.</p> http://mathoverflow.net/questions/13410/degree-of-divisors-and-degrees-of-the-corresponding-maps-to-projective-space/13489#13489 Answer by Fei YE for Degree of divisors and degrees of the corresponding maps to projective space Fei YE 2010-01-30T17:53:31Z 2010-01-30T17:53:31Z <p>Here's how I think about it.</p> <p>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. </p> <p>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.</p> <p>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 $D\cdot A^{\dim D}$. </p>