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The degree of a variety $X$ of dimension $r$ is defined by $r!$ times the leading coefficient of its Hilbert polynomial. This is the defination given in Hartshorne, but I find it is very hard to handle with. If $X$ is a curve, one can cut it by generic hyperplane, and count the number of intersection points. This gives the degree of curve. However, for the surface, I want to understand the following claim used by Hartshorne:

(1) $X$ is a surface with embedding $X \to \mathbb{P}^n$, and $D=O_X(1)$, then the degree of $X$ in $\mathbb{P}^n$ is $D^2$.

(2) $X$ is a surface with embedding $X \to \mathbb{P}^n$, and $D=O_X(1)$. Suppose $h$ is a divisor in $X$, then the degree of $h$ in $\mathbb{P}^n$(viewed as curves) is $D.h$

I certainly wish to see the rigorous proves, but presumably they will be long and not very enlightening. So any intuitive argument is also greatly welcome!

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Could you cite where in Hartshorne he makes these claims? – Fredrik Meyer Oct 3 '12 at 19:49
I change the notation a little bit. But you will find the first claim in page400, in the proof of Corollary4.7. For the second one, it is from page 401 to 402, it is in the proof of Proposition4.8(e). – Li Zhan Oct 3 '12 at 20:45
up vote 1 down vote accepted

(2) Let $H$ be a generic hyperplane in $P=\mathbb P^n$, then, as invertible sheaves, we have $O_X(1)\simeq O_P(H)|_X$. So $$O_X(1).h:=\deg O_X(1)|_h=\deg O_P(H)|_h$$ is the intersection number of $h$ with $H$.

(1) The degree of $X$ in $P$ is the intersection number of $X$ with a generic linear space of codimension $2$. Or, the intersection number of $X\cap H$ with $H'$, where $H$ and $H'$ are generic. So it is the degree of $X\cap H$ (scheme theoretic intersection) in $P$.

As $O_X(1)$ is generated by its global sections, $O_X(d)\simeq O_X(h)$ for some curve $h$ in $X$ (in fact $h=X\cap H$ is OK for any $H$ which does not contain $X$). So $$D^2=O_X(1).h=\deg O_X(1)|_h=\deg h=\deg X.$$

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Dear QiL, Thank you very very much! It makes a lot sense for me! However, I have one more question: you claim "The degree of $X$ in $P$ is the intersection number of $X$ with a generic linear space of codimension 2." How could I find the proof of this is equivalent to the definition using Hilbert polynomial? Besides, there seems a typo in $O_X(d) \cong O_X(h)$. – Li Zhan Oct 4 '12 at 20:04
@LiZhan By Bertini, you can choose a hyperplane (suppose we are over an infinite field to simplify) $H$ such that $H\cap X$ is smooth and irreducible. This implies by Hartshorne I.7.7 that $\deg X=\deg X\deg H=\deg (H\cap X)$. Now $H\cap X$ is a curve, and its degree is its intersection number with a generic hyperplane $H'$. Does this clarifies the situation ? – user18119 Oct 5 '12 at 21:42
Yes, I see the reason! Thank you again! – Li Zhan Oct 5 '12 at 21:55

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