5
$\begingroup$

Let $ C \subset \mathbb{P}^2 $ be a smooth curve defined by a homogeneous polynomial $ f $ of degree $ d $.

How to establish that the line bundle $ \mathcal{O} (1) $ restricted to $ C $ is of degree $ d $ ?

Thanks in advance for your help.

$\endgroup$
5
  • $\begingroup$ What's your definition of degree? In the end, the answer must somehow reflect the fact that a line and $C$ meet in $d$ points. $\endgroup$
    – Hoot
    Commented Sep 4, 2016 at 15:19
  • $\begingroup$ The degree of a divisor $ D = \displaystyle \sum n_i [x_i ] $ on a curve $ C $ is the integer $ \mathrm{deg} (D) = \sum n_i $. $ \mathcal{O} ([x_i] ) $ is the line bundle associated to the divisor $ [x_i ] $. :-) $\endgroup$
    – YoYo
    Commented Sep 4, 2016 at 15:39
  • $\begingroup$ All thing i know is that : $ \mathcal{O} (1) \simeq \mathcal{O} ( H ) $ for all hyperplans $ H \subset \mathbb{P}^2 $. right ? How to conclude the answer ? $\endgroup$
    – YoYo
    Commented Sep 4, 2016 at 15:54
  • $\begingroup$ Well $\mathcal O(1)|_C$ is not really a divisor yet. How do you get from there to a $D$? $\endgroup$
    – Hoot
    Commented Sep 4, 2016 at 20:07
  • $\begingroup$ by the map : $ D \in \mathrm{Div} ( \mathbb{P}^2 ) = H^0 ( \mathbb{P}^2 , \mathcal{K}_{\mathbb{P}^2 } / \mathcal{O}_{\mathbb{P}^{2} }^* ),\to \mathcal{O}_{ \mathbb{P}^{2} } (D) \in \mathrm{Pic} ( \mathbb{P}^n ) $, no ? $\endgroup$
    – YoYo
    Commented Sep 4, 2016 at 22:04

1 Answer 1

6
$\begingroup$

I don't understand if the answer to this question is clear or no, but it seems to me that it isn't.

First of all, note that in our situation talking about Cartier or Weil divisors is the same, so I suggest to look at the latter which is a bit more intuitive. A Weil divisor is basically a subscheme of codimension $1$ (plus all the hypotheses which are needed). The group of divisors is $Div(X)$. On the other hand we have line bundles, or better invertible sheaves. The group they form is $Pic(X)$. As you said there is a map $Div(X)\to Pic(X)$. Moreover if $Cl(X)$ is the quotient group of $Div(X)$ by linear equivalence, then $Cl(X)\cong Pic(X)$. (Everything here is in Hartshorne).

Now, the invertible sheaf $\mathcal{O}(1)\in Pic(\mathbb{P}^2)$ corresponds, under the above isomorphism, to the class of a hyperplane $[H]\in Cl(\mathbb{P}^2)$ (this is basically the definition of $\mathcal{O}(1)$). The restriction of the invertible sheaf $\mathcal{O}(1)$ to $C$ is an invertible sheaf on $C$, i.e. $\mathcal{O}(1)|_C\in Pic(C)$, and its (inverse) image in $Cl(C)$ is the ''restriction" $[H|_C]$, which is nothing more than the (class of the) intersection of $C$ and $H$. Here you may probably want to mess up with the definition of restriction map to convince yourself that it is true. Since $C$ is defined by a degree $d$ polynomial and $H$ is defined by a linear polynomial, their intersection consists of at most $d$ points, counted with the right multiplicities, and the sum of their multiplicities is $d$. This is basically Bezout theorem. Hence $H|_C=\sum_{p\in H\cap C}mult(p)\cdot p$ and its degree is $d$. Note that here we are working with a fix hyperplane $H$, but in projective space all hyperplanes are linear equivalent each other and we know that the degree is invariant on each linear equivalence class. Finally the degree of the invertible sheaf $\mathcal{O}(1)|_C$ is also $d$ by definition.

$\endgroup$
2
  • $\begingroup$ Thank you very much Cla. :) $\endgroup$
    – YoYo
    Commented Sep 6, 2016 at 12:30
  • $\begingroup$ You are very welcome! $\endgroup$
    – User3773
    Commented Sep 6, 2016 at 12:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .